MCQ 11 Mark
If $A$ is an invertible matrix of order $2,$ then det $(A–1)$ is equal to:
AnswerCorrect option: B. $\frac{1}{\text{det}\ (\text{A})}$
Since $AA^{-1} = I$
$\therefore\bigg|\text{AA}^{-1}\bigg|=\bigg|\text{I}\bigg|$
$\Rightarrow\bigg|\text{A}\bigg|\bigg|\text{A}^{-1}\bigg|=1$
$\Rightarrow\bigg|\text{A}^{-1}\bigg|=\frac{1}{|\text{A}|}$
Therefore, option $(b)$ is correct.
View full question & answer→MCQ 21 Mark
Choose the correct answer from given four options in each of the Exercise:
The value of $\begin{vmatrix}\text{a}-\text{b}&\text{b}+\text{c}&\text{a}\\\text{b}-\text{a}&\text{c}+\text{a}&\text{b}\\\text{c}-\text{a}&\text{a}+\text{b}&\text{c}\end{vmatrix}$ is:
- A
$a^3 + b^3 + c^3$
- B
$3bc$
- C
$a^3 + b^3 + c^3 - 3abc$
- ✓
AnswerWe have,
$\begin{vmatrix}\text{a}-\text{b}&\text{b}+\text{c}&\text{a}\\\text{b}-\text{a}&\text{c}+\text{a}&\text{b}\\\text{c}-\text{a}&\text{a}+\text{b}&\text{c}\end{vmatrix}=\begin{vmatrix}\text{a}+\text{c}&\text{b}+\text{c}+\text{a}&\text{a}\\\text{b}+\text{c}&\text{c}+\text{a}+\text{b}&\text{b}\\\text{c}+\text{b}&\text{a}+\text{b}+\text{c}&\text{c}\end{vmatrix}$ $\big[\because\ \text{C}_1\rightarrow\text{C}_1+\text{C}_2\text{ and C}_2\rightarrow\text{C}_2+\text{C}_3\big]$
$(\text{a}+\text{b}+\text{c})\begin{vmatrix}\text{a}+\text{c}&1&\text{a}\\\text{b}+\text{c}&1&\text{b}\\\text{c}+\text{b}&1&\text{c}\end{vmatrix}$ $\big[\text{Taking }(\text{a}+\text{b}+\text{c})\text{ common from C}_2\big]$
$=(\text{a}+\text{b}+\text{c})\begin{vmatrix}\text{a}-\text{b}&0&\text{a}-\text{c}\\0&0&\text{b}-\text{c}\\\text{c}+\text{b}&1&\text{c}\end{vmatrix}$ $\big[\because\ \text{R}_2\rightarrow\text{R}_2-\text{R}_3\text{ and R}_1\rightarrow\text{R}_1-\text{R}_3\big]$
$=(\text{a}+\text{b}+\text{c})\big[-(\text{b}-\text{c}).(\text{a}-\text{b})\big] [$expanding along $R_2]$
$=(\text{a}+\text{b}+\text{c})(\text{c}-\text{b})(\text{a}-\text{b})$
View full question & answer→MCQ 31 Mark
Choose the correct answer from given four options in each of the Exercise: If $\text{A}=\begin{vmatrix}2&\lambda&-3\\0&2&5\\1&1&3\end{vmatrix},$ then $A^{-1}$ exists, if:
- A
$\lambda=2$
- B
$\lambda\neq2$
- C
$\lambda\neq-2$
- ✓
$\text{None of these}$
AnswerCorrect option: D. $\text{None of these}$
We have, $\text{A}=\begin{vmatrix}2&\lambda&-3\\0&2&5\\1&1&3\end{vmatrix}$
Expanding along $R_1,$ we get
$\text{A}=2(6-5)-\lambda(-5)-3(-2)$
$=2+5\lambda+6$
$=5\lambda+8$
We know that, $A^{-1}$ exists, if $A$ is non $-$ singular matrix
i.e., $|\text{A}|\neq0.$
$\therefore\ 5\lambda+8\neq0$
$\Rightarrow\ 5\lambda\neq-8$
$\therefore\ \lambda\neq\frac{-8}{5}$
Thus, $A^{-1}$ exists for all values of $\lambda\text{ except }\frac{-8}{5}.$
View full question & answer→MCQ 41 Mark
The value of $\begin{vmatrix}5^2&5^3&5^4\\5^3&5^4&5^5\\5^4&5^5&5^6\end{vmatrix}$ is:
- A
$5^2$
- ✓
$0$
- C
$5^{13}$
- D
$5^9$
Answer$\begin{vmatrix}5^2&5^3&5^4\\5^3&5^4&5^5\\5^4&5^5&5^6\end{vmatrix}$
$=5^2\times5^3\times5^4\begin{vmatrix}1&5&5^2\\1&5&5^2\\1&5&5^2\end{vmatrix} [$Taking out common factors from $R_1, R_2, R_3]$
$=5^2\times5^3\times5^4\times5 \begin{vmatrix}1&1&5^2\\1&1&5^2\\1&1&5^2\end{vmatrix}$
$=5^2\times5^3\times5^4\times0$
$=0$
View full question & answer→MCQ 51 Mark
The value of $\text{(adj } A)$ is equal to
AnswerThe value of $(\text{adj} A)$ is equal to $2A$.
Option $A$ is correct answer.
View full question & answer→MCQ 61 Mark
Find the value of x, if $\begin{bmatrix}1&-1\\3&-5\end{bmatrix}=\begin{bmatrix}\text{x}&\text{x}^2\\3&-5\end{bmatrix}$ is:
- ✓
$\text{x}=2,-\frac{1}{3}$
- B
$\text{x}=-1,-\frac{1}{3}$
- C
$\text{x}=-2,-\frac{1}{3}$
- D
$\text{x}=0,-\frac{1}{3}$
AnswerCorrect option: A. $\text{x}=2,-\frac{1}{3}$
Given that, $\begin{bmatrix}1&-1\\3&-5\end{bmatrix}=\begin{bmatrix}\text{x}&\text{x}^2\\3&-5\end{bmatrix} -5—(-3)=5\text{x}-3\text{x}^2 $
$-2=5\text{x}-3\text{x}^2$
$3\text{x}^2-5\text{x}-2=0$
Solving for x, we get
$\text{x}=2,-\frac{1}{3}$
View full question & answer→MCQ 71 Mark
$\begin{vmatrix}\log_3512&\log_43\\\log_38&\log_49\end{vmatrix}\times\begin{vmatrix}\log_23&\log_83\\\log_34&\log_34\end{vmatrix}$
Answer$\begin{vmatrix}\log_3512&\log_43\\\log_38&\log_49\end{vmatrix}\times\begin{vmatrix}\log_23&\log_83\\\log_34&\log_34\end{vmatrix}$
$=\begin{vmatrix}\log_32^9&\log_{2^{2}}3\\\log_32^3&\log_{2^2}3^3\end{vmatrix}\times\begin{vmatrix}\log_23&\log_{2^{3}}3\\\log_32^3&\log_32^2\end{vmatrix}$
$=\begin{vmatrix}9\log_32&\frac{1}{2}\log_23\\3\log_32&\frac{1}{2}\times2\log_23\end{vmatrix}\times\begin{vmatrix}\log_23&\frac{1}{3}\log_23\\2\log_32&2\log_32\end{vmatrix}$
$=\Big(\big(9\log_32\times\log_23\big)-\big(3\log_32\times\frac{1}{2}\log_23\big)\Big)\times\Big(\big(\log_23\times2\log_32\big)\\-\Big(\frac{1}{3}\log_23\times2\log_32\Big)\Big)$
$=\Big(9-\frac{3}{2}\Big)\times\Big(2-\frac{2}{3}\Big)$
$=\frac{15}{2}\times\frac{4}{3}$
$=10$
View full question & answer→MCQ 81 Mark
Choose the correct answer from given four options in each of the Exercise: The determinant $\begin{vmatrix}\text{b}^2-\text{ab}&\text{b}-\text{c}&\text{bc}-\text{ac}\\\text{ab}-\text{a}^2&\text{a}-\text{b}&\text{b}^2-\text{ab}\\\text{bc}-\text{ac}&\text{c}-\text{a}&\text{ab}-\text{a}^2\end{vmatrix}$ equals to:
Answer$\begin{vmatrix}\text{b}^2-\text{ab}&\text{b}-\text{c}&\text{bc}-\text{ac}\\\text{ab}-\text{a}^2&\text{a}-\text{b}&\text{b}^2-\text{ab}\\\text{bc}-\text{ac}&\text{c}-\text{a}&\text{ab}-\text{a}^2\end{vmatrix}$
$=\begin{vmatrix}\text{b}(\text{b}-\text{a})&\text{b}-\text{c}&\text{c}(\text{b}-\text{a})\\\text{a}(\text{b}-\text{a})&\text{a}-\text{b}&\text{b}(\text{b}-\text{a})\\\text{c}(\text{b}-\text{a})&\text{c}-\text{a}&\text{a}(\text{b}-\text{a})\end{vmatrix}$
$=(\text{b}-\text{a})^2\begin{vmatrix}\text{b}&\text{b}-\text{c}&\text{c}\\\text{a}&\text{a}-\text{b}&\text{b}\\\text{c}&\text{c}-\text{a}&\text{a}\end{vmatrix}$
$[$on taking $(b - a)$ common from $C_1$ and $C_3$ each$]$
$=(\text{b}-\text{a})^2\begin{vmatrix}\text{b}-\text{c}&\text{b}-\text{c}&\text{c}\\\text{a}-\text{b}&\text{a}-\text{b}&\text{b}\\\text{c}-\text{a}&\text{c}-\text{a}&\text{a}\end{vmatrix}$ $\big[\because\ \text{C}_1\rightarrow\text{C}_1-\text{C}_3\big]$
$=0$
$[$Since, two columns $C_1$ and $C_2$ are identical, so the value of determinant is zero$]$
View full question & answer→MCQ 91 Mark
Which of the following matrices will not have a determinant?
- A
$\begin{bmatrix}4&2\\5&4\end{bmatrix}$
- B
$\begin{bmatrix}1&5&3\\3&6&2\\4&8&7\end{bmatrix}$
- ✓
$\begin{bmatrix}5&8&9\\3&4&6\end{bmatrix}$
- D
$\begin{bmatrix}1&2\\5&5\end{bmatrix}$
AnswerCorrect option: C. $\begin{bmatrix}5&8&9\\3&4&6\end{bmatrix}$
Determinant of the matrix $\text{A}=\begin{bmatrix}5&8&9\\3&4&6\end{bmatrix}$is not possible as it is a rectangular matrix and not a square matrix.
Determinants can be calculated only if the matrix is a square matrix.
View full question & answer→MCQ 101 Mark
If $\text{D}_\text{k}=\begin{vmatrix}1&\text{n}&\text{n}\\2\text{k}&\text{n}^2+\text{n}+2&\text{n}^2+\text{n}\\2\text{k}-1&\text{n}^2&\text{n}^2+\text{n}+2\end{vmatrix} $ and $\sum\limits_{\text{k}=1}^\text{n}\text{D}_\text{k}=48,$ then n equals:
Answer$\text{D}_\text{k}=\begin{vmatrix}1&\text{n}&\text{n}\\2\text{k}&\text{n}^2+\text{n}+2&\text{n}^2+\text{n}\\2\text{k}-1&\text{n}^2&\text{n}^2+\text{n}+2\end{vmatrix} $
$\text{D}_\text{k}=\begin{vmatrix}1&\text{n}&\text{n}\\1&\text{n}+2&-2\\2\text{k}-1&\text{n}^2&\text{n}^2+\text{n}+2\end{vmatrix}\ \text{Applying}\ \text{R}_2\rightarrow\text{R}_2-\text{R}_3$
$\text{D}_\text{k}=\begin{vmatrix}1&\text{n}&\text{n}\\0&2&-2-\text{n}\\2\text{k}-1&\text{n}^2&\text{n}^2+\text{n}+2\end{vmatrix}\ \text{Applying}\ \text{R}_2\rightarrow\text{R}_2-\text{R}_1$
Now,
$\sum\limits_{\text{k}=1}^\text{n}\text{D}_\text{k}=\begin{vmatrix}1&\text{n}&\text{n}\\0&2&-2-\text{n}\\1&\text{n}^2&\text{n}^2+\text{n}+2\end{vmatrix}+\begin{vmatrix}1&\text{n}&\text{n}\\0&2&-2-\text{n}\\3&\text{n}^2&\text{n}^2+\text{n}+2\end{vmatrix}+.......+\begin{vmatrix}1&\text{n}&\text{n}\\0&2&-2-\text{n}\\2\text{n}-1&\text{n}^2&\text{n}^2+\text{n}+2\end{vmatrix}$
$\sum\limits_{\text{k}=1}^\text{n}\text{D}_\text{k}=\Big[1\Big(2(\text{n}^2+\text{n}+2)+\text{n}^2(2+\text{n})\Big)+\text{n}\Big(0+1(-2-\text{n})\Big)+\text{n}\Big(0+2\Big)\Big]\\+\Big[1\Big(2(\text{n}^2+\text{n}+2)+\text{n}^2(2+\text{n})\Big)+\text{n}\Big(0+3(-2-\text{n})\Big)+\text{n}\Big(0+6\Big)\Big]+......\\+\Big[1\Big(2(\text{n}^2+\text{n}+2)+\text{n}^2(2+\text{n})\Big)+\text{n}\Big(0+(2\text{n}-1)(-2-\text{n})\Big)+\text{n}\Big(0+2(2\text{n}-1)\Big)\Big]$
$\sum\limits_{\text{k}=1}^\text{n}\text{D}_\text{k}=\Big[1\Big(2(\text{n}^2+\text{n}+2)+\text{n}^2(2+\text{n})\Big)+\text{n}\Big(-2-\text{n}\Big)-2\text{n}\Big]\Big(1+3+5+.....+\text{n}\Big)$
$\sum\limits_{\text{k}=1}^\text{n}\text{D}_\text{k}=\Big[1\Big(2(\text{n}^2+\text{n}+2)+\text{n}^2(2+\text{n})\Big)+\text{n}\Big(-2-\text{n}\Big)-2\text{n}\Big]\Big(\text{n}^2\Big)$
$\Rightarrow\ 2\text{n}^2+4\text{n}=48$
$\Rightarrow\ (\text{n}+6)(\text{n}-4)=0$
$\Rightarrow\ \text{n}=4$
View full question & answer→MCQ 111 Mark
Find the determinant of the matrix $\text{A}=\begin{bmatrix}9&8\\7&6\end{bmatrix}$ is:
AnswerGiven that, $\text{A}=\begin{bmatrix}9&8\\7&6\end{bmatrix}$
$\Rightarrow\triangle=\begin{bmatrix}9&8\\7&6\end{bmatrix}$
$=9(6)-7(8)=54-56=-2$
View full question & answer→MCQ 121 Mark
If the points $(\text{k} + 1, 1), (2\text{k} + 1, 3)$ and $(2\text{k} + 2, 2\text{k})$ are collinear, then the value of $\text{k}$ is:
- ✓
$2$
- B
$-2$
- C
$\frac{1}{2}$
- D
$1$
View full question & answer→MCQ 131 Mark
If $A$ is a matrix of order $3$ and $|A| = 8,$ then $|\text{adj} \ A| =$
Answer$|A| = d$
$\text{|adj} A| = |A|^{n-1}$
Here, $n = 3, |A| = 8$
$|\text{adj } A| = 8^2$
$|\text{adj A}|={(2^3)}^2=2^6$
View full question & answer→MCQ 141 Mark
What is the determinant of the matrix $\left [\begin{matrix} 3&\text{amp; } 6\\ -1 &\text{amp; } 2\end {matrix} \right]$ ?
AnswerGiven, $\left [\begin{matrix} 3&\text{amp; } 6\\ -1 &\text{amp; } 2\end {matrix} \right]$
Let determinent be $ \left| \text{d} \right|$
Value of $ \left| \text{d} \right|$ wil be
$\left| \text{d} \right|∣\text{d}∣=3\times 2-\left( 6\times -1 \right)$
$=6+6=12$
View full question & answer→MCQ 151 Mark
For which of the following elements in the determinant $\triangle\begin{bmatrix}2&8\\4&7\end{bmatrix},$ the minor of the element is 2:
AnswerConsider the element 7 in the determinant $\triangle\begin{bmatrix}2&8\\4&7\end{bmatrix}$
The minor of the element 7 can be obtained by deleting $\text{R}_2$ and $\text{C}_2$
$\therefore\text{M}_{22}=2$
Hence, the minor of the element 7 is 2.
View full question & answer→MCQ 161 Mark
If A is a singular matrix, then adj A is:
AnswerIf A is singular matrix then adjoint of A is also singular.
View full question & answer→MCQ 171 Mark
If $\begin{vmatrix}x&2\\18&x\end{vmatrix}=\begin{vmatrix}6&2\\18&6\end{vmatrix},$ then x is equal to:
AnswerCorrect option: B. $\pm6$
$\begin{vmatrix}x&2\\18&x\end{vmatrix}=\begin{vmatrix}6&2\\18&6\end{vmatrix}$
$\Rightarrow x^2-36=36-36$
$\Rightarrow x^2-36=0$
$\Rightarrow x^2=36$
$\Rightarrow x=\pm6$
Hence, the correct answer is (b).
View full question & answer→MCQ 181 Mark
Choose the correct answer.
Let A be a square matrix of order 3 × 3, then | kA| is equal to:
- A
$k\left|\text{A}\right|$
- B
$k^2\left|\text{A}\right|$
- ✓
$k^3\left|\text{A}\right|$
- D
$3k\left|\text{A}\right|$
AnswerCorrect option: C. $k^3\left|\text{A}\right|$
$\text{Let A}=\begin{bmatrix}a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&a_{23}\\a_{31}&a_{32}&a_{33}\end{bmatrix}$ be a square matrix of order $3\times3 \dots\dots(1)$
$\therefore\ \ k\text{A}=\begin{bmatrix}ka_{11}&ka_{12}&ka_{13}\\ka_{21}&ka_{22}&ka_{23}\\ka_{31}&ka_{32}&ka_{33}\end{bmatrix}$
$ \Rightarrow\ |k\text{A}|=\begin{vmatrix}ka_{11}&ka_{12}&ka_{13}\\ka_{21}&ka_{22}&ka_{23}\\ka_{31}&ka_{32}&ka_{33}\end{vmatrix}$
$\Rightarrow\ |k\text{A}|=k^3\begin{bmatrix}a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&a_{23}\\a_{31}&a_{32}&a_{33}\end{bmatrix}=k^3|\text{A}|$ [From eq. (1)]
Therefore, option (c) is correct.
View full question & answer→MCQ 191 Mark
Choose the correct answer from given four options in each of the Exercise : The number of distinct real roots of $\begin{vmatrix}\sin\text{x}&\cos\text{x}&\cos\text{x}\\\cos\text{x}&\sin\text{x}&\cos\text{x}\\\cos\text{x}&\cos\text{x}&\sin\text{x}\end{vmatrix}=0$ in the interval $-\frac{\pi}{4}\leq\text{x}\leq\frac{\pi}{4}$ is:
AnswerWe have,
$\begin{vmatrix}\sin\text{x}&\cos\text{x}&\cos\text{x}\\\cos\text{x}&\sin\text{x}&\cos\text{x}\\\cos\text{x}&\cos\text{x}&\sin\text{x}\end{vmatrix}=0$
$\text{Applying C}_1\rightarrow\text{C}_1+\text{C}_2+\text{C}_3,$
$\Rightarrow\ \begin{vmatrix}2\cos\text{x}+\sin\text{x}&\cos\text{x}&\cos\text{x}\\2\cos\text{x}+\sin\text{x}&\sin\text{x}&\cos\text{x}\\2\cos\text{x}+\sin\text{x}&\cos\text{x}&\sin\text{x}\end{vmatrix}=0$
$\Rightarrow\ (2\cos\text{x}+\sin\text{x})\begin{vmatrix}1&\cos\text{x}&\cos\text{x}\\1&\sin\text{x}&\cos\text{x}\\1&\cos\text{x}&\sin\text{x}\end{vmatrix}=0$
$\big[\text{Applying R}_2\rightarrow\text{R}_2-\text{R}_1\text{ and R}_3\rightarrow\text{R}_3-\text{R}_1\big]$
$\Rightarrow\ (2\cos\text{x}+\sin\text{x}) \begin{vmatrix}1&\cos\text{x}&\cos\text{x}\\0&\sin\text{x}-\cos\text{x}&0\\0&0&\sin\text{x}-\cos\text{x}\end{vmatrix}=0$
$\Rightarrow\ (2\cos\text{x}+\sin\text{x})[1\cdot(\sin\text{x}-\cos\text{x})^2]=0 ($expanding along $C_1)$
$\Rightarrow\ (2\cos\text{x}+\sin\text{x})(\sin\text{x}-\cos\text{x})^2=0$
$\Rightarrow\ 2\cos\text{x}=-\sin\text{x or }\sin\text{x}=\cos\text{x}$
$\Rightarrow\ \tan\text{x}=-2,$ which is not possible as for $-\frac{\pi}{4}\leq\text{x}\leq\frac{\pi}{4}$
or $\tan\text{x}=1$
$\therefore\ \ \text{x}=\frac{\pi}{4}$
So, only me one real root exist.
View full question & answer→MCQ 201 Mark
For non $-$ singular square matrix $A, B$ and $C$ of the same order $(AB^{-1} C) =$
- A
$A^{-1} BC^{-1}$
- B
$C^{-1} B^{-1} A^{-1}$
- C
$\text{CBA}^{-1}$
- ✓
$C^{-1} BA^{-1}$
AnswerCorrect option: D. $C^{-1} BA^{-1}$
We know that $(AB)^{-1} = B^{-1} A^{-1}$
Hence, $(AB^{-1}C)^{-1} = C^{-1}BA^{-1}$
View full question & answer→MCQ 211 Mark
Let $\begin{vmatrix}\text{x}^2+3\text{x}&\text{x}-1&\text{x}+ 3\\\text{x}+1&-2\text{x}&\text{x}-4\\\text{x}-3&\text{x}+4&3\text{x}\end{vmatrix}=\text{ax}^4+\text{bx}^3+\text{cx}^2+\text{dx}+\text{e}$ be an identity in $x,$ where $a, b, c, d, e$ are independent of $x$. Then the value of $e$ is:
AnswerLet $\begin{vmatrix}\text{x}^2+3\text{x}&\text{x}-1&\text{x}+ 3\\\text{x}+1&-2\text{x}&\text{x}-4\\\text{x}-3&\text{x}+4&3\text{x}\end{vmatrix}$
$=(\text{x}^2+3\text{x})\begin{vmatrix}-2\text{x}&\text{x}-4\\\text{x}+4&3\text{x}\end{vmatrix}-(\text{x}-1)\begin{vmatrix}\text{x}+1&\text{x}-4\\\text{x}-3&3\text{x}\end{vmatrix}+(\text{x}+3)\begin{vmatrix}\text{x}+1&-2\text{x}\\\text{x}-3&\text{x}+4\end{vmatrix}$
$= (x^2 + 3x)(-6x - x^2 + 16) - (x - 1)(3x^2 + 3x - x^2 + 7x - 12) + (x + 3)(x^2 + 5x + 4 + 2x^2 - 6x)$
$= -7x^4 + 16x^2 + 48x + 21x^3 + 8x^2 - 22x - 2x^3 - 12 + 8x^2 + x + 3x^3 + 12$
$= -7x^4 + 22x^3 + 32x^2 + 27x + 0$
But $x$ is a root of $ax^4 + bx^3 + cx^2 + dx + e$
$e= 0$
View full question & answer→MCQ 221 Mark
Evaluate $\begin{bmatrix}\sqrt{3}&\sqrt{2}\\-1&2\sqrt{3}\end{bmatrix}$ is:
- A
$6-3\sqrt{2}$
- B
$6-\sqrt{2}$
- C
$6+3\sqrt{2}$
- ✓
$6+\sqrt{2}$
AnswerCorrect option: D. $6+\sqrt{2}$
$\triangle=\begin{bmatrix}\sqrt{3}&\sqrt{2}\\-1&2\sqrt{3}\end{bmatrix}$
$\triangle=(\sqrt{3}\times2\sqrt{3})+\sqrt{2}$
$\triangle=6+\sqrt{2}$
View full question & answer→MCQ 231 Mark
If $\begin{vmatrix}\text{a}&\text{p}&\text{x}\\\text{b}&\text{q}&\text{y}\\\text{c}&\text{r}&\text{z}\end{vmatrix}=16,$ then the value of $\begin{vmatrix}\text{p}+\text{x}&\text{a}+\text{x}&\text{a}+\text{p}\\\text{q}+\text{y}&\text{b}+\text{y}&\text{b}+\text{q}\\\text{r}+\text{z}&\text{c}+\text{z}&\text{c}+\text{r}\end{vmatrix}$ is:
Answer$\begin{vmatrix}\text{p}+\text{x}&\text{a}+\text{x}&\text{a}+\text{p}\\\text{q}+\text{y}&\text{b}+\text{y}&\text{b}+\text{q}\\\text{r}+\text{z}&\text{c}+\text{z}&\text{c}+\text{r}\end{vmatrix}=\begin{vmatrix}\text{p}&\text{a}&\text{a}\\\text{q}&\text{b}&\text{b}\\\text{r}&\text{c}&\text{c}\end{vmatrix}+\begin{vmatrix}\text{p}&\text{a}&\text{p}\\\text{q}&\text{b}&\text{q}\\\text{r}&\text{c}&\text{r}\end{vmatrix}\begin{vmatrix}\text{p}&\text{x}&\text{a}\\\text{q}&\text{y}&\text{b}\\\text{r}&\text{z}&\text{c}\end{vmatrix}\\+\begin{vmatrix}\text{p}&\text{x}&\text{p}\\\text{q}&\text{y}&\text{q}\\\text{r}&\text{z}&\text{r}\end{vmatrix}+\begin{vmatrix}\text{x}&\text{a}&\text{a}\\\text{y}&\text{b}&\text{b}\\\text{r}&\text{c}&\text{c}\end{vmatrix}+\begin{vmatrix}\text{x}&\text{a}&\text{p}\\\text{y}&\text{b}&\text{q}\\\text{r}&\text{c}&\text{r}\end{vmatrix}+\begin{vmatrix}\text{x}&\text{x}&\text{a}\\\text{y}&\text{y}&\text{b}\\\text{z}&\text{z}&\text{c}\end{vmatrix}+\begin{vmatrix}\text{x}&\text{x}&\text{p}\\\text{y}&\text{y}&\text{q}\\\text{z}&\text{z}&\text{r}\end{vmatrix}$
$=0+0+\begin{vmatrix}\text{p}&\text{x}&\text{a}\\\text{q}&\text{y}&\text{b}\\\text{r}&\text{z}&\text{c}\end{vmatrix}+0+0+\begin{vmatrix}\text{x}&\text{a}&\text{p}\\\text{y}&\text{b}&\text{q}\\\text{z}&\text{c}&\text{r}\end{vmatrix}+0+0$
$=\begin{vmatrix}\text{p}&\text{x}&\text{a}\\\text{q}&\text{y}&\text{b}\\\text{r}&\text{z}&\text{c}\end{vmatrix}+\begin{vmatrix}\text{x}&\text{a}&\text{p}\\\text{y}&\text{b}&\text{q}\\\text{z}&\text{c}&\text{r}\end{vmatrix}$
$=2\begin{vmatrix}\text{a}&\text{p}&\text{x}\\\text{b}&\text{q}&\text{y}\\\text{c}&\text{r}&\text{z}\end{vmatrix}$
$=2\times16=32$
Hence, the correct option is (b)
View full question & answer→MCQ 241 Mark
Given that $A$ is a square matrix of order $3$ and $|A| = -4,$ then $\text{|adj A|}$ is equal to:
AnswerGiven that $A$ is a square matrix of order $3$ and $|A| = -4.$
We know that $\text{|adj A|} = |A|^{n−1},$ where $n$ is the order of matrix $A.$
So, $\text{|adj A|} = (−4)^{3-1} = (-4)^2 = 16$
View full question & answer→MCQ 251 Mark
If $\text{f(x)}=\begin{vmatrix}0&\text{x}-\text{a}&\text{x}-\text{b}\\\text{x}+\text{a}&0&\text{x}-\text{c}\\\text{x}+\text{b}&\text{x}+\text{c}&0\end{vmatrix},$ then:
AnswerLet $\text{f(x)}=\begin{vmatrix}0&\text{x}-\text{a}&\text{x}-\text{b}\\\text{x}+\text{a}&0&\text{x}-\text{c}\\\text{x}+\text{b}&\text{x}+\text{c}&0\end{vmatrix}$
Now, $\text{f(a)}=\begin{vmatrix}0&\text{a}-\text{a}&\text{a}-\text{b}\\\text{a}+\text{a}&0&\text{a}-\text{c}\\\text{a}+\text{b}&\text{a}+\text{c}&0\end{vmatrix}$
$=\begin{vmatrix}0&0&\text{a}-\text{b}\\2\text{a}&0&\text{a}-\text{c}\\\text{a}+\text{b}&\text{a}+\text{c}&0\end{vmatrix}$
$=(\text{a}-\text{b})(2\text{a}^2+2\text{ac})\neq0$
$\text{f(b)}=\begin{vmatrix}0&\text{b}-\text{a}&\text{b}-\text{b}\\\text{b}+\text{a}&0&\text{b}-\text{c}\\\text{b}+\text{b}&\text{b}+\text{c}&0\end{vmatrix}$
$=\begin{vmatrix}0&\text{b}-\text{a}&0\\\text{b}+\text{a}&0&\text{b}-\text{c}\\2\text{a}&\text{b}+\text{c}&0\end{vmatrix}$
$=(\text{b}-\text{a})(2\text{ab}-2\text{ac})\neq0$
$\text{f(0)}=\begin{vmatrix}0&\text{0}-\text{a}&\text{0}-\text{b}\\\text{0}+\text{a}&0&\text{0}-\text{c}\\\text{0}+\text{b}&\text{0}+\text{c}&0\end{vmatrix}$
$=\begin{vmatrix}0&-\text{a}&-\text{b}\\\text{a}&0&\text{c}\\\text{b}&\text{c}&0\end{vmatrix}$
$=\text{a}(\text{bc})-\text{b}(\text{ac})=0$
Hence, the correct option is (c)
View full question & answer→MCQ 261 Mark
Let $\text{A}=\begin{bmatrix} 1 & 2 \\ 3 & -5 \end{bmatrix}\text{ and B}=\begin{bmatrix} 1 & 0 \\ 0 & 2 \end{bmatrix}$ and $X$ be a matrix such that $A = BX,$ then $ X$ is equal to:
- ✓
$\frac{1}{2}\begin{bmatrix} 2 & 4 \\ 3 & -5 \end{bmatrix}$
- B
$\frac{1}{2}\begin{bmatrix} -2 & 4 \\ 3 & 5 \end{bmatrix}$
- C
$\begin{bmatrix} 2 & 4 \\ 3 & -5 \end{bmatrix}$
- D
AnswerCorrect option: A. $\frac{1}{2}\begin{bmatrix} 2 & 4 \\ 3 & -5 \end{bmatrix}$
$A = BX$
$B^{-1}A = B^{-1}BX$
$X = B^{-1}A$
Using adjoint method of inverse
$\text{B}^{-1}=\frac{1}{2}\begin{bmatrix} 2 & 0 \\ 0 & 1 \end{bmatrix}$
$\text{X}=\text{B}^{-1}\text{A}$
$\text{X}=\frac{1}{2}\begin{bmatrix} 2 & 0 \\ 0 & 1 \end{bmatrix}\begin{bmatrix} 1 & 2 \\ 3 & -5 \end{bmatrix}$
$\text{x}=\frac{1}{2}\begin{bmatrix} 2 & 4 \\ 3 & -5 \end{bmatrix}$
View full question & answer→MCQ 271 Mark
Choose the correct answer from given four options in each of the Exercise: There are two values of a which makes determinant $\begin{bmatrix}1 & -2 & 5 \\2 & \text{a} & 1\\0 & 4 & 2\text{a} \end{bmatrix} = 86,$ then sum of these number is:
AnswerWe have,
$\begin{bmatrix}1 & -2 & 5 \\2 & \text{a} & 1\\0 & 4 & 2\text{a} \end{bmatrix} = 86,$
$\Rightarrow\ 1(2\text{a}^{2} + 4) - 2(-4\text{a} - 20) + 0 = 86 [$Expanding along $C_1]$
$\Rightarrow\ \text{a}^{2} + 4\text{a} - 21 = 0$
$\Rightarrow\ \text{(a + 7)} (\text{a} - 3) = 0$
$\Rightarrow\ \text{a} = -7 $ and $ 3$
$\therefore$ Required sum $= -7 + 3 = -4$
View full question & answer→MCQ 281 Mark
If A is a skew symmetric matrix, then ∣A∣ is:
AnswerSince the skew symmetric matrix consist of elements of opposite sign at opposite side of matrix diagonal with all.
the diagonal elements as zero therefore the determinant of skew symmetric matrix is zero.
View full question & answer→MCQ 291 Mark
If $w$ is a non $-$ real cube root of unity and $n$ is not a multiple of $3,$ then $\begin{vmatrix}1&\omega^{\text{n}}&\omega^{2\text{n}}\\\omega^{2\text{n}}&1&\omega^{\text{n}}\\\omega^{\text{n}}&\omega^{2\text{n}}&1\end{vmatrix}$ is equal to:
- ✓
$0$
- B
$\omega$
- C
$\omega^2$
- D
$1$
Answer$\triangle=\begin{vmatrix}1&\omega^{\text{n}}&\omega^{2\text{n}}\\\omega^{2\text{n}}&1&\omega^{\text{n}}\\\omega^{\text{n}}&\omega^{2\text{n}}&1\end{vmatrix}$
$=\begin{vmatrix}1+\omega^{\text{n}}+\omega^{2\text{n}}&\omega^{\text{n}}&\omega^{2\text{n}}\\\omega^{2\text{n}}+1+\omega^{\text{n}}&1&\omega^{\text{n}}\\\omega^{\text{n}}+\omega^{2\text{n}}+1&\omega^{2\text{n}}&1\end{vmatrix}\ [$Applying $C_1 → C_1+ C_2 + C_3]$
Now, $1+\omega+\omega^2=0$ $[\because$ is a complex cube root of unity$]$
$1+\omega^{\text{n}}+\omega^{2\text{n}}=0 \ [\because \ n$ is not a multiple of $3]$
$\triangle=\begin{vmatrix}1&\omega^{\text{n}}&\omega^{2\text{n}}\\\omega^{2\text{n}}&1&\omega^{\text{n}}\\\omega^{\text{n}}&\omega^{2\text{n}}&1\end{vmatrix}=0$
View full question & answer→MCQ 301 Mark
Choose the correct answer from given four options in each of the Exercise:
If A, B and C are angles of a triangle, then the determinant $ \begin{vmatrix}-1&\cos\text{C}&\cos\text{B}\\\cos\text{C}&-1&\cos\text{A}\\\cos\text{B}&\cos\text{A}&-1\end{vmatrix}$ is equal to:
AnswerWe have, $ \begin{vmatrix}-1&\cos\text{C}&\cos\text{B}\\\cos\text{C}&-1&\cos\text{A}\\\cos\text{B}&\cos\text{A}&-1\end{vmatrix}$
$ =\begin{vmatrix}-\text{a}+\text{b}\cos\text{C}+\cos\text{B}&\cos\text{C}&\cos\text{B}\\\text{a}\cos\text{C}-\text{b}+\text{c}\cos\text{A}&-1&\cos\text{A}\\\text{a}\cos\text{B}+\text{b}\cos\text{A}-\text{C}&\cos\text{A}&-1\end{vmatrix}$ $\big[\text{C}_1\rightarrow\text{a C}_1+\text{b C}_2+\text{c C}_3\big]$
We know that, $\text{a}=\text{b}\cos\text{C}+\text{c}\cos\text{B, b}=\text{c}\cos\text{A}+\text{a}\cos\text{C and c}=\text{a}\cos\text{B}+\text{b}\cos\text{A}$
Substituting these values we get
$\begin{bmatrix}-\text{a}+\text{a}&\cos\text{C}&\cos\text{B}\\\text{b}-\text{b}&-1&\cos\text{A}\\\text{c}-\text{c}&\cos\text{A}&-1\end{bmatrix}$
$\begin{bmatrix}0&\cos\text{C}&\cos\text{B}\\0&-1&\cos\text{A}\\0&\cos\text{A}&-1\end{bmatrix}$
$=0$
View full question & answer→MCQ 311 Mark
The determinant $\begin{vmatrix}\text{b}^2-\text{ab}&\text{b}-\text{c}&\text{bc}-\text{ac}\\\text{ab}-\text{a}^2&\text{a}-\text{b}&\text{b}^2-\text{ab}\\\text{bc}-\text{ac}&\text{c}-\text{a}&\text{ab}-\text{a}^2\end{vmatrix}$ equals:
Answer$\begin{vmatrix}\text{b}^2-\text{ab}&\text{b}-\text{c}&\text{bc}-\text{ac}\\\text{ab}-\text{a}^2&\text{a}-\text{b}&\text{b}^2-\text{ab}\\\text{bc}-\text{ac}&\text{c}-\text{a}&\text{ab}-\text{a}^2\end{vmatrix}$
$=\begin{vmatrix}\text{b}(\text{b}-\text{a})&\text{b}-\text{c}&\text{c}(\text{b}-\text{a})\\\text{a}(\text{b}-\text{a})&\text{a}-\text{b}&\text{b}(\text{b}-\text{a})\\\text{c}(\text{b}-\text{a})&\text{c}-\text{a}&\text{a}(\text{b}-\text{a})\end{vmatrix}$
$=(\text{b}-\text{a})^2\begin{vmatrix}\text{b}&\text{b}-\text{c}&\text{c}\\\text{a}&\text{a}-\text{b}&\text{b}\\\text{c}&\text{c}-\text{a}&\text{a}\end{vmatrix} [$Taking $(b - a)$ common from $C_1$ and $C_3]$
$=(\text{b}-\text{a})^2\begin{vmatrix}0&\text{b}-\text{c}&\text{c}\\0&\text{a}-\text{b}&\text{b}\\0&\text{c}-\text{a}&\text{a}\end{vmatrix} [$Applying $C_1\rightarrow C_1 - C_2 - C_3]$
$=0$
Hence, the correct option is $(d)$
View full question & answer→MCQ 321 Mark
Let $\text{A}=\begin{bmatrix} 2 & 3 \\ 5 & -2 \end{bmatrix}$ be such that $A^{-1} = kA,$ then k equals:
- A
$19$
- ✓
$\frac{1}{19}$
- C
$-19$
- D
$-\frac{1}{19}$
AnswerCorrect option: B. $\frac{1}{19}$
$\text{adj A}=\begin{bmatrix} -2 & -3 \\ -5 & 2 \end{bmatrix}$
$|\text{A}|=-19$
$\therefore\ \text{A}^{-1}=-\frac{1}{|\text{A}|}\text{ adj A}$
$\Rightarrow\ \text{A}^{-1}=-\frac{1}{19}\begin{bmatrix} -2 & -3 \\ -5 & 2 \end{bmatrix}$
Now,
$A^{-1} = kA$
$\Rightarrow-\frac{1}{19}\begin{bmatrix} -2 & -3 \\ -5 & 2 \end{bmatrix}=\text{kA}$
$\Rightarrow\frac{1}{19}\begin{bmatrix} 2 & 3 \\ 5 & -2 \end{bmatrix}=\text{kA}$
$\Rightarrow\frac{1}{19}\text{A}=\text{kA}$
$\Rightarrow\text{k}=\frac{1}{19}$
View full question & answer→MCQ 331 Mark
Find the values of x, if: $\begin{vmatrix} 2 &\text{amp; } 4 \\ 5 &\text{amp; } 1 \end{vmatrix}= \begin{vmatrix} 2\text{x} &\text{amp; }4 \\ 6 &\text{amp; x} \end{vmatrix}$
- ✓
$\pm \sqrt{3}$
- B
$3$
- C
$-3$
- D
$\text{None of these}$
AnswerCorrect option: A. $\pm \sqrt{3}$
We have,
$\begin{vmatrix} 2 &\text{amp; } 4 \\ 5 &\text{amp; } 1 \end{vmatrix}= \begin{vmatrix} 2\text{x} &\text{amp; }4 \\ 6 &\text{amp; x} \end{vmatrix}\Rightarrow2-20=2\text{x}^2-24\Rightarrow2\text{x}^2=6$
$\Rightarrow\pm\sqrt{3}$
View full question & answer→MCQ 341 Mark
If A is any skew-symmetric matrix of odd order then ∣A∣ equals
Answerif A is skew symmetric matrix
then A = -AT
Therefore, ∣A∣ = -∣AT∣ = -∣A∣
⇒ 2∣A∣ =0
⇒ ∣A∣ = 0
View full question & answer→MCQ 351 Mark
Evaluate $\begin{bmatrix}2&5\\-1&-1\end{bmatrix}$
AnswerExpanding along $\text{R}_1,$ we get
$\triangle=2(-1)-5(-1)=2+5$
$=3$
View full question & answer→MCQ 361 Mark
Find the cofactor of element -3 in the determinant $\triangle=\begin{bmatrix}1&4&4\\-3&5&9\\2&1&2\end{bmatrix}$ is:
AnswerThe minor of element -3 is given by
$\text{M}_{21}=\begin{bmatrix}4&4\\1&2\end{bmatrix}=4(2)-4=4$ (Obtained by eliminating $\text{R}_2$ and $\text{C}_1$)
$\therefore\text{A}_{21}=(-1)^{2+1} \text{M}_{21}=(-1)^3 4=-4.$
View full question & answer→MCQ 371 Mark
If a matrix $A$ is such that $3A^3 + 2A^2 + 5A + I = 0,$ then $A^{-1}$ equal to:
- A
$-(3A^2 + 2A + 5)$
- B
$3A^2 + 2A + 5$
- C
$3A^2 - 2A - 5$
- ✓
Answer$3A^3 + 2A^2 + 5A + I = 0$
$\Rightarrow 3A^{-1} A^3 + 2A^{-1}A^2 + 5A^{-1}A + A^{-1}I = A^{-1}0$
$\Rightarrow 3A^2 + 2A + 5I + A^{-1} = 0$
$\Rightarrow A^{-1} = -(3A^2 + 2A + 5I)$
View full question & answer→MCQ 381 Mark
If A and B are square matrices of order 2, then det (A + B) = 0 is possible only when:
- A
Det (A) = 0 or det (B) = 0
- B
- C
Det (A) = 0 and det (B) = 0
- ✓
AnswerLet $\text{A}=[\text{a}_{\text{ij}}]$ and $\text{B}=[\text{b}_{\text{ij}}]$ be a square matrix of order 2
As their orders are same, A + B is defined as
$\text{A}+\text{B}=[\text{a}_{\text{ij}}+\text{b}_\text{ij}]$
$\Rightarrow|\text{A}+\text{B}|=|\text{a}_{\text{ij}}+\text{b}_\text{ij}|$
Now,
$|\text{A}+\text{B}|=0$
$\Rightarrow|\text{a}_{\text{ij}}+\text{b}_\text{ij}|=0$
$\Rightarrow[\text{a}_{\text{ij}}+\text{b}_\text{ij}]=0$
[corrsponding term is 0]
$\Rightarrow\text{A}+\text{B}=0$
View full question & answer→MCQ 391 Mark
Choose the correct answer from given four options in each of the Exercise : The area of a triangle with vertices $(-3, 0), (3, 0)$ and $(0, k)$ is $9$ sq. units. The value of $k$ will be:
AnswerWe know that, area of a triangle with vertices $(x_1, y_1), (x_2, y_2)$ and $(x_3, y_3)$ is given by
$\triangle=\begin{vmatrix}\frac{1}{2} \begin{vmatrix}\text{x}_1&\text{y}_1&1\\\text{x}_2&\text{y}_2&1\\\text{x}_3&\text{y}_3&1\end{vmatrix}\end{vmatrix}$
$\therefore$ Area of triangle with vertices $(-3, 0), (3, 0) $and$ (0, k)$ is
$\therefore\ \ \triangle=\begin{vmatrix}\frac{1}{2} \begin{vmatrix}-3&0&1\\3&0&1\\0&\text{k}&1\end{vmatrix}\end{vmatrix}=9 \ ($given$)$
$\Rightarrow\ [-3(-\text{k)}-0+1(3\text{k})]=\pm18$
$\Rightarrow\ 6\text{k}=\pm18$
$\therefore\ \ \text{k}=\pm\frac{18}{6}=\pm3$
View full question & answer→MCQ 401 Mark
If one of the roots of $\left |\begin{matrix} 3 &\text{amp; 5} &\text{amp; x}\\ 7 &\text{amp; x} &\text{amp; 7}\\ \text{x} &\text{amp; 5} &\text{amp; 3}\end{matrix}\right |=0$ is $-10,$ the other roots are:
AnswerGiven, $\left |\begin{matrix} 3 &\text{amp; 5} &\text{amp; x}\\ 7 &\text{amp; x} &\text{amp; 7}\\ \text{x} &\text{amp; 5} &\text{amp; 3}\end{matrix}\right |=0$
$\Rightarrow3(3\text{x}-35)-5(21-7\text{x})+\text{x}(35-\text{x}^2)=0$
$\Rightarrow9\text{x}-105-105+35\text{x}+35\text{x}-\text{x}^3=0$
$\Rightarrow\text{x}^3-79\text{x}+210=0$
$\Rightarrow(\text{x}+10)(\text{x}-3)(\text{x}-7)=0$
$\Rightarrow\text{x}=10, 3, 7$
View full question & answer→MCQ 411 Mark
If $ \begin{vmatrix} \text{a} &\text{amp; a} &\text{amp; x}\\ \text{m} &\text{amp; m} &\text{amp; m}\\ \text{b} &\text{amp; x} &\text{amp; b}\end{vmatrix}=0$ then $\text{x}=$
AnswerCorrect option: C. $a$ or $b$
Determinant of a matrix is zero if $2$ rows or columns are same.
Hence, if $x = a$ we get $1^{st}$ and $3^{rd}$ column sameAlso
if $x = b$ we get $1^{st}$ and $2^{nd}$ column same.
View full question & answer→MCQ 421 Mark
Choose the correct answer from given four options in each of the Exercise:
The maximum value of $\Delta=\begin{vmatrix}1 & 1&1 \\1 &1+\sin\theta&1\\1+\cos\theta &1&1\end{vmatrix}$ is ($\theta$ is real number):
- ✓
$\frac{1}{2}$
- B
$\frac{\sqrt{3}}{2}$
- C
$\sqrt{2}$
- D
$\frac{2\sqrt{3}}{4}$
AnswerCorrect option: A. $\frac{1}{2}$
Since,
$\Delta=\begin{vmatrix}1 & 1&1 \\1 &1+\sin\theta&1\\1+\cos\theta &1&1\end{vmatrix}$
$=\begin{vmatrix}0 &0&0 \\0 &\sin\theta&1\\\cos\theta &0&1\end{vmatrix}$ $\big[\because\ \text{C}_1\rightarrow\text{C}_2-\text{C}_3\text{ and C}_2\rightarrow\text{C}_2-\text{C}_3\big]$
$=-(\sin\theta.\cos\theta)$
$=\frac{1}{2}.2\sin\cos\theta=\frac{1}{2}\sin2\theta$
Since, the maximum value of $\sin2\theta$ is 1. So, for maximum value of $\theta$ should be 45°
$\therefore\ \Delta-\frac{1}{2}\sin2.45^\circ$
$=\frac{1}{2}\sin90^\circ=\frac{1}{2}.1=\frac{1}{2}$
View full question & answer→MCQ 431 Mark
The value of the determinant $\begin{vmatrix}\text{a}-\text{b}&\text{b}+\text{c}&\text{c}\\\text{b}-\text{c}&\text{c}+\text{b}&\text{b}\\\text{c}+\text{a}&\text{a}+\text{b}&\text{c}\end{vmatrix}$ is :
- A
$a^3 + b^3 + c^3$
- B
$3bc$
- ✓
$a^3 + b^3 + c^3 - 3abc$
- D
AnswerCorrect option: C. $a^3 + b^3 + c^3 - 3abc$
$\begin{vmatrix}\text{a}-\text{b}&\text{b}+\text{c}&\text{a}\\\text{b}-\text{c}&\text{c}+\text{b}&\text{b}\\\text{c}-\text{a}&\text{a}+\text{b}&\text{c}\end{vmatrix}$
$=\begin{vmatrix}-\text{b}&\text{b}+\text{c}+\text{a}&\text{a}\\-\text{c}&\text{c}+\text{a}+\text{b}&\text{b}\\-\text{a}&\text{a}+\text{b}+\text{c}&\text{c}\end{vmatrix}$ [Applying $C_1 \rightarrow C_1 - C_3$ and $C_2 \rightarrow C_2 + C_3]$
$=(-1)(\text{a}+\text{b}+\text{c})\begin{vmatrix}\text{b}&1&\text{a}\\\text{c}&1&\text{b}\\\text{a}&1&\text{c}\end{vmatrix} [$Taking $(-1)$ common from $C_1$ and $(a + b + c)$ common from $C_2]$
$=(-1)(\text{a}+\text{b}+\text{c})\begin{vmatrix}\text{b}&1&\text{a}\\\text{c}-\text{b}&0&\text{b}-\text{a}\\\text{a}-\text{b}&0&\text{c}-\text{a}\end{vmatrix} [$Applying $R_2 \rightarrow R_2 - R_1$ and $R_3\rightarrow R_3 - R_1]$
$= (-1)(a + b + c)[-(c - b)(c - a) + (b - a)(a - b)]$
$= (-1)(a + b + c)[-c^2 + ac + bc - ab + ba - b^2 - a^2 + ab]$
$= (-1)(a + b + c)(-a^2 - b^2 - c^2 + ab + bc + ac)$
$= (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ac)$
$= a^3 + ab^2 + ac^2 - a^2b - abc - a^2c + ba^2 + b^3 + bc^2 - ab^2-b^2c - abc + ca^2 + cb^2 + c^3 - acb - bc^2 - ac^2$
$= a^3 + b^3 + c^3- 3abc$
Hence, the correct option is $(c)$
View full question & answer→MCQ 441 Mark
Using the factor theorem it is found that a + b, b + c and c + a are three factors of the determinant $\begin{vmatrix}-2\text{a}&\text{a}+\text{b}&\text{a}+\text{c}\\\text{b}+\text{a}&-2\text{b}&\text{b}+\text{c}\\\text{c}+\text{a}&\text{c}+\text{b}&-2\text{c}\end{vmatrix}$ the other factor in the value of the determinant is:
Answer$\triangle=\begin{vmatrix}-2\text{a}&\text{a}+\text{b}&\text{a}+\text{c}\\\text{b}+\text{a}&-2\text{b}&\text{b}+\text{c}\\\text{c}+\text{a}&\text{c}+\text{b}&-2\text{c}\end{vmatrix}$
Let a + b = 2C, b + c = 2A and c + a = 2B
⇒ a + b + b + c + c + a = 2A + 2B + 2C
⇒ 2(a + b + c) = (A + B + C)
Also, a = (a + b + c) - (b + c) = (A + B + C) - 2A = B + C - A
Similarly, b = C + A - B, c = A + B - C
Hence, 4 is the order factor of the determinant.
View full question & answer→MCQ 451 Mark
Which of the following is not correct in a given determinant of $A,$ where $A = [a_{ij}]_{3\times 3}$:
AnswerCorrect option: B. Minor of an element can never be equal to cofactor of the same element.
$C_{ij} = (-1)^{i+j}M_{ij}$
So, for even values of $i + j, C_{ij} = M_{ij}$.
View full question & answer→MCQ 461 Mark
Choose the correct answer If $a, b, c,$ are in $A.P,$ then the determinant:
$\begin{vmatrix}\text{x}+2&\text{x}+3&\text{x}+2\text{a}\\\text{x}+3&\text{x}+4&\text{x}+2\text{b}\\\text{x}+4&\text{x}+5&\text{x}+2\text{c}\end{vmatrix}$ is
Answer$\triangle=\begin{vmatrix}\text{x}+2&\text{x}+3&\text{x}+\text{2a}\\\text{x}+3&\text{x}+4&\text{x}+2\text{b}\\\text{x}+4&\text{x}+5&\text{x}+2\text{c}\end{vmatrix}$
$\triangle=\begin{vmatrix}\text{x}+2&\text{x}+3&\text{x}+\text{2a}\\\text{x}+3&\text{x}+4&\text{x}+(\text{a}+\text{c})\\\text{x}+4&\text{x}+5&\text{x}+2\text{c}\end{vmatrix} (2b = a + c$ as $a,b$ and $c$ are in $A.P.)$
Applying $R_1→ R_1 - R_2$ and $R_3 → R_3 - R_2$, we have:
$\triangle=\begin{vmatrix}-1&-1&\text{a}-\text{c}\\\text{x}+3&\text{x}+4&\text{x}+(\text{a}+\text{c})\\1&1&\text{c}-\text{a}\end{vmatrix}$
Applying $R_1 → R_1 + R_3$, we have:
$\triangle=\begin{vmatrix}0&0&0\\\text{x}+3&\text{x}+4&\text{x}+(\text{a}+\text{c})\\1&1&\text{c}-\text{a}\end{vmatrix}$
View full question & answer→MCQ 471 Mark
If $A$ and $B$ are invertible matrices, which of the following statement is not correct.
- A
$\operatorname{adj} A=|A| A^{-1}$
- B
$\operatorname{det}\left(A^{-1}\right)=(\operatorname{det} A)^{-1}$
- ✓
$(A+B)^{-1}=A^{-1}+B^{-1}$
- D
$(A B)^{-1}=B^{-1} A^{-1}$
AnswerCorrect option: C. $(A+B)^{-1}=A^{-1}+B^{-1}$
We have$, adj A = |A|A^{-1}, det (A^{-1}) = (det A)^{-1}$ and $(AB)^{-1} = B^{-1}A^{-1}$ all are the properites of inverse of a matrix.
View full question & answer→MCQ 481 Mark
If $\text{A}=\begin{bmatrix} 1 & 0 & 1 \\ 0 & 0 & 1 \\ \text{a} & \text{b} & 2 \end{bmatrix},$ then $aI + bA + 2 A^2$ equals:
Answer$\text{A}=\begin{bmatrix} 1 & 0 & 1 \\ 0 & 0 & 1 \\ \text{a} & \text{b} & 2 \end{bmatrix}$
$\text{A}^2=\begin{bmatrix} 1+\text{a} & \text{b} & 3 \\ \text{a} & \text{b} & 2 \\ 3\text{a} & 2\text{b} & \text{a}+\text{b}+4 \end{bmatrix}$
$\Rightarrow\text{aI}+\text{bA}+2\text{A}^2$
$=\begin{bmatrix} 3\text{a}+2+\text{b} & 2\text{b} & 6+\text{b} \\ 2\text{a} & \text{a}+2\text{b} & \text{b}+4 \\ \text{ab}6\text{a} & 6\text{b}+\text{b}^2 & 3\text{a}+4\text{b}+8 \end{bmatrix}$
View full question & answer→MCQ 491 Mark
If x, y, z are non-zero real numbers, then the inverse, then the inverse of the matrix $\begin{bmatrix}\text{x} & 0 & 0\\ 0 & \text{y} & 0 \\ 0 & 0 & \text{z}\end{bmatrix}$, is:
- ✓
$\begin{bmatrix}\text{x}^{-1} & 0 & 0\\ 0 & \text{y}^{-1} & 0 \\ 0 & 0 & \text{z}^{-1}\end{bmatrix}$
- B
$\text{xyz}\begin{bmatrix}\text{x}^{-1} & 0 & 0\\ 0 & \text{y}^{-1} & 0 \\ 0 & 0 & \text{z}^{-1}\end{bmatrix}$
- C
$\frac{1}{\text{xyz}}\begin{bmatrix}\text{x} & 0 & 0\\ 0 & \text{y} & 0 \\ 0 & 0 & \text{z}\end{bmatrix}$
- D
$\frac{1}{\text{xyz}}\begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$
AnswerCorrect option: A. $\begin{bmatrix}\text{x}^{-1} & 0 & 0\\ 0 & \text{y}^{-1} & 0 \\ 0 & 0 & \text{z}^{-1}\end{bmatrix}$
A = IA
$\Rightarrow\begin{bmatrix}\text{x} & 0 & 0\\ 0 & \text{y} & 0 \\ 0 & 0 & \text{z}\end{bmatrix}=\begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}\text{A}$
$\Rightarrow\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}=\begin{bmatrix}\text{x}^1 & 0 & 0\\ 0 & \text{y}^1 & 0 \\ 0 & 0 & \text{z}^1\end{bmatrix}\text{A}$
$\Big[\text{Applying R}_1=\frac{1}{\text{x}}\text{R}_1,\text{R}_2=\frac{1}{\text{y}}\text{R}_2\text{ and R}_3=\frac{1}{\text{z}}\text{R}_3\Big]$
$\Rightarrow\text{A}^{-1}=\begin{bmatrix}\text{x}^{-1} & 0 & 0\\ 0 & \text{y}^{-1} & 0 \\ 0 & 0 & \text{z}^{-1}\end{bmatrix}$
View full question & answer→MCQ 501 Mark
If $\text{A}=\begin{bmatrix} 2 & -1 \\ 3 & -2 \end{bmatrix},$ then $A^n=$
- ✓
$\text{A}=\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix},$ if $n$ is an even natural number
- B
$\text{A}=\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix},$ if $n$ is an odd natural number
- C
$\text{A}=\begin{bmatrix} -1 & 0 \\ 0 & 1 \end{bmatrix},$ if ${n}\in\text{N}$
- D
AnswerCorrect option: A. $\text{A}=\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix},$ if $n$ is an even natural number
$\text{A}=\begin{bmatrix} 2 & -1 \\ 3 & -2 \end{bmatrix}$
$\text{A}^2=\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}=1$
If $n$ is an natural number.
View full question & answer→MCQ 511 Mark
Find the value of the following determinant: $\begin{vmatrix}\displaystyle \frac{-4}{7} &\text{amp; } \displaystyle \frac{-6}{35}\\ 5 &\text{amp; } \displaystyle \frac{-2}{5}\end{vmatrix}$
- A
$\displaystyle \frac{15}{34}$
- B
$\displaystyle \frac{32}{45}$
- C
$\displaystyle \frac{25}{33}$
- ✓
$\displaystyle \frac{38}{35}$
AnswerCorrect option: D. $\displaystyle \frac{38}{35}$
The value of $\begin{vmatrix}\displaystyle \frac{-4}{7} &\text{amp; } \displaystyle \frac{-6}{35}\\ 5 &\text{amp; } \displaystyle \frac{-2}{5}\end{vmatrix}$ is $\bigg(\frac{-4}{7}\times\frac{-2}{5}\bigg)-\bigg(\frac{-4}{7}\times5\bigg)$
$=\frac{8}{25}+\frac{30}{35}=\frac{38}{35}$
View full question & answer→MCQ 521 Mark
Which of the following is not a property of determinant:
- ✓
The value of determinant changes if all of its rows and columns are interchanged
- B
The value of determinant changes if any two rows or columns are interchanged
- C
The value of determinant is zero if any two rows and columns are identical
- D
The value of determinant gets multiplied by k, if each element of row or column is multiplied by k
AnswerCorrect option: A. The value of determinant changes if all of its rows and columns are interchanged
The value of determinant remains unchanged if all of its rows and columns are interchanged i.e. |A| = |A’|
where A is a square matrix and A’ is the transpose of the matrix A.
View full question & answer→MCQ 531 Mark
If A is a singular matrix, then adj A is.
AnswerGiven ∣A∣ = 0
We know ∣adjA∣ = ∣A∣ n - 1
∴ ∣adjA∣ = 0
Hence, adj A is singular
View full question & answer→MCQ 541 Mark
Evaluate $\begin{bmatrix}3&-1&3\\6&-5&4\\3&-2&3\end{bmatrix}$ is:
AnswerExpanding along $\text{R}_1,$ we get
$\triangle=\begin{bmatrix}3&-1&3\\6&-5&4\\3&-2&3\end{bmatrix}$
$\triangle=3\begin{bmatrix}-5&45\\-2&3\end{bmatrix}-(-1)\begin{bmatrix}6&4\\3&3\end{bmatrix}+3\begin{bmatrix}6&-5\\-3&-2\end{bmatrix}$
$\triangle=3(-15+90)+(18-12)+3(-12+15) $
$\triangle=3(75)+6+9=240. $
View full question & answer→MCQ 551 Mark
Choose the correct answer from given four options in each of the Exercise:
If $\text{f(x)}==\begin{vmatrix}0&\text{x}-\text{a}&\text{x}-\text{b} \\\text{x}+\text{a} &0&\text{x}-\text{c}\\\text{x}+\text{b}&\text{x}+\text{c}&0\end{vmatrix},$ then:
Answer$\text{f(x)}=\begin{vmatrix}0&\text{x}-\text{a}&\text{x}-\text{b} \\\text{x}+\text{a} &0&\text{x}-\text{c}\\\text{x}+\text{b}&\text{x}+\text{c}&0\end{vmatrix}$
$\Rightarrow\ \text{f}(0)=\begin{vmatrix}0&-\text{a}&-\text{b}\\\text{a}&0&-\text{c}\\\text{b}&\text{c}&0\end{vmatrix},$ Which is skew-symmetric determinant of order 3
Hence f(0) = 0.
View full question & answer→MCQ 561 Mark
If $A$ and $B$ are square matrices such that $B = -A^{-1} BA,$ then $(A + B)^2 =$
- A
$O$
- ✓
$A^2 + B^2$
- C
$A^2 + 2AB + B^2$
- D
$A + B$
AnswerCorrect option: B. $A^2 + B^2$
$B = -A^{-1} BA$
$\Rightarrow AB = -AA^{-1}BA$
$\Rightarrow Ab = -IBA$
$\Rightarrow AB = -BA$
$\Rightarrow AB + BA = 0 .....(i)$
Consider,
$(A + B)^2 = A^2 + AB + BA + B^2$
$(\because\text{AB}\neq\text{BA})$
$(A + B)^2 = A^2 + O + B^2$
$(A + B)^2 = A^2 + B^2$
View full question & answer→MCQ 571 Mark
If $\triangle_1=\begin{vmatrix}1&1&1\\\text{a}&\text{b}&\text{c}\\\text{a}^2&\text{b}^2&\text{c}^2\end{vmatrix},\triangle_2=\begin{vmatrix}1&\text{bc}&\text{a}\\1&\text{ca}&\text{b}\\1&\text{ab}&\text{c}\end{vmatrix},$ then:
- ✓
$\triangle_1+\triangle_2=0$
- B
$\triangle_1+2\triangle_2=0$
- C
$\triangle_1=\triangle_2$
- D
AnswerCorrect option: A. $\triangle_1+\triangle_2=0$
$\triangle_2=\begin{vmatrix}1&\text{bc}&\text{a}\\1&\text{ca}&\text{b}\\1&\text{ab}&\text{c}\end{vmatrix}$
$=\frac{1}{\text{abc}}\begin{vmatrix}1&\text{abc}&\text{a}^2\\1&\text{bca}&\text{b}^2\\1&\text{cab}&\text{c}^2\end{vmatrix}\ [R_1, R_2, R_3$ are multiplies by $a, b$ and $c$ respectively, therefore we divide by $abc]$
$=\frac{\text{abc}}{\text{abc}}\begin{vmatrix}1&1&\text{a}^2\\1&1&\text{b}^2\\1&1&\text{c}^2\end{vmatrix} [$Taking abc common from $C_2]$
$=-\begin{vmatrix}1&\text{a}&\text{a}^2\\1&\text{b}&\text{b}^2\\1&\text{c}&\text{c}^2\end{vmatrix}$ $\text{C}_1\leftrightarrow\text{C}_2$
We know that the value of a determinant remains unchanged if its rows and columns are interchanged.
so,
$\triangle_2=-\begin{vmatrix}1&1&1\\\text{a}&\text{b}&\text{c}\\\text{a}^2&\text{b}^2&\text{c}^2\end{vmatrix} $
$=-\triangle_1$
$\triangle_1+\triangle_2=0$
View full question & answer→MCQ 581 Mark
If $\text{A}=\begin{bmatrix} 3 & 4 \\ 2 & 4 \end{bmatrix},\text{B}=\begin{bmatrix} -2 & -2 \\ 0 & -1 \end{bmatrix}$ then $(A + B)^{-1} =$
AnswerWe have
$(\text{A}+\text{B})=\begin{bmatrix} 1 & 2 \\ 2 & 3 \end{bmatrix}$
$\therefore|\text{A}+\text{B}| = -1\neq0$
Thus, $(A + B)^{-1}$ exists.
Now,
$(\text{A}+\text{B})^\text{T}=\begin{bmatrix} 1 & 2 \\ 2 & 3 \end{bmatrix}$
Here,
$(\text{A}+\text{B})^\text{T}\neq-(\text{A}+\text{B})$
Hence, it is not a akew symmetric matrix.
We also know that $A^{-1} + B^{-1}$ is not the same as $(A + B)^{-1}$.
View full question & answer→MCQ 591 Mark
Maximum value of a second order determinant whose every element is either 0, 1 or 2 only is:
AnswerSo, $\text{A}=\begin{bmatrix}\text{a}&\text{b}\\\text{c}&\text{d}\end{bmatrix}$
Given a, b, c & D can only be 0, 1, 2
det A = ad-bc
So for max. value of A,
a = 2 and d = 2 and b, c $\in$ 0, 0
So, Max value of det $\text{A}=\begin{bmatrix}2&0\\0&2\end{bmatrix}=4$
View full question & answer→MCQ 601 Mark
The equations in terms of $x$ and $y$ are:
- A
$x – y = 50, 2x – y = 550$
- ✓
$x – y = 50, 2x + y = 550$
- C
$x + y = 50, 2x + y = 550$
- D
$x + y = 50, 2x – y = 550$
AnswerCorrect option: B. $x – y = 50, 2x + y = 550$
View full question & answer→MCQ 611 Mark
If $a > 0$ and discriminant of $ax^2 + 2bx + c$ is negative, then $\triangle=\begin{vmatrix}\text{a}&\text{b}&\text{ax}+\text{b}\\\text{b}&\text{c}&\text{bx}+\text{c}\\\text{ax}+\text{b}&\text{bx}+\text{c}&0\end{vmatrix}$ is:
AnswerDiscriminant $D$ of $ax^2 + 2bx + c = (2b)^2- 4ac < 0 [$Given$]$
$\Rightarrow 4b^2 - 4ac < 0$
$\Rightarrow b^2 - ac < 0,$ where $a > 0 .....(i)$
$\Rightarrow\triangle=\begin{vmatrix}\text{a}&\text{b}&\text{ax}+\text{b}\\\text{b}&\text{c}&\text{bx}+\text{c}\\\text{ax}+\text{b}&\text{bx}+\text{c}&0\end{vmatrix}$
$\Rightarrow\triangle=\begin{vmatrix}\text{ax}&\text{bx}&\text{ax}^2+\text{bx}\\\text{b}&\text{c}&\text{bx}+\text{c}\\\text{ax}+\text{b}&\text{bx}+\text{c}&0\end{vmatrix} [$Applying $R_1 \rightarrow xR_1]$
$\Rightarrow\triangle=\frac{1}{\text{x}}\begin{vmatrix}\text{ax}+\text{b}&\text{bx}+\text{c}&\text{ax}^2+\text{bx}+\text{bx}+\text{c}\\\text{b}&\text{c}&\text{bx}+\text{c}\\\text{ax}+\text{b}&\text{bx}+\text{c}&0\end{vmatrix} [$Applying $R_1 \rightarrow R_1 + R_2]$
$\Rightarrow\triangle=\frac{1}{\text{x}}\begin{vmatrix}0&0&\text{ax}^2+2\text{bx}+\text{c}\\\text{b}&\text{c}&\text{bx}+\text{c}\\\text{ax}+\text{b}&\text{bx}+\text{c}&0\end{vmatrix} [$Applying $R_1 \rightarrow R_1 - R_3]$
$\Rightarrow\triangle=\frac{1}{\text{x}}\begin{Bmatrix}\text{ax}^2+2\text{bx}+\text{c}\begin{vmatrix}\text{b}&\text{c}\\\text{ax}+\text{b}&\text{bx}+\text{c}\end{vmatrix}\end{Bmatrix} [$Expanding along $R_1]$
$\Rightarrow\triangle=\frac{1}{\text{x}}(\text{ax}^2+2\text{bx}+\text{c})(\text{b}^2\text{x}+\text{bc}-\text{acx}-\text{bc})$
$\Rightarrow\triangle=\frac{1}{\text{x}}(\text{ax}^2+2\text{bx}+\text{c})\text{ x }(\text{b}^2-\text{ac})$
$\Rightarrow\triangle=(\text{ax}^2+2\text{bx}+\text{c})(\text{b}^2\text{ac})<0 \ [$From eq. $(i)]$
$\Rightarrow\triangle<0$
View full question & answer→MCQ 621 Mark
If $A$ is an invertible matrix of order $3,$ then which of the following is not true:
- A
$|\text{adj A}|=|\text{A}|^2$
- B
$(\text{A}^{-1})^{-1}=\text{A}$
- ✓
If $BA = CA,$ than $\text{B}\neq\text{C},$ where $B$ and $C$ are square matrices of order $3$
- D
$(\text{AB})^{-1}=\text{B}^{-1}\text{A}^{-1},$ where $\text{B}\neq\big[\text{b}_{\text{ij}}\big]_{3\times3}$ and $\text{|B|}\neq0$
AnswerCorrect option: C. If $BA = CA,$ than $\text{B}\neq\text{C},$ where $B$ and $C$ are square matrices of order $3$
$BA = CA$
$\Rightarrow BAA^{-1} = CAA^{-1}$
$\Rightarrow BI = CI$
$\Rightarrow B = C$
Hence$, (c)$ is not correct.
View full question & answer→MCQ 631 Mark
If $A$ is a square matrix such that $A^2 = I$, then $A^{-1}$ is equal to:
Answer$A^2=1$
$A^{-1} A^2=A^{-1} I$
$A=A^{-1}$
View full question & answer→MCQ 641 Mark
If $A$ is an invertible matrix, then which of the following is not true:
- ✓
$(\text{A}^2)^\text{-1}=(\text{A}^{-1})^2$
- B
$|\text{A}^{-1}|=|\text{A}|^{-1}$
- C
$(\text{A}^\text{T})^\text{-1}=(\text{A}^{-1})^\text{T}$
- D
$|\text{A}|\neq0$
AnswerCorrect option: A. $(\text{A}^2)^\text{-1}=(\text{A}^{-1})^2$
We have, $\left|A^{-1}\right|=|A|^{-1},(A T)^{-1}=\left(A^{-1}\right)^{\top}$ and $|\text{A}|\neq0$ all are the properties of the inverse of a matrix $A.$
View full question & answer→MCQ 651 Mark
If a, b, c are distinct, then the value of x satisfying $\begin{vmatrix}0&\text{x}^2-\text{a}&\text{x}^3-\text{b}\\\text{x}^2+\text{a}&0&\text{x}^2+\text{c}\\\text{x}^4+\text{b}&\text{x}-\text{c}&0\end{vmatrix}=0$ is:
AnswerWhen we put x = 0 in the given matrix, then it turns out to be the skew symmetric matrix of order 3 and the determinant of the skew symmetric matrix of odd order is always 0.
View full question & answer→MCQ 661 Mark
Choose the correct answer from given four options in each of the Exercise:
If A and B are invertible matrices, then which of the following is not correct?
- A
$\text{adj A} = |\text{A}|.\text{A}^{-1}$
- B
$\text{det (A)}^{-1}=[\text{det(A)}]^{-1}$
- C
$(\text{AB})^{-1}=\text{B}^{-1}\text{A}^{-1}$
- ✓
$(\text{A}+\text{B})^{-1}=\text{B}^{-1}+\text{A}^{-1}$
AnswerCorrect option: D. $(\text{A}+\text{B})^{-1}=\text{B}^{-1}+\text{A}^{-1}$
Since, A and B are invertible matrices, So, we can say that
$(\text{AB})^{-1}=\text{B}^{-1}\text{A}^{-1}\ \dots(\text{i})$
Also, $\text{A}^{-1}=\frac{1}{|\text{A}|}(\text{adj A})$
$\Rightarrow\ \text{adj A}=|\text{A}|.\text{A}^{-1}\ \ \dots(\text{ii})$
Also, $\text{det (A)}^{-1}=[\text{det (A)}]^{-1}$
$\Rightarrow\ \text{det (A)}^{-1}=\frac{1}{\big[\text{det (A)}\big]}$
$\Rightarrow\ \text{det (A)}.\text{det (A)}^{-1}=1\ \ \dots(\text{iii})$
Which is true.
Again, $(\text{A}+\text{B})^{-1}=\frac{1}{\big|(\text{A}+\text{B})\big|}\text{ adj }(\text{A}+\text{B})$
$\Rightarrow\ (\text{A}+\text{B})^{-1}\neq\text{B}^{-1}+\text{A}^{-1}\ \ \dots(\text{iv})$
So, only option (d) is incorrect.
View full question & answer→MCQ 671 Mark
If $\text{A}=\begin{bmatrix}\alpha &\text{amp; 2} \\2 &\text{amp; }\alpha \end{bmatrix}$and $ |\text{A}^3|=125,$ then $\alpha$ is equal to:
- ✓
$\pm 3$
- B
$\pm 2$
- C
$\pm 5$
- D
$0$
AnswerCorrect option: A. $\pm 3$
Given, $ |\text{A}^3| =|\text{A}|^3 = 125$
$\Rightarrow|\text{A}|=5\Rightarrow\alpha^2-4=5\Rightarrow\alpha^2=9$
$\Rightarrow \alpha=\pm 3$
View full question & answer→MCQ 681 Mark
If $ \begin{vmatrix} 6\text{i} &\text{amp;} -3\text{i} &\text{amp;} 1\\ 4 &\text{amp; } 3\text{i} &\text{amp;} -1 \\ 20 &\text{amp; } 3 &\text{amp; i}\end{vmatrix}=\text{x}+\text{iy},$ then?
Answer$ \begin{vmatrix} 6\text{i} &\text{amp;} -3\text{i} &\text{amp;} 1\\ 4 &\text{amp; } 3\text{i} &\text{amp;} -1 \\ 20 &\text{amp; } 3 &\text{amp; i}\end{vmatrix}=\text{x}+\text{iy},$
$\Rightarrow6\text{i}(3\text{i}^2+3)+3\text{i}(4\text{i}+20)+1(12-60\text{i})=\text{x}+\text{iy}\Rightarrow0=\text{x}+\text{iy}$
$\therefore \text{x}=\text{y}=0$
View full question & answer→MCQ 691 Mark
A determinant of second order is made with the elements 0 and 1. The number of determinants with non - negative values is:
AnswerThere are only three determinants of second order with negative value,
$\begin{bmatrix}0&\text{amp; }1\\1&\text{amp; }0\end{bmatrix}, \begin{bmatrix}0&\text{amp; }1\\1&\text{amp; }1\end{bmatrix}, \begin{bmatrix}1&\text{amp; }1\\1&\text{amp; }0\end{bmatrix}$
Number of possible determinants with elements $0$ and$1$ are ${ 2 }^{ 4 }=16$
therefore, number of determinants with non - negative values is $13.$
View full question & answer→MCQ 701 Mark
If $A$ is a $3 \times 3$ matrix and det$(3A) = k($det $A),$ then $k =$
View full question & answer→MCQ 711 Mark
The system of linear equations:
x + y + z = 2
2x + y − z = 3
3x + 2y + kz = 4
has a unique solution if
Answerx + y + z = 22x + y − z = 3
3x + 2y + kz = 4
The determination of the coefficient matrix $\begin{bmatrix}1&1&1\\2&1&-1\\3&2&\text{k}\end{bmatrix}$ is
= k + 2 -2k - 3 + 1
=-k
To have a unique solution the determinant ≠ 0
⇒ k ≠ 0
View full question & answer→MCQ 721 Mark
If A is a skew symmetric matrix, then ∣A∣ is
AnswerSINCE THE SKEW SYMMETRIC MATRIX CONSIST OF ELEMENTS OF OPPOSITE SIGN AT OPPOSITE SIDE OF MATRIX DIAGONAL WITH ALL THE DIAGONAL ELEMENTS AS ZERO THEREFORE THE DETERMINANT OF SKEW SYMMETRIC MATRIX IS ZERO.
View full question & answer→MCQ 731 Mark
For which of the following element in the determinant $\triangle=\begin{bmatrix}5&-5&8\\6&2&-1\\5&-6&8\end{bmatrix},$ the minor and the cofactor both are zero.
AnswerConsider the element 2 in the determinant $\triangle=\begin{bmatrix}5&-5&8\\6&2&-1\\5&-6&8\end{bmatrix}$
The minor of the element 2 is given by
$\therefore\text{M}_{22}=\begin{bmatrix}5&8\\5&8\end{bmatrix}=40-40=0$
$\Rightarrow\text{A}^{22}=(-1)^2+2 (0)=0.$
View full question & answer→MCQ 741 Mark
If $A$ is a singular matrix, then $\text{A (adj A)}$ is a
AnswerGiven $A$ is a singular matrix.
$\Rightarrow ∣A∣ = 0$
$\text{A (adj A) = ∣A∣I = 0I = O}$
$\therefore \text{A (adj A)}$ is a zero matrix.
View full question & answer→MCQ 751 Mark
The number of solutions of the system of equations$:\ 2x + y − z = 7;x − 3y + 2z = 1;x + 4y − 3z = 5$
AnswerThe given system of equations can be written in matrix form as follows:
$\begin{bmatrix}2&1&-1\\1&-3&2\\1&4&-3\end{bmatrix}\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}=\begin{bmatrix}7\\1\\5\end{bmatrix}$
$\text{AX}=\text{B}$
Here,
$\text{A}=\begin{bmatrix}2&1&-1\\1&-3&2\\1&4&-3\end{bmatrix},\text{X}=\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}$ and $\text{B}=\begin{bmatrix}7\\1\\5\end{bmatrix}$
Now,
$|\text{A}|=2(9-8)-1(-3-2)-1(4+3)$
$=2+5-7$
$=0$
Let $c_{ij}$ be the $co-$factors of the elements $a_{ij}$ in $A = [a_{ij}]$. Then,
$\text{C}_{11}=(-1)^{1+1}\begin{vmatrix}-3&2\\4&-3\end{vmatrix}=1,$ $\text{C}_{12}=(-1)^{1+2}\begin{vmatrix}1&2\\1&-3\end{vmatrix}=5,$ $\text{C}_{13}=(-1)^{1+3}\begin{vmatrix}1&-3\\1&4\end{vmatrix}=7$
$\text{C}_{21}=(-1)^{2+1}\begin{vmatrix}1&-1\\4&-3\end{vmatrix}=-1,$ $\text{C}_{22}=(-1)^{2+2}\begin{vmatrix}2&-1\\1&-3\end{vmatrix}=-5,$ $\text{C}_{23}=(-1)^{2+3}\begin{vmatrix}2&1\\1&4\end{vmatrix}=-7$
$\text{C}_{31}=(-1)^{3+1}\begin{vmatrix}1&-1\\-3&2\end{vmatrix}=5,$ $\text{C}_{32}=(-1)^{3+2}\begin{vmatrix}2&-1\\1&2\end{vmatrix}=-5,$ $\text{C}_{33}=(-1)^{3+3}\begin{vmatrix}2&1\\1&-3\end{vmatrix}=-7$
$\text{adj }\text{A}=\begin{bmatrix}1&5&7\\-1&-5&-7\\5&-5&-7\end{bmatrix}^\text{T}=\begin{bmatrix}1&-1&5\\5&-5&-5\\7&-7&-7\end{bmatrix}$
$\Rightarrow(\text{adj }\text{A})\text{B}=\begin{bmatrix}1&-1&5\\5&-5&-5\\7&-7&-7\end{bmatrix}\begin{bmatrix}7\\1\\5\end{bmatrix}$
$=\begin{bmatrix}7-1+25\\35-5-25\\49-7-35\end{bmatrix}=\begin{bmatrix}32\\5\\6\end{bmatrix}\neq0$
The given system of equations is inconsistent.
Thus, it has no solution.
View full question & answer→MCQ 761 Mark
The number of solutions of the system of equations
2x + y − z = 7
x − 3y + 2z = 1
x + 4y − 3z = 5
AnswerFrom the given equation we get,
$\triangle=\begin{vmatrix}2&1&-1\\1&-3&2\\1&4&-3\end{vmatrix}$
⇒ 2(9 - 8) -1(-3 - 2) - 1(4 + 3)
⇒ 2(1) - 1(-5) - 1(7)
⇒ 2 + 5 - 7
⇒ 2 + 5 -7
⇒ 0
$\triangle_1=\begin{vmatrix}7&1&-1\\1&-3&2\\5&4&-3\end{vmatrix}$
⇒ 7(9 - 8) - 1(-3 - 10) - 1(4 + 15)
⇒ 7(1) - 1(-13) - 1(19)
⇒ 7 + 13 - 19
⇒ 20 - 19
$\Rightarrow1\neq0$
Hence the gvien system no solution.
View full question & answer→MCQ 771 Mark
Choose the correct answer from given four options in each of the Exercise:
If $\begin{vmatrix}2\text{x}&5\\8&\text{x}\end{vmatrix}=\begin{vmatrix}6&-2\\7&3\end{vmatrix},$ then value of x is:
AnswerWe have, $\begin{vmatrix}2\text{x}&5\\8&\text{x}\end{vmatrix}=\begin{vmatrix}6&-2\\7&3\end{vmatrix}$
$\Rightarrow\ 2\text{x}^2-40=18+17$
$\Rightarrow\ 2\text{x}^2=32+40$
$\Rightarrow\ \text{x}^2=\frac{72}{2}=36$
$\Rightarrow\ \text{x}^2=36$
$\Rightarrow\ \text{x}=\pm6$
View full question & answer→MCQ 781 Mark
The number of distinct real roots of $\begin{vmatrix}\text{cosec}&\sec\text{x}&\sec\text{x}\\\sec\text{x}&\text{cosec}\text{x}&\sec\text{x}\\\sec\text{x}&\sec\text{x}&\text{cosecx}\end{vmatrix}=0$ lies in the interval $-\frac{\pi}{4}\leq\text{x}\leq\frac{\pi}{4}$ is:
AnswerLet $\triangle=\begin{vmatrix}\text{cosec x}&\sec\text{x}&\sec\text{x}\\\sec\text{x}&\text{cosec}\text{x}&\sec\text{x}\\\sec\text{x}&\sec\text{x}&\text{cosec x}\end{vmatrix}$
$=(\text{cosec x})^3\begin{vmatrix}1&\frac{\sec\text{x}}{\text{cosecx}}&\frac{\sec\text{x}}{\text{cosecx}}\\\frac{\sec\text{x}}{\text{cosecx}}&1&\frac{\sec\text{x}}{\text{cosecx}}\\\frac{\sec\text{x}}{\text{cosecx}}&\frac{\sec\text{x}}{\text{cosecx}}&1\end{vmatrix}$
$=(\text{cosec x})^3\begin{vmatrix}1&\tan\text{x}&\tan\text{x}\\\tan\text{x}&1&\tan\text{x}\\\tan\text{x}&\tan\text{x}&1\end{vmatrix}$
$=(\text{cosec x})^3\begin{vmatrix}1-\tan\text{x}&\tan\text{x}-1&0\\0&1-\tan\text{x}&\tan\text{x}-1\\\tan\text{x}&\tan\text{x}&1\end{vmatrix} [$Applying $R_1 \rightarrow R_1 - R_2, R_2 \rightarrow R_2 - R_3]$
$=(\text{cosec x})^3(1-\tan\text{x})^2\begin{vmatrix}1&-1&0\\0&1&-1\\\tan\text{x}&\tan\text{x}&1\end{vmatrix} [$Taking out $(1-\tan\text{x})$ common from $R_1$ and $R_2]$
$=(\text{cosec x})^3(1-\tan\text{x})^2\begin{Bmatrix}1\begin{vmatrix}1&-1\\\tan\text{x}&1\end{vmatrix}+\tan\text{x}\begin{vmatrix}-1&0\\1&-1\end{vmatrix}\end{Bmatrix} [$Expanding along $C_1]$
$=(\text{cosec x})^3(1-\tan\text{x})^2\{1+\tan\text{x}+\tan\text{x}\}$
$=(\text{cosec x})^3(1-\tan\text{x})^2\{1+2\tan\text{x}\}$
$\triangle=0$
$=(\text{cosec x})^3(1-\tan\text{x})^2(1+2\tan\text{x})=0$
$(1-\tan\text{x})=0,(\text{coses x})^3=0$ and $(1+2\tan\text{x})=0$
Or $\tan\text{x}=1,\text{cosec x}=0$ and $\tan\text{x}=\frac{-1}{2}$
$\Rightarrow-\frac{\pi}{4}\leq\text{x}\leq\frac{\pi}{4}$ $\Big[\tan\text{x}=1,\text{x}=\frac{-1}{2}$ are $2$ real roots as $\text{cosec x}=0$ has no solution$\Big]$
Thus, these are $2$ solutions.
View full question & answer→MCQ 791 Mark
Which of the following is correct?
- A
Determinant is a square matrix
- B
Determinant is a number associated to a matrix
- ✓
Determinant is a number associated to a square matrix
- D
AnswerCorrect option: C. Determinant is a number associated to a square matrix
Determinant is defined only for a square matrix.
and its denotes the value of that square matrix.
View full question & answer→MCQ 801 Mark
The system of equations:
x + y + z = 5
x + 2y + 3z = 9
x + 3y + λz = µ
has a unique solution, if
Answerx + y + z = 5
x + 2y + 3z = 9
x + 3y + λz = µ
The determinant of the coefficient matrix $\begin{bmatrix}1&1&1\\1&2&3\\1&3&\lambda\end{bmatrix}$ is
= 2λ - 9 - λ + 3 + 1
= λ - 5
For unique solution determinant ≠ 0
⇒ λ ≠ 5
The right hand side is non zero what so ever be the value of µ.
View full question & answer→MCQ 811 Mark
If A is a square matrix of order 3 and |A| = 5, then the value of |2A′| is:
AnswerAccording to the property of transpose of a matrix,
(kA′) = kA′
Also, from the property of determinant of a matrix,
|A′| = |A|
Thus, |2A′| = 2|A|
= 2 × 5
= 10
View full question & answer→MCQ 821 Mark
If $A$ is an invertible matrix, then det $(A^{-1})$ is equal to:
AnswerCorrect option: B. $\frac{1}{\text{det(A)}}$
We know that $\big|\text{A}^{-1}\big|=\frac{1}{|\text{A}|}$
View full question & answer→MCQ 831 Mark
If area of triangle is 35 sq units with vertices (2, – 6), (5, 4) and (k, 4). Then k is
Answer$\therefore\ \text{Given: Area of triangle}=\text{Modulus of}\ \frac{1}{2}\begin{vmatrix}x_1&y_1&1\\x_2&y_2&1\\x_3&y_3&1\end{vmatrix}=35$
$\Rightarrow\ \text{Modulus of}\ \frac{1}{2}\begin{vmatrix}2&-6&1\\5&4&1\\k&4&1\end{vmatrix}=35$
$\Rightarrow\ \bigg|\frac{1}{2}\left[2(4-4)-(-6)(5-k)+1(20-4k)\right]\bigg|=35$
$\Rightarrow\ \begin{vmatrix}\frac{1}{2}\left[0+30-6k+20-4k\right]\end{vmatrix}=35\ \Rightarrow\ \begin{vmatrix}\frac{1}{2}\left[50-10k\right]\end{vmatrix}=35$
$\Rightarrow\ \begin{vmatrix}25-5k\end{vmatrix}=35\ \Rightarrow\ \ 25-5k=\pm35$
Taking positive sign, 25 - 5k = 35 $\ \ \Rightarrow k=-2$
Taking negative sign, 25 - 5k = -35 $\Rightarrow k=12$
Therefore, option (d) is correct.
View full question & answer→MCQ 841 Mark
If $\text{A}_{\text{r}}=\begin{vmatrix}1&\text{r}&2^{\text{r}}\\2&\text{n}&\text{n}^{2}\\\text{n}&\frac{\text{n}(\text{n}+1)}{2}&2^{\text{n}+1} \end{vmatrix},$ then the value of $\sum\limits_{\text{r}=1}^\text{n}\text{A}_\text{r}$ is:
AnswerCorrect option: C. $-2n^3$
$\text{A}_\text{r}=\begin{vmatrix}1&\text{r}&2^{\text{r}}\\2&\text{n}&\text{n}^{2}\\\text{n}&\frac{\text{n}(\text{n}+1)}{2}&2^{\text{n}+1} \end{vmatrix}$
$\Rightarrow\sum\limits_{\text{r}=1}^\text{n}\text{A}_\text{r}=\begin{vmatrix}\sum\limits_{\text{r}=1}^\text{n}1&\sum\limits_{\text{r}=1}^\text{n}\text{r}&\sum\limits_{\text{r}=1}^\text{n}2\text{r}\\\sum\limits_{\text{r}=1}^\text{n}2&\text{n}&\text{n}^{2}\\\text{n}&\frac{\text{n}(\text{n}+1)}{2}&2^{\text{n}+1} \end{vmatrix}$
As $\sum\limits_{\text{r}=1}^\text{r}1=1+1+1\ ......+1(n$ times$)=\text{n}$
$\Rightarrow\sum\limits_{\text{r}=1}^\text{r}\text{r}=1+2+3+\ .....+\text{n}=\frac{\text{n}(\text{n}+1)}{2}$
Let $\text{S}=\sum\limits_{\text{r}=1}^\text{r}2^\text{r}=2+2^2+2^3=\ .....+2^{\text{n}}$
$\Rightarrow2\text{S}=2^2+3^2=\ ....+2^{\text{n}}+2^{\text{n}+1}$
$\Rightarrow2\text{S}-\text{S}$
$\Rightarrow\text{S}=\sum\limits_{\text{r}=1}^\text{n}2^{\text{r}}=2^{\text{n+1}}-2$
$\Rightarrow\sum\limits_{\text{r}=1}^\text{n}\text{A}_\text{r}=\begin{vmatrix}\text{n}&\frac{\text{n}(\text{n}+1)}{2}&2^{\text{n}+1}-2\\2\text{n}&\text{n}&\text{n}^{2}\\\text{n}&\frac{\text{n}(\text{n}+1)}{2}&2^{\text{n}+1} \end{vmatrix}$
$[$Applying $R_1 \rightarrow R_1 - R_2]$
$\Rightarrow\sum\limits_{\text{r}=1}^\text{n}\text{A}_\text{r}=\begin{vmatrix}\text{n}-\text{n}&\frac{\text{n}(\text{n}+1)}{2}-\frac{\text{n}(\text{n}+1)}{2}&2^{\text{n}+1}-2-2^{\text{n}+1}\\2\text{n}&\text{n}&\text{n}^{2}\\\text{n}&\frac{\text{n}(\text{n}+1)}{2}&2^{\text{n}+1} \end{vmatrix}$
$=\begin{vmatrix}0&0&-2\\2\text{n}&\text{n}&\text{n}^{2}\\\text{n}&\frac{\text{n}(\text{n}+1)}{2}&2^{\text{n}+1} \end{vmatrix}$
$=-2\times\begin{vmatrix}2\text{n}&\text{n}\\\text{n}&\frac{\text{n}(\text{n}+1)}{2}\end{vmatrix}$
$=-2\big[\text{n}^{3}+\text{n}^2-\text{n}^2\big]$
$=-2\text{n}^3$
View full question & answer→MCQ 851 Mark
The value of $\begin{vmatrix}1&1&1\\^\text{n}\text{C}_1&^{\text{n}+2}\text{C}_1&^{\text{n}+4}\text{C}_1\\^\text{n}\text{C}_2&^{\text{n}+2}\text{C}_2&^{\text{n}+4}\text{C}_2\end{vmatrix}$ is:
Answer$\begin{vmatrix}1&1&1\\^\text{n}\text{C}_1&^{\text{n}+2}\text{C}_1&^{\text{n}+4}\text{C}_1\\^\text{n}\text{C}_2&^{\text{n}+2}\text{C}_2&^{\text{n}+4}\text{C}_2\end{vmatrix}$
$=\begin{vmatrix}1&1&1\\\text{n}&\text{n}+2&\text{n}+3\\\frac{\text{n}(\text{n}-1)}{2}&\frac{(\text{n}+2)(\text{n}+1)}{2}&\frac{(\text{n}+4)(\text{n}+3)}{2}\end{vmatrix}$
$=\begin{vmatrix}1&0&0\\\text{n}&2&4\\\frac{\text{n}(\text{n}+1)}{2}&\frac{4\text{n}+2}{2}&\frac{8\text{n}+12}{2}\end{vmatrix} [$Applying $C_2 \rightarrow C_2 - C_1$ and $C_3 \rightarrow C_3 - C_1]$
$=\begin{vmatrix}1&0&0\\\text{n}&2&4\\\frac{\text{n}(\text{n}+1)}{2}&(2\text{n}+1)&(4\text{n}+6)\end{vmatrix}$
$=8\text{n}+12-8\text{n}-4$
$=8$
Hence, the correct option is $(c)$
View full question & answer→MCQ 861 Mark
If $\text{A}=\begin{bmatrix} \text{a} & 0 & 0 \\ 0 & \text{a} & 0 \\ 0 & 0 &\text{a} \end{bmatrix},$ then the value of $\text{|adj A|}$ is:
- A
$a^{27}$
- B
$a^9$
- ✓
$a^6$
- D
$a^2$
Answer$\text{A}=\begin{bmatrix} \text{a} & 0 & 0 \\ 0 & \text{a} & 0 \\ 0 & 0 &\text{a} \end{bmatrix}$
$\therefore|\text{A}|=\begin{bmatrix} \text{a} & 0 & 0 \\ 0 & \text{a} & 0 \\ 0 & 0 &\text{a} \end{bmatrix}=\text{a}^3\neq0$
and
$n = 3$
Thus, we have
$\text{|adj A|} = |A|^{n-1} = (a^3)^2 = a^6.$
View full question & answer→MCQ 871 Mark
If $\text{x},\text{ y}\in\text{R},$ then the determinant $\triangle=\begin{vmatrix}\cos\text{x}&-\sin\text{x}&1\\\sin\text{x}&\cos\text{x}&1\\\cos(\text{x}+\text{y})&-\sin(\text{x}+\text{y})&0\end{vmatrix}$ lies in the interval:
AnswerCorrect option: A. $\Big[-\sqrt{2},\sqrt{2}\Big]$
$\triangle=\begin{vmatrix}\cos\text{x}&-\sin\text{x}&1\\\sin\text{x}&\cos\text{x}&1\\\cos(\text{x}+\text{y})&-\sin(\text{x}+\text{y})&0\end{vmatrix}$
$=\begin{vmatrix}\cos\text{x}&-\sin\text{x}&1\\\sin\text{x}&\cos\text{x}&1\\0&0&\sin\text{y}-\cos\text{y}\end{vmatrix} [$Applying $R_3 \rightarrow R_3 - \text{cosy} R_1 + \text{siny} R_2]$
$=(\sin\text{y}-\cos\text{y})(\cos^2\text{x}+\sin^2\text{x})$
$=\sin\text{y}-\cos\text{y}$
$=\sqrt{2}\Big(\frac{1}{\sqrt{2}}\sin\text{y}-\frac{1}{\sqrt{2}}\cos\text{y}\Big)$
$=\sqrt{2}\Big(\cos\frac{\pi}{4}\sin\text{y}-\sin\frac{\pi}{4}\cos\text{y}\Big)$
$=\sqrt{2}\sin\Big(\text{y}-\frac{\pi}{4}\Big)$
Therefore, $-\sqrt{2}\leq\triangle\leq\sqrt{2}$
Hence, the correct option is $(a)$
View full question & answer→MCQ 881 Mark
If $ A^5 = 0$ Such that $\text{A}^{\text{n}}\neq\text{I for }1\leq\text{n}\leq4,\text{ then}(\text{I}-\text{A})^{-1}$ equals:
Answer$A^5 = 0$
Using $a^5 - b^5 = (a - b)(a^4 + a^3b + a^2b^2 + ab^3 + b^4)$
$I - A^5 = (I - A)(I + A + A^2 + A^3 + A^4)$
$I = (I - A)(I + A + A^2 + A^3 + A^4)$
$(I - A)^{-1}I = (I - A)^{-1}(I - A)(I + A + A^2 + A^3 + A^4)$
$(I - A)-1 = I + A + A^2 + A^3 + A^4$
View full question & answer→MCQ 891 Mark
If two rows of a determinant are identical, then what is the value of the determinant ?
AnswerLet determinant of this matrix is x, if we interchange the two identical rows of the matrix then by property the determinant of the new matrix is - x, but overall the matrix will be same as we have interchanged only the two identical rows.
So, x = -x, we have x = 0.
Hence, the determinant is zero.
View full question & answer→MCQ 901 Mark
Find the cofactor of element -3 in the determinant $\triangle=\begin{bmatrix}1&4&4\\-3&5&9\\2&1&2\end{bmatrix}$ is:
AnswerThe minor of element -3 is given by
$\text{M}_{21}=\begin{bmatrix}4&4\\1&2\end{bmatrix}=4(2)-4=4$ (Obtained by eliminating $\text{R}_2$ and $\text{C}_1$)
$\therefore\text{A}_{21}=(-1)^{2+1} \text{M}_{21}=(-1)^3 4=-4.$
View full question & answer→MCQ 911 Mark
A and B are two points and C is any point collinear with A and B. IF AB=10, BC=5, then AC is equal to:
AnswerSince C is collinear with A and B,C lies either
(i) to the left of point B or
(ii) to the right of point B
∴ In case (i) AC = AB - BC = 10 - 5 = 5
In case (ii) AC = AB + BC = 10 + 5 = 15
View full question & answer→MCQ 921 Mark
Evaluate $\begin{bmatrix}4&8&12\\6&12&18\\7&14&21\end{bmatrix}$ is:
Answer$\triangle=\begin{bmatrix}4&8&12\\6&12&18\\7&14&21\end{bmatrix}$
Taking $4, 6$ and $7$ from $\text{R}_1, \text{ R}_2,\text{ R}_3$ respectively
$\triangle=4\times6\times7\begin{bmatrix}4&8&12\\6&12&18\\7&14&21\end{bmatrix}$
Since the elements of all rows are identical, the determinant is zero.
View full question & answer→MCQ 931 Mark
Find the value of x if $\begin{bmatrix}3&\text{3}\\2&\text{x}^2\end{bmatrix}=\begin{bmatrix}5&3\\3&2\end{bmatrix}.$
- ✓
$\text{x}=1,-\frac{1}{3}$
- B
$\text{x}=-1,-\frac{1}{3}$
- C
$\text{x}=1,\frac{1}{3}$
- D
$\text{x}=-1,\frac{1}{3}$
AnswerCorrect option: A. $\text{x}=1,-\frac{1}{3}$
Given that $\begin{bmatrix}3&\text{3}\\2&\text{x}^2\end{bmatrix}=\begin{bmatrix}5&3\\3&2\end{bmatrix}$
$\Rightarrow3\text{x}^2-2\text{x}=5(2)-3(3)$
$⇒3\text{x}^2-2\text{x}=1$
solving for x, we get
$\text{x}=1,-\frac{1}{3}$
View full question & answer→MCQ 941 Mark
Consider the system of equations: $a_1x + b_1y + c_1z = 0 , a_2x + b_2y + c_2z = 0 , a_3x + b_3y + c_3z = 0,$
if$\begin{vmatrix}\text{a}_1&\text{b}_1&\text{c}_1\\\text{a}_2&\text{b}_2&\text{c}_2\\\text{a}_3&\text{b}_3&\text{c}_3\end{vmatrix}=0$, then the system has
- ✓
- B
One trivial and one non $-$ trivial solutions.
- C
- D
Only trivial solution $(0, 0, 0).$
AnswerHere, $|A| = 0$ and $B = 0 \ ($Given$)$
If $|A| = 0 $ and $(\text{adj}\ A)B = 0,$ then the system is consistent and has infinitely many solutions.
Clearly, it has more than two solutions.
View full question & answer→MCQ 951 Mark
Find the value of x, if $\begin{bmatrix}2&5\\3&\text{x}\end{bmatrix}=\begin{bmatrix}\text{x}&-1\\5&3\end{bmatrix}$ is:
Answer$\begin{bmatrix}2&5\\3&\text{x}\end{bmatrix}=\begin{bmatrix}\text{x}&-1\\5&3\end{bmatrix}$
$\Rightarrow2\text{x}-15=3\text{x}+5 $
$\Rightarrow\text{x}=-20$
View full question & answer→MCQ 961 Mark
If $\begin{bmatrix} 1 & -\tan\theta \\ \tan\theta & 1 \end{bmatrix}\begin{bmatrix} 1 & \tan\theta \\ -\tan\theta & 1 \end{bmatrix}-1=\begin{bmatrix} \text{a} & -\text{b} \\ \text{b} & \text{a} \end{bmatrix},$ then:
- A
$\text{a}=1,\text{b}=1$
- ✓
$\text{a}=\cos2\theta,\text{b}=\sin2\theta$
- C
$\text{a}=\sin2\theta,\text{b}=\cos2\theta$
- D
AnswerCorrect option: B. $\text{a}=\cos2\theta,\text{b}=\sin2\theta$
$\begin{bmatrix} 1 & -\tan\theta \\ \tan\theta & 1 \end{bmatrix}^{-1}=\frac{1}{\sec^2\theta}\begin{bmatrix} 1 & \tan\theta \\ -\tan\theta & 1 \end{bmatrix}^{-1}$
$\begin{bmatrix} 1 & -\tan\theta \\ \tan\theta & 1 \end{bmatrix}\begin{bmatrix} 1 & \tan\theta \\ -\tan\theta & 1 \end{bmatrix}^{-1}=\begin{bmatrix} \text{a} & -\text{b} \\ \text{c} & \text{a} \end{bmatrix}$
$\Rightarrow\begin{bmatrix} 1 & -\tan\theta \\ \tan\theta & 1 \end{bmatrix}\frac{1}{\sec^2\theta}\begin{bmatrix} 1 & \tan\theta \\ -\tan\theta & 1 \end{bmatrix}^{-1}=\begin{bmatrix} \text{a} & -\text{b} \\ \text{b} & \text{a} \end{bmatrix}$
$\Rightarrow\frac{1}{\sec^2\theta}\begin{bmatrix} 1 & -\tan\theta \\ \tan\theta & 1 \end{bmatrix}\begin{bmatrix} 1 & \tan\theta \\ -\tan\theta & 1 \end{bmatrix}^{-1}=\begin{bmatrix} \text{a} & -\text{b} \\ \text{b} & \text{a} \end{bmatrix}$
$\Rightarrow\begin{bmatrix} \frac{1-\tan^2\theta}{\sec^2\theta} & \frac{-2\tan\theta}{\sec^2\theta} \\ \frac{2\tan\theta}{\sec^2\theta} & \frac{1-\tan^2\theta}{\sec^2\theta} \end{bmatrix}=\begin{bmatrix}\text{a} & -\text{b} \\ \text{b} & \text{a} \end{bmatrix}$
On comparing, we get
$\text{a}=\frac{1-\tan^2\theta}{\sec^2\theta}\text{ and b}=\frac{2\tan\theta}{\sec^2\theta}$
$\Rightarrow\text{a}=\cos^2\theta-\sin^2\theta\text{ and b}=2\sin\theta\cos\theta$
$\Rightarrow\text{a}=\cos2\theta\text{ and b}=\sin2\theta$
View full question & answer→MCQ 971 Mark
Let $\text{X}=\begin{bmatrix}\text{x}_1\\\text{x}_2\\\text{x}_3\end{bmatrix},\text{A}=\begin{bmatrix}1&-1&2\\2&0&1\\3&2&1\end{bmatrix}$ and $\text{B}=\begin{bmatrix}3\\1\\4\end{bmatrix}$. If $AX = B,$ then $X$ is equal to:
- ✓
$\begin{bmatrix}1\\2\\3\end{bmatrix}$
- B
$\begin{bmatrix}-1\\-2\\-3\end{bmatrix}$
- C
$\begin{bmatrix}-1\\2\\3\end{bmatrix}$
- D
$\begin{bmatrix}0\\2\\1\end{bmatrix}$
AnswerCorrect option: A. $\begin{bmatrix}1\\2\\3\end{bmatrix}$
Here,
$\text{A}=\begin{bmatrix}1&-1&2\\2&0&1\\3&2&1\end{bmatrix},\text{X}=\begin{bmatrix}\text{x}_1\\\text{x}_2\\\text{x}_3\end{bmatrix}$ and $\text{B}=\begin{bmatrix}3\\1\\4\end{bmatrix}$
$|\text{A}|=1(0-2)+1(2-3)+2(4-0)$
$=-2-1+8$
$=5$
Let $C_{ij}$ be the $co-$factors of the elements $a_{ij}$ in $A = [a_{ij}]$. Then,
$\text{C}_{11}=(-1)^{1+1}\begin{vmatrix}0&1\\2&1\end{vmatrix}=-2,$ $\text{C}_{12}=(-1)^{1+2}\begin{vmatrix}2&1\\3&1\end{vmatrix}=1,$ $\text{C}_{13}=(-1)^{1+3}\begin{vmatrix}2&0\\3&2\end{vmatrix}=4$
$\text{C}_{21}=(-1)^{2+1}\begin{vmatrix}-1&2\\2&1\end{vmatrix}=5,$ $\text{C}_{22}=(-1)^{2+2}\begin{vmatrix}1&2\\3&1\end{vmatrix}=-5,$ $\text{C}_{23}=(-1)^{2+3}\begin{vmatrix}1&-1\\3&2\end{vmatrix}=-5$
$\text{C}_{31}=(-1)^{3+1}\begin{vmatrix}-1&2\\0&1\end{vmatrix}=-1,$ $\text{C}_{32}=(-1)^{3+2}\begin{vmatrix}1&2\\2&1\end{vmatrix}=3,$ $\text{C}_{33}=(-1)^{3+3}\begin{vmatrix}1&-1\\2&0\end{vmatrix}=2$
$\text{adj }\text{A}=\begin{bmatrix}-2&5&-1\\5&-5&-5\\-1&2&3\end{bmatrix}^\text{T}$
$=\begin{bmatrix}-2&5&-1\\1&-5&3\\4&-5&2\end{bmatrix}$
$\Rightarrow\text{A}^{-1}=\frac{1}{|\text{A}|}\text{adj }\text{A}$
$=\frac{1}{5}\begin{bmatrix}-2&5&-1\\1&-5&3\\4&-5&2\end{bmatrix}$
$\therefore\ \text{X}=\text{A}^{-1}\text{B}$
$\Rightarrow\text{X}=\frac{1}{5}\begin{bmatrix}-2&5&-1\\1&-5&3\\4&-5&2\end{bmatrix}\begin{bmatrix}3\\1\\4\end{bmatrix}$
$\Rightarrow\text{X}=\frac{1}{5}\begin{bmatrix}-6+5-4\\3-5+12\\12-5+8\end{bmatrix}$
$\Rightarrow\begin{bmatrix}\text{x}_1\\\text{x}_2\\\text{x}_3\end{bmatrix}=\frac{1}{5}\begin{bmatrix}-5\\10\\15\end{bmatrix}$
$\Rightarrow\begin{bmatrix}\text{x}_1\\\text{x}_2\\\text{x}_3\end{bmatrix}=\begin{bmatrix}-1\\2\\3\end{bmatrix}$
View full question & answer→MCQ 981 Mark
Evaluate $\begin{bmatrix}\text{x}^2&\text{x}^3&\text{x}^4\\\text{x}&\text{y}&\text{z}\\\text{x}^2&\text{x}^3&\text{x}^4\end{bmatrix}$ is:
Answer$\begin{bmatrix}\text{x}^2&\text{x}^3&\text{x}^4\\\text{x}&\text{y}&\text{z}\\\text{x}^2&\text{x}^3&\text{x}^4\end{bmatrix}$
If the elements of any two rows or columns are identical, then the value of determinant is zero.
Here, the elements of row $1$ and row $3$ are identical.
Hence, its determinant is $0.$
View full question & answer→MCQ 991 Mark
The existence of the unique solution of the system of equations:
$x + y + z = \lambda $
$5x − y + µz = 10$
$2x + 3y − z = 6$
depends on
AnswerCorrect option: A. $µ$ only.
For a unique solution, $|\text{A}|\neq0$
$\Rightarrow\begin{vmatrix}1 1 15 -1 \mu2 3 -1\end{vmatrix}\neq0$
$\Rightarrow1(1-3\mu)-1(-5-2\mu)+1(15+2)\neq0$
$\Rightarrow1-3\mu+5+2\mu+17\neq0$
$\Rightarrow-\mu+23\neq0$
$\Rightarrow\mu\neq23$
So, existence of a unique solution depends only on $\mu$.
View full question & answer→MCQ 1001 Mark
If $\begin{vmatrix}2\text{x}&5\\8&\text{x}\end{vmatrix}=\begin{vmatrix}6&-2\\7&3\end{vmatrix},$ then x =
AnswerCorrect option: C. $\pm6$
$\begin{vmatrix}2\text{x}&5\\8&\text{x}\end{vmatrix}=\begin{vmatrix}6&-2\\7&3\end{vmatrix}$
$\Rightarrow2\text{x}^2-40=18+14$
$\Rightarrow2\text{x}^2-40=32$
$\Rightarrow2\text{x}^2=72$
$\Rightarrow\text{x}^2=36$
$\Rightarrow\text{x}=\pm6$
Hence, the correct option is (C)
View full question & answer→MCQ 1011 Mark
The value of $y ($breadth of rectangular field$)$ is:
- ✓
$150m$
- B
$200m$
- C
$430m$
- D
$350m$
AnswerCorrect option: A. $150m$
View full question & answer→MCQ 1021 Mark
$\begin{bmatrix}2\text{x}&5\\8&\text{x}\end{bmatrix}=\begin{bmatrix}6\text{x}&-2\\7&3\end{bmatrix},$ then the value of $\text{x}$ is:
AnswerCorrect option: C. $\pm6$
View full question & answer→MCQ 1031 Mark
If $a, b, c$ are in $A.P.$, then the determinant $\begin{vmatrix}\text{x}+2&\text{x}+3&\text{x}+2\text{a}\\\text{x}+3&\text{x}+4&\text{x}+2\text{b}\\\text{x}+4&\text{x}+5&\text{x}+2\text{c}\end{vmatrix}$
Answer$\begin{vmatrix}\text{x}+2&\text{x}+3&\text{x}+2\text{a}\\\text{x}+3&\text{x}+4&\text{x}+2\text{b}\\\text{x}+4&\text{x}+5&\text{x}+2\text{c}\end{vmatrix}$
$=\begin{vmatrix}0&0&2(\text{a}+\text{c}-2\text{b})\\\text{x}+3&\text{x}+4&\text{x}+2\text{b}\\\text{x}+4&\text{x}+5&\text{x}+2\text{c}\end{vmatrix}$
$[$Applying $R_1 \rightarrow R_1 + R_3- R_2, R_1 \rightarrow R_1 - R_2]$
$=\begin{vmatrix}0&0&0\\\text{x}+3&\text{x}+4&\text{x}+2\text{b}\\\text{x}+4&\text{x}+5&\text{x}+2\text{c}\end{vmatrix}$ $[\because a, b, c$ are in $A.P.]$
$=0$
View full question & answer→MCQ 1041 Mark
If A is a singular matrix, then A (adj A) is a.
AnswerGiven A is a singular matrix.
⇒ ∣A∣ = 0
A(adjA) = ∣A∣ I = 0I = O
∴ A(adjA) is a zero matrix.
View full question & answer→MCQ 1051 Mark
Let $\text{A}=\begin{vmatrix}1&\sin\theta&1\\-\sin\theta&1&\sin\theta\\-1&-\sin\theta&1\end{vmatrix},$ where $0\leq\theta\leq2\pi.$ Then:
AnswerCorrect option: D. $\text{Det (A)}\in[2,4]$
$\begin{vmatrix}1&\sin\theta&1\\-\sin\theta&1&\sin\theta\\-1&-\sin\theta&1\end{vmatrix}$
$=\begin{vmatrix}1&\sin\theta&2\\-\sin\theta&1&\sin\theta\\-1&-\sin\theta&1\end{vmatrix} [$Applying $C_3 \rightarrow C_3 + C_1]$
$=2\times\begin{vmatrix}-\sin\theta&1\\-1&-\sin\theta\end{vmatrix} [$Expanding along $C_3]$
$=2(\sin^2\theta+1)$
Given, $0\leq\theta\leq2\pi$
$-1\leq\sin\theta\leq1$
$0\leq\sin^2\theta\leq1$
$|\text{A}|=2(\sin^2\theta+1)$
$|\text{A}|=2\times1=2$ $[\theta=0]$
$|\text{A}|=2\times2=4$ $[\theta=2\pi]$
$\text{Det (A)}\in[2,4]$
View full question & answer→MCQ 1061 Mark
If $\text{A}=\begin{bmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{bmatrix},$ then $A^5 =$
Answer$\text{A}=\begin{bmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{bmatrix}$
$\Rightarrow\text{A}=2\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$
$\Rightarrow\text{A}=2\text{I}$
$\Rightarrow\text{A}^5=(2\text{I})^5$
$\Rightarrow\text{A}^5=16\times2\text{I}$
$\Rightarrow\text{A}^5=16\begin{bmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{bmatrix}$
$\Rightarrow\text{A}^5=16\text{A}$
View full question & answer→MCQ 1071 Mark
If the determinant $\begin{vmatrix}\text{a}&\text{b}&2\text{a}\alpha+3\text{b}\\\text{b}&\text{c}&2\text{b}\alpha+3\text{c}\\2\text{a}\alpha+ 3\text{b}&2\text{b}\alpha+3\text{c}&0\end{vmatrix}=0,$ then:
- A
$a, b, c$ are in $H.P.$
- ✓
$\alpha$ is a root of $4ax^2 + 12bx + 9c = 0$ or $a, b, c$ are in $G.P$.
- C
$a, b, c$ are in $G.P$. only.
- D
$a, b, c$ are in $A.P$.
AnswerCorrect option: B. $\alpha$ is a root of $4ax^2 + 12bx + 9c = 0$ or $a, b, c$ are in $G.P$.
Let $\triangle=\begin{vmatrix}\text{a}&\text{b}&2\text{a}\alpha+3\text{b}\\\text{b}&\text{c}&2\text{b}\alpha+3\text{c}\\2\text{a}\alpha+3\text{b}&2\text{b}\alpha+3\text{c}&0\end{vmatrix}$
$=\begin{vmatrix}\text{a}-\text{b}&\text{b}&2\text{a}\alpha+3\text{b}\\\text{b}-\text{c}& \text{c}&2\text{b}\alpha+3\text{c}\\2\text{a}\alpha+ 3\text{b}-2\text{b}\alpha-3&2\text{b}\alpha+3\text{c}&0\end{vmatrix}$ $[$Applying $C_1 \rightarrow C_1- C_2]$
$=\begin{vmatrix}\text{a}-\text{b}&\text{b}&2\text{a}\alpha+3\text{b}\\\text{b}-\text{c}& \text{c}&2\text{b}\alpha+3\text{c}\\2(\text{a}-\text{b})\alpha+3(\text{b}-\text{c})&2\text{b}\alpha+3\text{b}&0\end{vmatrix}$
$=2\alpha(2\text{a}\alpha+3\text{b})-3(2\text{b}\alpha+3\text{c})\begin{vmatrix}\text{a}-\text{b}&\text{b}\\\text{b}-\text{c} \text{c}\end{vmatrix}$ [Expanding along $R_3$]
$=-(4\text{a}\alpha^2+12\text{b}\alpha+9\text{c})(\text{ac}-\text{b}^2)$
But $\triangle=0\ [$Given$]$
$\Rightarrow-(4\text{a}\alpha^2+12\text{b}\alpha+9\text{c})(\text{ac}-\text{b}^2)=0$
$\Rightarrow(4\text{a}\alpha^2+12\text{b}\alpha+9\text{c})=0$
Or $(\text{ac}-\text{b}^2)=0$
$\Rightarrow\alpha$ is a root of $4ax^2 + 12bx + 9c = 0$ or $ac = b^2$, i.e. $a, b, c$ are in $G.P.$
View full question & answer→MCQ 1081 Mark
If $\text{A}+\text{B}+\text{C}=\pi,$ then the value of $\begin{vmatrix}\sin(\text{A}+\text{B}+\text{C})&\sin(\text{A}+\text{C})&\cos\text{C}\\-\sin\text{B}&0&\tan\text{A}\\\cos(\text{A}+\text{B})&\tan(\text{B}+\text{C})&0\end{vmatrix}$ is equal to:
Answer$\text{A}+\text{B}+\text{C}=\pi$
$\text{A}+\text{C}=\pi-\text{B},\text{A}+\text{B}=\pi-\text{C}$ and $\text{B}+\text{C}=\pi-\text{A}$
Thus the determinant becomes
$\begin{vmatrix}\sin\pi&\sin(\pi-\text{B})&\cos\text{C}\\-\sin\text{B}&0&\tan\text{A}\\\cos(\pi-\text{C})&\tan(\pi-\text{A})&0\end{vmatrix}$
$=\begin{vmatrix}0&\sin\text{B}&\cos\text{C}\\-\sin\text{B}&0&\tan\text{A}\\-\cos\text{C}&-\tan\text{A}&0\end{vmatrix}$
$[\sin\pi=0,\sin(\pi-\text{B}),\cos(\pi-\text{C})=-\cos\text{C},\tan(\pi-\text{A})=-\tan\text{A}]$
It is a skew symmetric matrix of the odd order 3. Thus by property of determinants, we get
$|\triangle|=0$
$\begin{vmatrix}0&\sin\text{B}&\cos\text{C}\\-\sin\text{B}&0&\tan\text{A}\\-\cos\text{C}&-\tan\text{A}&0\end{vmatrix}$
View full question & answer→MCQ 1091 Mark
If $\text{A}=\begin{vmatrix} 10 & 2 \\ 30 & 6 \end{vmatrix}$ then $|\text{A}|$
Answer$\text{A}=\begin{vmatrix} 10 & 2 \\ 30 & 6 \end{vmatrix}$
$=10\times6-30\times2=60-60=0$
View full question & answer→MCQ 1101 Mark
If A is any skew - symmetric matrix of odd order then ∣A∣ equals:
Answerif A is skew symmetric matrix
then $\text{A} = \text{-A}^\text{T}$
$\therefore |\text{A}|=-|\text{A}^\text{T}|=-|\text{A}|$
$\Rightarrow 2|\text{A}|=0$
$\Rightarrow|\text{A}|=0$
View full question & answer→MCQ 1111 Mark
Evaluate $\begin{bmatrix}5&-4\\1&\sqrt{3}\end{bmatrix}$
- A
$4\sqrt{3}+4$
- B
$4\sqrt{3}+5$
- ✓
$5\sqrt{3}+4$
- D
$4\sqrt{3}-4$
AnswerCorrect option: C. $5\sqrt{3}+4$
Evaluating along $\text{R}_1$,we get
$\triangle5(\sqrt3)-(-4)^1=5\sqrt{3}+4$
View full question & answer→MCQ 1121 Mark
Let $\text{f(x)}=\begin{vmatrix}\cos\text{x}&\text{x}&1\\2\sin\text{x}&\text{x}&2\text{x}\\\sin\text{x}&\text{x}&\text{x}\end{vmatrix},$ then $\lim_\limits{\text{x}\rightarrow0}\frac{\text{f(x)}}{\text{x}^2}$ is equal to:
Answer$\text{f(x)}=\begin{vmatrix}\cos\text{x}&\text{x}&1\\2\sin\text{x}&\text{x}&2\text{x}\\\sin\text{x}&\text{x}&\text{x}\end{vmatrix}$
$=\begin{vmatrix}\cos\text{x}&\text{x}&1\\\sin\text{x}&0&\text{x}\\\sin\text{x}&\text{x}&\text{x}\end{vmatrix} [$Applying $R_2 \rightarrow R_2 - R_3]$
$=\begin{vmatrix}\cos\text{x}&\text{x}&1\\\sin\text{x}&0&\text{x}\\\sin\text{x}-\cos\text{x}&0&\text{x}-1\end{vmatrix} [$Applying $R_3 \rightarrow R_3 - R_1]$
$=-\text{x}[\text{x}\sin\text{x}-\sin\text{x}-\text{x}\sin\text{x}+\text{x}\cos\text{x}]$
$=-\text{x}(\text{x}\cos\text{x}-\sin\text{x})$
$\therefore\ \lim_\limits{\text{x}\rightarrow0}\frac{\text{f(x)}}{\text{x}^2}=\lim_\limits{\text{x}\rightarrow0}\frac{\text{x}(\sin\text{x}-\text{x}\cos\text{x})}{\text{x}^2}$
$=\lim_\limits{\text{x}\rightarrow0}\frac{\sin\text{x}}{\text{x}^2}-\lim_\limits{\text{x}\rightarrow0}\cos\text{x}$
$=1-1=0$
Hence, the correct option is $(a)$
View full question & answer→MCQ 1131 Mark
Find the minor of the element 1 in the determinant $\triangle=\begin{bmatrix}1&5\\3&8\end{bmatrix}$ is:
AnswerThe minor of the element 1 can be obtained by deleting the first row and the first column
$\therefore\text{ M}_{11}=8$
View full question & answer→MCQ 1141 Mark
The value of the determinant $\begin{vmatrix} 5 &\text{amp; } 1 \\ 3 &\text{amp; } 2 \end{vmatrix}$
View full question & answer→MCQ 1151 Mark
Which of the following is not correct?
- A
$|\text{A}|=|\text{A}^{\text{T}}|,$ where $\text{A}=[\text{a}_{\text{ij}}]_{3\times3}$
- B
$|\text{kA}|=|\text{k}^3|,$ where $\text{A}=[\text{a}_{\text{ij}}]_{3\times3}$
- C
If a is a skew$-$symmetric of odd order, then $|A| = 0$
- ✓
$\begin{vmatrix}\text{a}&\text{c}\\\text{e}&\text{g} \end{vmatrix}+\begin{vmatrix}\text{b}&\text{c}\\\text{f}&\text{g} \end{vmatrix}+\begin{vmatrix}\text{a}&\text{d}\\\text{e}&\text{h}\end{vmatrix}+\begin{vmatrix}\text{b}&\text{d}\\\text{f}&\text{h}\end{vmatrix}$
AnswerCorrect option: D. $\begin{vmatrix}\text{a}&\text{c}\\\text{e}&\text{g} \end{vmatrix}+\begin{vmatrix}\text{b}&\text{c}\\\text{f}&\text{g} \end{vmatrix}+\begin{vmatrix}\text{a}&\text{d}\\\text{e}&\text{h}\end{vmatrix}+\begin{vmatrix}\text{b}&\text{d}\\\text{f}&\text{h}\end{vmatrix}$
$\begin{vmatrix}\text{a}+\text{b}&\text{c}+\text{d}\\\text{e}+\text{f}&\text{g}+\text{h} \end{vmatrix}=\begin{vmatrix}\text{a}+\text{b}&\text{c}\\\text{e}+\text{f}&\text{h} \end{vmatrix}+\begin{vmatrix}\text{a}+\text{b}&\text{d}\\\text{e}+\text{f}&\text{h}\end{vmatrix}$
$=\begin{vmatrix}\text{a}&\text{c}\\\text{e}&\text{g} \end{vmatrix}+\begin{vmatrix}\text{b}&\text{c}\\\text{f}&\text{g} \end{vmatrix}+\begin{vmatrix}\text{a}&\text{d}\\\text{e}&\text{h}\end{vmatrix}+\begin{vmatrix}\text{b}&\text{d}\\\text{f}&\text{h}\end{vmatrix}$
View full question & answer→MCQ 1161 Mark
Let $\text{P}$ and $\text{Q}$ be $3\times3$ matrices with $\text{P}\neq\text{Q}.$ If $\text{P}^3=\text{Q}^3$ and $\text{P}^2\text{Q}=\text{Q}^2\text{P}$ then determinant of $(\text{P}^2+\text{Q}^2)$ is equal to:
Answer$\text{P}^3=\text{Q}^3$
$\Rightarrow\text{P}^3- \text{P}^2\text{Q}=\text{Q}^3- \text{Q}^2\text{P}$
$\Rightarrow\text{P}^2(\text{P- Q})=\text{Q}^2(\text{Q- P})$
$\Rightarrow \text{P}^2(\text{P- Q})-\text{Q}^2(\text{Q- P})=0$
$\Rightarrow (\text{P}^2+\text{Q}^2)(\text{P}-\text{Q})=0\Rightarrow|\text{P}^2+\text{Q}^2|=0$
View full question & answer→MCQ 1171 Mark
Evaluate $\begin{bmatrix}1&0&1\\0&0&1\\1&0&1\end{bmatrix}$ is:
Answer$\triangle=\begin{bmatrix}1&0&1\\0&0&1\\1&0&1\end{bmatrix}$
$\triangle=1\begin{bmatrix}0&1\\0&1\end{bmatrix}-0\begin{bmatrix}0&1\\1&1\end{bmatrix}+1\begin{bmatrix}0&0\\1&0\end{bmatrix}$
$\triangle=1(0-0)-0(0-1)+1(0-0)$
$\triangle=0-0+0=0.$
View full question & answer→MCQ 1181 Mark
For the system of equations:
x + 2y + 3z = 1
2x + y + 3z = 2
5x + 5y + 9z = 4
- ✓
There is only one solution.
- B
There exists infinitely many solution.
- C
- D
AnswerCorrect option: A. There is only one solution.
x + 2y + 3z = 1
2x + y + 3z = 2
5x + 5y + 9z = 4
The determinant of the coefficient matrix $\begin{bmatrix}1&2&3\\2&1&3\\5&5&9\end{bmatrix}$ is
$= -6 - 2(18 - 15) + 3(10 - 5)$
$= -6 - 6 + 15$
$=3\neq0$
The right hand side is also non zero.
The system has a unique solution.
View full question & answer→MCQ 1191 Mark
If the coordinates of the vertices of a triangle are (0, 0), (0, 2) and(3, 1), then area of the triangle is:
AnswerArea of triangle $=\frac{1}{2} \begin{vmatrix} 0 &\text{amp; }0 &\text{amp; 1} \\ 0&\text{amp; 2} &\text{amp; 1} \\3 &\text{amp;1} &\text{amp; 1} \end{vmatrix}= \frac{1}{2}\times|-6|=3$
View full question & answer→MCQ 1201 Mark
If for the matrix $A, A^3 = I,$ than $A^{-1} =$
Answer$A^3 = Ia$
$\Rightarrow A^{-1}A^3 = A^{-1}I$
$\Rightarrow IA^2 = A^{-1}I$
$\Rightarrow A^2 = A^{-1}$
View full question & answer→MCQ 1211 Mark
If $x, y, z$ are different from zero and $\begin{vmatrix}1+\text{x}&1&1\\1&1+\text{y}&1\\1&1&1+\text{z}\end{vmatrix}=0,$ then the value $x^{-1} + y^{-1} + z^{-1}$ is:
Answer$\begin{vmatrix}1+\text{x}&1&1\\1&1+\text{y}&1\\1&1&1+\text{z}\end{vmatrix}=0$
$\Rightarrow\begin{vmatrix}\text{x}&0&-\text{z}\\0&\text{y}&-\text{z}\\1&1&1+\text{z}\end{vmatrix}=0 [$Applying R$_2\rightarrow R_2 - R_3$ and $R_1 \rightarrow R_1 - R_3]$
$\Rightarrow\text{x}\big[\text{y}(1+\text{z})+\text{z}\big]+1(\text{yz})=0\ [$Expanding along first column$]$
$\Rightarrow\text{x}[\text{y}+\text{yz}+\text{z}]+\text{yz}=0$
$\Rightarrow\text{xy}+\text{xyz}+\text{xz}+\text{yz}=0$
$\Rightarrow\text{xy}+\text{yz}+\text{zx}=-\text{xyz}$
$\Rightarrow\frac{\text{xy}}{\text{xyz}}+\frac{\text{yz}}{\text{xyz}}+\frac{\text{zx}}{\text{xyz}}=-\frac{\text{xyz}}{\text{xyz}}$
$\Rightarrow\frac{1}{\text{z}}+\frac{1}{\text{x}}+\frac{1}{\text{y}}=-1$
$\Rightarrow\text{x}^{-1}+\text{y}^{-1}+\text{z}^{-1}=-1$
Hence, the correct option is $(d)$
View full question & answer→MCQ 1221 Mark
Choose the correct answer.
Which of the following is correct:
- A
Determinant is a square matrix.
- B
Determinants is a number associated to a matrix.
- ✓
Determinants is a number associated to a square matrix.
- D
AnswerCorrect option: C. Determinants is a number associated to a square matrix.
Since, Determinants is a number associated to a square matrix.
Therefore, option (c) is correct.
View full question & answer→MCQ 1231 Mark
Evaluate $\begin{bmatrix}5&4&3\\3&4&1\\5&6&1\end{bmatrix}$is:
AnswerExpanding along the first row, we get
$\triangle=5\begin{bmatrix}4&1\\6&1\end{bmatrix}-4\begin{bmatrix}3&1\\5&1\end{bmatrix}+3\begin{bmatrix}3&4\\5&6\end{bmatrix}$
$=5(4-6)-4(3-5)+3(18-20)$
$=5(-2)-4(-2)+3(-2)=-10+8-6=-8.$
View full question & answer→MCQ 1241 Mark
Find the determinant of the matrix $\text{A}=\begin{bmatrix}-\cos\theta&-tan\theta\\\cot\theta&\cos\theta\end{bmatrix}$ is:
- ✓
$\sin^2\theta$
- B
$\sin\theta$
- C
$-\sin\theta$
- D
$-\sin^2\theta$
AnswerCorrect option: A. $\sin^2\theta$
Given that, $\text{A}=\begin{bmatrix}-\cos\theta&-tan\theta\\\cot\theta&\cos\theta\end{bmatrix}$
$|\text{A}|=\begin{bmatrix}-\cos\theta&-tan\theta\\\cot\theta&\cos\theta\end{bmatrix}$
$|\text{A}|=-\cos\theta (\cos\theta )-\cot\theta(-\tan\theta) $
$|\text{A}|=-\cos^2\theta+1=\sin^2\theta.$
View full question & answer→MCQ 1251 Mark
If $x = – 4$ is a root of $\triangle=\begin{bmatrix}\text{x}&2&3\\1&\text{x}&1\\3&2&\text{x}\end{bmatrix}=0,$ then the other roots are:
- ✓
$1, 3$
- B
$0, 2$
- C
$-1, 1$
- D
$2, 4$
AnswerCorrect option: A. $1, 3$
View full question & answer→MCQ 1261 Mark
Evaluate $\begin{bmatrix}\sqrt{3}&\sqrt{2}\\-1&2\sqrt{3}\end{bmatrix}$ is:
- A
$6-3\sqrt{2}$
- B
$6-\sqrt{2}$
- C
$6+3\sqrt{2}$
- ✓
$6+\sqrt{2}$
AnswerCorrect option: D. $6+\sqrt{2}$
$\triangle=\begin{bmatrix}\sqrt{3}&\sqrt{2}\\-1&2\sqrt{3}\end{bmatrix}$
$\triangle=(\sqrt{3}\times2\sqrt{3})+\sqrt{2}$
$\triangle=6+\sqrt{2}$
View full question & answer→MCQ 1271 Mark
The system of equation x + y + z = 2, 3x − y + 2z = 6 and 3x + y + z = −18 has:
- ✓
- B
- C
An infinite number of solutions.
- D
Zero solution as the only solution.
AnswerThe given system of equations can be written in matrix form as follows:
$\begin{bmatrix}1&1&1\\3&-1&2\\3&1&1\end{bmatrix}\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}=\begin{bmatrix}2\\6\\-18\end{bmatrix}$
$\text{AX}=\text{B}$
Here,
$\text{A}=\begin{bmatrix}1&1&1\\3&-1&2\\3&1&1\end{bmatrix},\text{X}=\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}\text{and }\text{B}=\begin{bmatrix}2\\6\\-18\end{bmatrix}$
$|\text{A}|=1(-1-2)-1(3-6)+1(3+3)$
$=-3+3+6$
$=6\neq0$
So, the given system of equations has a unique solution.
View full question & answer→MCQ 1281 Mark
The maximum value of $\triangle=\begin{vmatrix}1&1&1\\1&1+\sin\theta&1\\1+\cos\theta&1&1\end{vmatrix}$ is $(\theta$ is real$):$
- ✓
$\frac{1}{2}$
- B
$\frac{\sqrt{3}}{2}$
- C
$\sqrt{2}$
- D
$-\frac{\sqrt{3}}{2}$
AnswerCorrect option: A. $\frac{1}{2}$
$\triangle=\begin{vmatrix}1&1&1\\1&1+\sin\theta&1\\1+\cos\theta&1&1\end{vmatrix}$
$=\begin{vmatrix}1&1&1\\0&\sin\theta&0\\\cos\theta&0&0\end{vmatrix} [$Applying $R_2 \rightarrow R_2 - R_1$ and $R_3 \rightarrow R_3 - R_1]$
$=-\sin\theta\cos\theta$
$=-\frac{\sin2\theta}{2}$
Now, maximum and minimum value of $\sin\theta$ is $1$ and $-1.$
So, the maximum value of $-\sin\theta$ is $1.$
So, the maximum value of $-\sin2\theta$ is $1.$
Therefore, the maximum value of $-\frac{\sin2\theta}{2}$ is $\frac{1}{2}$
Hence, the correct option is $(a)$
View full question & answer→MCQ 1291 Mark
Evaluate $\begin{bmatrix}8\text{x}+1&2\text{x}-2\\\text{x}^2-1&3\text{x}+5\end{bmatrix}$ is:
- A
$-2x^3- 26x^2+ 45x + 3$
- ✓
$-2x^3+ 26x^2+ 45x + 3$
- C
$-2x^3+ 26x^2+ 45x - 3$
- D
$-2x^3- 26x^2- 45x + 3$
AnswerCorrect option: B. $-2x^3+ 26x^2+ 45x + 3$
Expanding along the first row, we get
$\triangle=8\text{x}+1(3\text{x}+5)-(2\text{x}-2)(\text{x}^2-1)$
$=(24\text{x}^2+43\text{x}+5)-(2\text{x}^3-2\text{x}^2-2\text{x}+2)$
$=-2\text{x}^3+26\text{x}^2+45\text{x}+3.$
View full question & answer→MCQ 1301 Mark
The value of the determinant $\begin{vmatrix}\text{x}&\text{x}+\text{y}&\text{x}+2\text{y}\\\text{x}+2\text{y}&\text{x}&\text{x}+\text{y}\\\text{x}+\text{y}&\text{x}+2\text{y}&\text{x}\end{vmatrix}$ is:
- A
$9x^2(x + y)$
- ✓
$9y^2(x + y)$
- C
$3y^2(x + y)$
- D
$7x^2(x + y)$
AnswerCorrect option: B. $9y^2(x + y)$
$\begin{vmatrix}\text{x}&\text{x}+\text{y}&\text{x}+2\text{y}\\\text{x}+2\text{y}&\text{x}&\text{x}+\text{y}\\\text{x}+\text{y}&\text{x}+2\text{y}&\text{x}\end{vmatrix}$
$=\begin{vmatrix}-2\text{y}&\text{y}&\text{y}\\\text{x}+2\text{y}&\text{x}&\text{x}+\text{y}\\-\text{y}&2\text{y}&-\text{y}\end{vmatrix} \ [$Applying $R_1 \rightarrow R_1 - R_2$ and $R_3\rightarrow R_3 - R_2]$
$=\text{y}^2\begin{vmatrix}-2&1&1\\\text{x}+2\text{y}&\text{x}&\text{x}+\text{y}\\-1&2&-1\end{vmatrix}\ [$Taking $(y)$ common from $R_1$ and from $R_3]$
$=\text{y}^2\begin{vmatrix}-2&-3&3\\\text{x}+2\text{y}&3\text{x}+4\text{y}&-\text{y}\\-1&0&0\end{vmatrix} \ [$Applying $C_2 \rightarrow C_2 + 2C_1$ and $C_3 \rightarrow C_3 - C_1]$
$=\text{y}^2\big[-1(3\text{y}-9\text{x}-12\text{y})\big]$
$=\text{y}^2[9\text{y}+9\text{x}]$
$=9\text{y}^2(\text{y}+\text{x})$
Hence, the correct option is $(b)$
View full question & answer→MCQ 1311 Mark
The value of the determinant $\begin{vmatrix}\text{a}^2&\text{a}&1\\\cos\text{nx}&\cos(\text{n}+1)\text{x}&\cos(\text{n}+2)\text{x}\\\sin\text{nx}&\sin(\text{n}+1)\text{x}&\sin(\text{n}+2)\text{x}\end{vmatrix}$ is independent of:
AnswerLet A = nx, B = (n - 1)x, C = (n + 2)x
⇒ C - B = x, B - A = x, C - A = 2x
Thus, the given determinant is
$\begin{vmatrix}\text{a}^2&\text{a}&1\\\cos\text{A}&\cos\text{B}&\cos\text{C}\\\sin\text{A}&\sin\text{B}&\sin\text{C}\end{vmatrix}$
$=\text{a}^2(\cos\text{B}\sin\text{C}-\cos\text{C}\sin\text{B})-\text{a}\times(\cos\text{A}\sin\text{C}-\cos\text{C}\sin\text{A})\\+1\times(\cos\text{A}\sin\text{B}-\sin\text{A}\cos\text{B})$
$=\text{a}^2\sin(\text{C}-\text{B})-\text{a}\sin(\text{C}-\text{A})+\sin(\text{B}-\text{A})$
$=\text{a}^2\sin{\text{x}}-\text{a}\sin2\text{x}+\sin\text{x}$ [Independent of n]
View full question & answer→MCQ 1321 Mark
Evaluate $\begin{bmatrix}5&0&5\\1&4&3\\0&8&6\end{bmatrix}$ is:
Answer$\triangle=\begin{bmatrix}5&0&5\\1&4&3\\0&8&6\end{bmatrix}$
Expanding along $\text{R}_1,$ we get:
$\triangle=5\begin{bmatrix}4&3\\8&6\end{bmatrix}-0\begin{bmatrix}1&3\\0&6\end{bmatrix}+5\begin{bmatrix}1&4\\0&8\end{bmatrix}$
$\triangle=5(24-24)-0+5(8-0)$
$\triangle=0-0+40=40.$
View full question & answer→MCQ 1331 Mark
If $\begin{bmatrix}\text{x} &\text{amp; } 1 &\text{amp; 1}\\ 2 &\text{amp; 3} &\text{amp; 4}\\ 1 &\text{amp; 1} &\text{amp; 1}\end{bmatrix}$ has no inverse, then $\text{x}=$
AnswerWe know that, If Dett = 0 there is no inverse
⇒ D = x(3 - 4) - 1(2 - 4) + 1(2 - 3) = 0
⇒ -x + 2 - 1 = 0
⇒ x = 1
View full question & answer→MCQ 1341 Mark
For any $2 \times 2$ matrix, if $\text{A(adj A)}=\begin{bmatrix} 10 & 0 \\ 0 & 10 \end{bmatrix},$ then $|A|$ is equal to:
Answer$\text{A(adj A)}\begin{bmatrix} 10 & 0 \\ 0 & 10 \end{bmatrix}$
By definition, we have
$\text{A(adj A) = |A|I = (adj A)A}\ ($Where $I$ is the identity matrix$)$
$\Rightarrow \text{|A|I = A(adj A)}$
$\Rightarrow|\text{A}|\text{I}=10\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$
$\Rightarrow|\text{A}|=10$
View full question & answer→MCQ 1351 Mark
If $A,\ B$ are two $n \times n$ non $-$ singular matrices, then
- ✓
$AB$ is non $-$ singular.
- B
$AB$ is singular.
- C
$(AB)^{-1} A^{-1} B^{-1}.$
- D
$(AB)^{-1}$ does not exist.
AnswerCorrect option: A. $AB$ is non $-$ singular.
$A $ and $B$ are non$-$ singular matrices of order $n \times n.$
$\therefore|\text{A}|\neq0$ and $|\text{B}|\neq0\ .....(\text{i})$
$A$ and $B$ are of the same order, so $AB$ is defined and is of the same order.
Thus,
$|AB| = |A||B|$
$\Rightarrow|\text{AB}|\neq0\ \big[\text{Using (1)}\big]$
Thus $, Ab$ is non $-$ singular.
View full question & answer→MCQ 1361 Mark
If $A^2 - A + I = 0,$ then the inverse of $A$ is:
- A
$A^{-2}$
- B
$A + I$
- ✓
$I - A$
- D
$A - I$
AnswerCorrect option: C. $I - A$
$A^2 - A + I = 0$
$\Rightarrow A^{-1}A^2 - A^{-1}A + A^{-1}I = A^{-1}0$
$\Rightarrow A - I + A^{-1} = 0$
$\Rightarrow A^{-1} = I - A$
View full question & answer→MCQ 1371 Mark
Which of the following is correct?
- ✓
Determinant is a square matrix.
- B
Determinant is a number associated with a matrix.
- C
Determinant is a number associated with a square matrix.
- D
AnswerCorrect option: A. Determinant is a square matrix.
View full question & answer→MCQ 1381 Mark
How much is the area of rectangular field:
- A
$60000 \ \text{sq.m}$
- ✓
$30000 \ \text{sq.m}$
- C
$30000m$
- D
$3000m$
AnswerCorrect option: B. $30000 \ \text{sq.m}$
View full question & answer→MCQ 1391 Mark
Evaluate $\begin{bmatrix}\text{i}&-1\\-1&\text{i}\end{bmatrix}$
AnswerExpanding along $\text{R}_1,$ we get.
$\triangle=\text{-i}(\text{i})-(-1)(-1)=-\text{i}^2-1=-(-1)-1=0.$
View full question & answer→MCQ 1401 Mark
The number of line segments possible with three collinear points is $........$
AnswerLet three collinear points be $A, B, C$
They can represent three line segments namely, $AB, BC, AC$.
Thus namely $3$ line segments are possible with three collinear points.
View full question & answer→MCQ 1411 Mark
Let$\begin{vmatrix}\text{x}&2&\text{x}\\\text{x}^2&\text{x}&6\\\text{x}&\text{x}&6\end{vmatrix}=\text{ax}^4+\text{bx}^3+\text{cx}^2+\text{dx}+\text{e}.$ Then, the value of $5a + 4b + 3c + 2d + e$ is equal to:
Answer$\triangle=\begin{vmatrix}\text{x}&2&\text{x}\\\text{x}^2&\text{x}&6\\\text{x}&\text{x}&6\end{vmatrix}$
$=\text{x}\begin{vmatrix}\text{x}&6\\\text{x}&6\end{vmatrix}-\text{x}^2\begin{vmatrix}2&\text{x}\\\text{x}&6\end{vmatrix}+\text{x}\begin{vmatrix}2&\text{x}\\\text{x}&6\end{vmatrix}$
$= 0 - x^2(12 - x^2) + x(12 - x^2)$
$= x^4 - 12x^2 + 12x - x^3$
$= ax^4 + bx^3 + cx^2 + dx + e$
$\Rightarrow x^4 - 12x^2 + 12x - x^3 $
$= ax^4 + bx^3 + cx^2 + dx + e$
$\Rightarrow a = 1, b = -1, c = -12, d = 12, e = 0$
Thus,
$5a + 4b + 3c + 2d + e $
$= 5 - 4 - 36 + 24 + 0 = -11$
View full question & answer→MCQ 1421 Mark
Find the minor of the element 2 in the determinant $\triangle=\begin{bmatrix}1&9\\2&3\end{bmatrix}$?
AnswerThe minor of the element 2 can be obtained by deleting the first row and the first column
$\therefore\text{M}_{11}=9$
View full question & answer→MCQ 1431 Mark
If $\triangle=\begin{vmatrix}a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&a_{23}\\a_{31}&a_{32}&a_{33}\end{vmatrix}$ and $A_{ij}$ is cofactors of $a_{ij},$ then value of $\triangle$ is given by:
- A
$a_{11}A_{31} + a_{12}A_{32} + a_{13}A_{33}$
- B
$a_{11}A_{11} + a_{12}A_{21} + a_{13}A_{31}$
- C
$a_{21}A_{11} + a_{22}A_{12} + a_{23}A_{13}$
- ✓
$a_{11}A_{11} + a_{21}A_{21} + a_{31}A_{31}$
AnswerCorrect option: D. $a_{11}A_{11} + a_{21}A_{21} + a_{31}A_{31}$
We know that:
$\triangle =$ Sum of the product of the elements of a column $($or a row$)$ with their corresponding cofactors
$\therefore\triangle = \text{a}_{11}\text{A}_{11} +\text{a}_{21}\text{A}_{21} + \text{a}_{31}\text{A}_{31}$
Hence, the value of $\triangle$ is given by the expression given in alternative $d$. the correct answer is $d$.
View full question & answer→MCQ 1441 Mark
The matrix $\begin{bmatrix} 5 & 10 & 3 \\ -2 & -4 & 6 \\ -1 & -2 & \text{b} \end{bmatrix}$ is a singular matrix, if the value of b is:
Answer$\begin{bmatrix} 5 & 10 & 3 \\ -2 & -4 & 6 \\ -1 & -2 & \text{b} \end{bmatrix}$ is singular matrix.
$\Rightarrow\begin{vmatrix} 5 & 10 & 3 \\ -2 & -4 & 6 \\ -1 & -2 & \text{b} \end{vmatrix}=0$
⇒ 5(-4b + 12) - 10(-2b + 6) + 3(4 - 4) = 0
⇒ -20b + 60 + 20b - 60 = 0
b does not exist.
View full question & answer→MCQ 1451 Mark
If$\text{A}=\begin{vmatrix}\text{a}_{11}&\text{a}_{12}&\text{a}_{13}\\\text{a}_{21}&\text{a}_{22}&\text{a}_{23}\\\text{a}_{31}&\text{a}_{32}&\text{a}_{33}\end{vmatrix}$ and $C_{ij}$ is cofactor of $a_{ij}$ in a, then value of $|A|$ is given by:
- A
$a_{11}C_{31} + a_{12}C_{32} + a_{13}C_{33}$
- B
$a_{11}C_{11} + a_{12}C_{21} + a_{13}C_{31}$
- C
$a_{21}C_{11} + a_{22}C_{12} + a_{23}C_{13}$
- ✓
$a_{11}C_{11} + a_{21}C_{21} + a_{13}C_{31}$
AnswerCorrect option: D. $a_{11}C_{11} + a_{21}C_{21} + a_{13}C_{31}$
Properties of determinants state that if a is a square matrix of the order $n,$
then Det $(A)$ is the sum of products of elements of a row $($or a column$)$ with the
corresponding cofactor of that element.
View full question & answer→MCQ 1461 Mark
If $x, y, z$ are nonzero real numbers, then the inverse of matrix $\text{A}=\begin{bmatrix}\text{x}&0&0\\0&\text{y}&0\\0&0&\text{z}\end{bmatrix}$ is
- ✓
$\begin{bmatrix}\text{x}^{-1}&0&0\\0&\text{y}^{-1}&0\\0&0&\text{z}^{-1}\end{bmatrix}$
- B
$\text{xyz}\begin{bmatrix}\text{x}^{-1}&0&0\\0&\text{y}^{-1}&0\\0&0&\text{z}^{-1}\end{bmatrix}$
- C
$\frac{1}{\text{xyz}}\begin{bmatrix}\text{x}&0&0\\0&\text{y}&0\\0&0&\text{z}\end{bmatrix}$
- D
$\frac{1}{\text{xyz}}\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}$
AnswerCorrect option: A. $\begin{bmatrix}\text{x}^{-1}&0&0\\0&\text{y}^{-1}&0\\0&0&\text{z}^{-1}\end{bmatrix}$
$\text{A}=\begin{bmatrix}\text{x}&0&0\\0&\text{y}&0\\0&0&\text{z}\end{bmatrix}$
$\therefore |A| = x (yz - 0) = xyz \neq 0$
Now, $A_{11} = yz, A_{12} = 0, A_{13}= 0$
$A_{21} = 0, A_{22} = xz, A_{23} = 0$
$A_{31} = 0, A_{32} = 0, A_{33} = xya$
$\therefore\text{adj. A}=\begin{bmatrix}\text{yz}&0&0\\0&\text{xz}&0\\0&0&\text{xy}\end{bmatrix}$
$\therefore\text{A}^{-1}=\frac{1}{|\text{A}|}\text{adj. A}$
$=\frac{1}{\text{xyz}}\begin{bmatrix}\text{yz}&0&0\\0&\text{xz}&0\\0&0&\text{xy}\end{bmatrix}$
$=\begin{bmatrix}\frac{\text{yz}}{\text{xyz}}&0&0\\0&\frac{\text{xz}}{\text{xyz}}&0\\0&0&\frac{\text{xy}}{\text{xyz}}\end{bmatrix}$
$=\begin{bmatrix}\frac{1}{\text{x}}&0&0\\0&\frac{1}{\text{y}}&0\\0&0&\frac{1}{\text{z}}\end{bmatrix}=\begin{bmatrix}\text{x}^{-1}&0&0\\0&\text{y}^{-1}&0\\0&0&\text{z}^{-1}\end{bmatrix}$
The correct answer is $a$.
View full question & answer→MCQ 1471 Mark
$ \begin{bmatrix}1 & \text{x} & \text{x}^2 \\1 & \text{y} & \text{y}^2 \\1 & \text{z} & \text{z}^2\end{bmatrix}$
AnswerCorrect option: D. $(x - y) (y - z) (z - x)$
View full question & answer→MCQ 1481 Mark
If $ \displaystyle \begin{vmatrix} 2 &\text{amp;} -4 \\ 9 &\text{amp; d}-3 \end{vmatrix}=4$ then $\text{d}=$
AnswerGiven, determinant of the matrix $ \displaystyle \begin{vmatrix} 2 &\text{amp;} -4 \\ 9 &\text{amp; d}-3 \end{vmatrix}=4$
$\Rightarrow2(\text{d}−3)−(9)(−4)=4$
$\Rightarrow2\text{d}-6+36 = 4$
$\Rightarrow 2\text{d}=-26$
$\Rightarrow\text{d} = -13$
View full question & answer→MCQ 1491 Mark
Choose the correct answer from given four options in each of the Exercise:
Let $\text{f}(\text{t})=\begin{vmatrix}\cos\text{t}&\text{t}&1\\2\sin\text{t}&\text{t}&2\text{t}\\\sin\text{t}&\text{t}&\text{t}\end{vmatrix} ,$ then $\lim\limits_{\text{t}\rightarrow0}\frac{\text{f(t)}}{\text{t}^2}$ is equal to:
AnswerWe have, $\text{f}(\text{t})=\begin{vmatrix}\cos\text{t}&\text{t}&1\\2\sin\text{t}&\text{t}&2\text{t}\\\sin\text{t}&\text{t}&\text{t}\end{vmatrix} $
$\Rightarrow\ \frac{\text{f(x)}}{\text{t}^2}=\frac{1}{\text{t}^2}\begin{vmatrix}\cos\text{t}&\text{t}&1\\2\sin\text{t}&\text{t}&2\text{t}\\\sin\text{t}&\text{t}&\text{t}\end{vmatrix} $
$=\begin{vmatrix}\cos\text{t}&\text{t}&1\\\frac{2\sin\text{t}}{\text{t}}&1&2\\\frac{\sin\text{t}}{\text{t}}&1&1\end{vmatrix} $ $\big[\text{Dividing R}_2\text{ and R}_3\text{ by }'\text{t}'\big]$
$\Rightarrow\ \lim\limits_{\text{t}\rightarrow0}\frac{\text{f(t)}}{\text{t}^2}=\begin{vmatrix} \lim\limits_{\text{t}\rightarrow0}\text{t}\cos\text{t}&\lim\limits_{\text{t}\rightarrow0}\text{t}&\lim\limits_{\text{t}\rightarrow0}1\\\lim\limits_{\text{t}\rightarrow0}\text{t}\frac{2\sin}{\text{t}}&\lim\limits_{\text{t}\rightarrow0}1&\lim\limits_{\text{t}\rightarrow0}2\\\lim\limits_{\text{t}\rightarrow0}\text{t}\frac{\sin\text{t}}{\text{t}}&\lim\limits_{\text{t}\rightarrow0}1&\lim\limits_{\text{t}\rightarrow0}1\end{vmatrix}$
$=\begin{vmatrix}1&0&1\\2&1&2\\1&1&1\end{vmatrix}$ $\bigg(\because\lim\limits_{\text{t}\rightarrow 0}\frac{\sin\text{t}}{\text{t}}=1\bigg)$
$=1(1-2)-0+1(2-1)$
$=0$
View full question & answer→MCQ 1501 Mark
$\text{Let A}=\begin{bmatrix}1&\sin\theta&1\\-\sin\theta&1&\sin\theta\\-1&-\sin\theta&1\end{bmatrix},$ where $0\leq\theta\leq2\pi.$ Then
AnswerCorrect option: D. Det $(A) \in [2, 4]$
$\text{A}=\begin{bmatrix}1&\sin\theta&1\\-\sin\theta&1&\sin\theta\\-1&-\sin\theta&1\end{bmatrix}$
$\therefore|\text{A}|=1(1+\sin^2\theta)-\sin\theta(-\sin\theta+\sin\theta)+1(\sin^2\theta+1)$
$=1+\sin^2\theta+\sin^2\theta+1$
$= 2 + 2 \sin^2 \theta$
$= 2 (1 + \sin^2\theta$)
Now, $0\leq\theta\leq2\pi$
$\Rightarrow 0\leq\sin\theta\leq1$
$\Rightarrow 0\leq1+\sin^2\theta\leq2$
$\Rightarrow2\leq2(1+\sin^2\theta)\leq4$
$\therefore$ Det $(A) \in [2, 4]$
The correct answer is $d.$
View full question & answer→MCQ 1511 Mark
If $\text{A}=\begin{bmatrix}2&\lambda&-3\\0&2&5\\1&1&3\end{bmatrix},$ then $\text{A}^{-1}$ exists if:
- A
$\lambda=2$
- B
$\lambda\neq2$
- C
$\lambda\neq-2$
- ✓
$\text{None of these}$
AnswerCorrect option: D. $\text{None of these}$
$\text{A}=\begin{bmatrix}2&\lambda&-3\\0&2&5\\1&1&3\end{bmatrix}$
The inverse of a matrix exists if its determinant is not equal to 0.
Consider,
$|\text{A}|=\begin{bmatrix}2&\lambda&-3\\0&2&5\\1&1&3\end{bmatrix}\neq0$
$\Rightarrow|\text{A}| = 2 (6 – 5) – \lambda (0 – 5) + (-3) (0 – 2)\neq0$
$\Rightarrow2 + 5\lambda + 6 \neq 0$
$\Rightarrow5\lambda + 8 \neq 0$
$\Rightarrow5\lambda \neq -8$
$\Rightarrow\lambda\neq\frac{-8}{5}$
View full question & answer→MCQ 1521 Mark
Let $A$ be a nonsingular square matrix of order $3 \times 3$. Then $|\text{adj.} \ A|$ is equal to:
- A
$|A|$
- ✓
$|A|^2$
- C
$|A|^3$
- D
$3|A|$
AnswerCorrect option: B. $|A|^2$
If $A$ is a non $-$ singular matrix of order $\text{n}\times\text{n},$ then $|\text{adj}. A| = |A|^{n-1}$
$\therefore$ Putting $n = 3, |\text{adj.} A| = |A|^2$
Therefore, option $(b) $ is correct.
View full question & answer→MCQ 1531 Mark
If $A$ satisfies the equation $\text{x}^2-5\text{x}^2+4\text{x}+\lambda=0$ then $A^{-1}$ exists if:
- A
$\lambda\neq1$
- B
$\lambda\neq2$
- C
$\lambda\neq-1$
- ✓
$\lambda\neq0$
AnswerCorrect option: D. $\lambda\neq0$
A satisfies $\text{x}^3-5\text{x}^2+4\text{x}+\lambda=0$
$\Rightarrow\text{A}^3-5\text{A}^2+4\text{A}=-\lambda$
Assuming $A^{-1}$ exists, we get
$\text{A}^{-1}(\text{A}^3-5\text{A}^2+4\text{A})=-\lambda\text{A}^{-1}$
$\Rightarrow\text{A}^2-5\text{A}+4=-\text{A}^{-1}\lambda$
$\Rightarrow\text{A}-1=\frac{-(\text{A}^2-5\text{A}+4)}{\lambda}$
Thus$, A^{-1}$ exists if $\lambda\neq0$.
View full question & answer→MCQ 1541 Mark
What is the value of a + b + c + d ?
Answer$\text{ax}^3+\text{bx}^2+\text{cx}+\text{d}=\begin{bmatrix}\text{x}+1&\text{amp;}2\text{x}&\text{amp; 3}\text{x}\\2\text{x}+3&\text{amp;}\text{x+1}&\text{amp;}\text{x}\\2-\text{x}&\text{amp;}3\text{x}+4&\text{amp;}5\text{x}-1\end{bmatrix}$ if
$\text{x}=1\text{a}+\text{b}+\text{c}+\text{d}=\begin{bmatrix}2&\text{amp;}2&\text{amp;3}\\5&\text{amp;}2&\text{amp;1}\\1&\text{amp;}7&\text{amp;4} \end{bmatrix}$
$\text{a}+\text{b}+\text{c}+\text{d}=2-38+99=101-38=63$
View full question & answer→MCQ 1551 Mark
Let a, b, c be positive real numbers. The following system of equations in x, y and z $\frac{\text{x}^2}{\text{a}^2}+\frac{\text{y}^2}{\text{b}^2}-\frac{\text{z}^2}{\text{c}^2}=1,$ $\frac{\text{x}^2}{\text{a}^2}-\frac{\text{y}^2}{\text{b}^2}+\frac{\text{z}^2}{\text{c}^2}=1,$ $-\frac{\text{x}^2}{\text{a}^2}+\frac{\text{y}^2}{\text{b}^2}+\frac{\text{z}^2}{\text{c}^2}=1$ has:
- A
- ✓
- C
Infinitely many solutions.
- D
AnswerThe given system of equations can be written in matrix form as follows:
$\begin{bmatrix}\frac{1}{\text{a}^2}&\frac{1}{\text{b}^2}&\frac{-1}{\text{c}^2}\\\frac{1}{\text{a}^2}&\frac{-1}{\text{b}^2}&\frac{1}{\text{c}^2}\\\frac{-1}{\text{a}^2}&\frac{1}{\text{b}^2}&\frac{1}{\text{c}^2}\end{bmatrix}\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}=\begin{bmatrix}1\\1\\1\end{bmatrix}$
Here,
$\text{A}=\begin{bmatrix}\frac{1}{\text{a}^2}&\frac{1}{\text{b}^2}&\frac{-1}{\text{c}^2}\\\frac{1}{\text{a}^2}&\frac{-1}{\text{b}^2}&\frac{1}{\text{c}^2}\\\frac{-1}{\text{a}^2}&\frac{1}{\text{b}^2}&\frac{1}{\text{c}^2}\end{bmatrix},\text{X}=\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}\text{and }\text{B}=\begin{bmatrix}1\\1\\1\end{bmatrix}$
Now,
$|\text{A}|=\begin{vmatrix}\frac{1}{\text{a}^2}&\frac{1}{\text{b}^2}&\frac{-1}{\text{c}^2}\\\frac{1}{\text{a}^2}&\frac{-1}{\text{b}^2}&\frac{1}{\text{c}^2}\\\frac{-1}{\text{a}^2}&\frac{1}{\text{b}^2}&\frac{1}{\text{c}^2}\end{vmatrix}$
$=\frac{1}{\text{a}^2\text{b}^2\text{c}^2}\begin{vmatrix}1&1&-1\\1&-1&1\\-1&1&1\end{vmatrix}$
$=\frac{1}{\text{a}^2\text{b}^2\text{c}^2}\times1(-1-1)-1(1+1)-1(1-1)$
$=\frac{1}{\text{a}^2\text{b}^2\text{c}^2}\times(-2-2)$
$=\frac{-4}{\text{a}^2\text{b}^2\text{c}^2}$
$\Rightarrow|\text{A}|\neq0$
So, the given system of equations has a unique solution.
View full question & answer→MCQ 1561 Mark
If $B$ is a non$-$singular matrix and $A$ is a square matrix, then $\text{det} (B^{-1} AB)$ is equal to:
- A
$\text{Det} (A^{-1})$
- B
$\text{Det} (B^{-1})$
- ✓
$\text{Det} (A)$
- D
$\text{Det} (B)$
AnswerCorrect option: C. $\text{Det} (A)$
$B$ is non$-$singular.
This implies that $|\text{B}|\neq0,$ that $B$ is invertible and that $B^{-1}$ exists.
Here $B$ is invertible.
$\therefore|\text{B}^{-1}|=|\text{B}|^{-1}=\frac{1}{|\text{B}|}$
$\Rightarrow|\text{B}^{-1}\text{AB}|=|\text{B}^{-1}||\text{AB}|$
$\Rightarrow|\text{B}^{-1}\text{AB}|=|\text{B}|^{-1}|\text{A}||\text{B}|$
$\Rightarrow|\text{B}^{-1}\text{AB}|=\frac{1}{|\text{B}|}|\text{A}||\text{B}|$
$\Rightarrow|\text{B}^{-1}\text{AB}|=|\text{A}|$
View full question & answer→MCQ 1571 Mark
There are two value of a which makes the determinant $\triangle=\begin{vmatrix}1&-2&5\\2&\text{a}&-1\\0&4&2\text{a}\end{vmatrix}$ equal to $86$. The sum of these two values is:
Answer$\triangle=\begin{vmatrix}1&-2&5\\2&\text{a}&-1\\0&4&2\text{a}\end{vmatrix}=86$
$\Rightarrow 1(2a^2 + 4) - 2(-4a - 20) = 86$
$\Rightarrow 2a^2 + 4 + 8a + 40 = 86$
$\Rightarrow 2a^2 + 8a - 42 = 0$
$\Rightarrow a^2 + 4a - 21 = 0$
$\Rightarrow a^2+ 7a - 3a - 21 = 0$
$\Rightarrow a(a + 7) - 3(a + 7) = 0$
$\Rightarrow a = -7, 3$
Sum of the two values of $a = -7 + 3 = -4$
Hence, the correct option is $(c)$
View full question & answer→MCQ 1581 Mark
If $\text{A}=\begin{bmatrix} 1 & 2 & -1 \\ -1 & 1 & 2 \\ 2 & -1 & 1 \end{bmatrix},$ then $\text{det (adj (adj A))}$ is:
- ✓
$14^4$
- B
$14^3$
- C
$14^2$
- D
$14$
AnswerCorrect option: A. $14^4$
$\text{A}=\begin{bmatrix} 1 & 2 & -1 \\ -1 & 1 & 2 \\ 2 & -1 & 1 \end{bmatrix}$
$|\text{A}|=14$
$\text{det(adj(adj A))}=|\text{A}|^{{\text{n}-1}^{2}}$
$\text{det(adj(adj A))}=|14|^{{3-1}^{2}}=14^4$
View full question & answer→MCQ 1591 Mark
If $\text{A}=\begin{vmatrix} 1 &\text{amp; 2} \\ 2 &\text{amp; 1} \end{vmatrix}$ and $\text{f}\text{(x)}=\frac{1+\text{x}}{1-\text{x}},$ then $\text{f}(|\text{A}|)$ is:
- ✓
$\dfrac{-1}{2}$
- B
$\dfrac{1}{2}$
- C
$\dfrac{-1}{3}$
- D
$\text{None of these}$
AnswerCorrect option: A. $\dfrac{-1}{2}$
Here, $|\text{A}| =1\times 1-2\times 2 = -3$
$\therefore\text{f}(|\text{A}|)=\cfrac{1+(-3)}{1+3}=-\cfrac{1}{2}$
View full question & answer→MCQ 1601 Mark
If $d$ is the determinant of a square matrix $A$ of order $n,$ then the determinant of its adjoint is:
- A
$d^n$
- ✓
$d^{n-1}$
- C
$d^{n+1}$
- D
$d$
AnswerCorrect option: B. $d^{n-1}$
We know,
$\text{|adj} A| = |A|^{n-1}$
$\Rightarrow \text{|adj} A| = d^{n-1}$
View full question & answer→MCQ 1611 Mark
Evaluate $ |\text{A}|^2-5|\text{A}|+1,$ if $\text{A}=\begin{bmatrix}7&4\\5&5\end{bmatrix}$ is:
AnswerGiven that, $\text{A}=\begin{bmatrix}7&4\\5&5\end{bmatrix}$
$|\text{A}|=(7(5)-5(4))=35-20=15$
$|\text{A}|^2-5|\text{A}|+1=(15)^2-5(15)+1=225-75+1=151.$
View full question & answer→MCQ 1621 Mark
If $\left|\begin{array}{lll}<\text{br}> &1 &\text{amp; } 0 &\text{amp; } 0\\ <\text{br}>&2 &\text{amp; } 3 &\text{amp; } 4\\ <\text{br}>&5 &\text{amp; } -6 &\text{amp; x}<\text{br}> \end{array}\right|=45$ then $\text{x}=$
AnswerGiven, $\left|\begin{array}{lll}<\text{br}> &1 &\text{amp; } 0 &\text{amp; } 0\\ <\text{br}>&2 &\text{amp; } 3 &\text{amp; } 4\\ <\text{br}>&5 &\text{amp; } -6 &\text{amp; x}<\text{br}> \end{array}\right|=45$
By operation of matrix (5),
$1(3\text{x}+24)=45$
$3\text{x}=21$
$\Rightarrow \text{x}=7$
View full question & answer→MCQ 1631 Mark
Choose the correct answer from given four options in each of the Exercise:If $x, y, z$ are all different from zero and $\begin{vmatrix}1+\text{x}&1&1\\1&1+\text{y}&1\\1&1&1+\text{z}\end{vmatrix}=0,$ then the value of $x^{-1} + y^{-1} + z^{-1}$ is:
AnswerWe have, $\begin{vmatrix}1+\text{x}&1&1\\1&1+\text{y}&1\\1&1&1+\text{z}\end{vmatrix}=0$
Applying $\text{C}_1\rightarrow\text{C}_1-\text{C}_3$ and $\text{C}_2\rightarrow\text{C}_2-\text{C}_3,$
$\Rightarrow\ \begin{vmatrix}\text{x}&0&1\\0&\text{y}&1\\-\text{z}&-\text{z}&1+\text{z}\end{vmatrix}=0$
Expanding along $R_1$,
$\text{x}\big[\text{y}(1+\text{z})+\text{z}\big]-0+1(\text{yz})=0$
$\Rightarrow\text{x}(\text{y}+\text{yz}+\text{z})+\text{zy}=0$
$\Rightarrow\text{xy}+\text{xyz}+\text{xz}+\text{xz}=0$
$\Rightarrow\frac{\text{xy}}{\text{xyz}}+\frac{\text{xyz}}{\text{xyz}}+\frac{\text{xz}}{\text{xyz}} +\frac{\text{yz}}{\text{xyz}} =0\ [$On dividing by $(xyz)$ from both sides$]$
$\Rightarrow\ \frac{1}{\text{x}}+\frac{1}{\text{y}}+\frac{1}{\text{z}}+1=0$
$\Rightarrow\ \frac{1}{\text{x}}+\frac{1}{\text{y}}+\frac{1}{\text{z}}=-1$
$\therefore\text{x}^{-1}+\text{y}^{-1}+\text{z}^{-1}=-1$
View full question & answer→MCQ 1641 Mark
Choose the correct answer from given four options in each of the Exercise:The value of the determinant $\begin{vmatrix}\text{x}&\text{x}+\text{y}&\text{x}+2\text{y}\\\text{x}+2\text{y}&\text{x}&\text{x}+\text{y}\\\text{x}+\text{y}&\text{x}+2\text{y}&\text{x}\end{vmatrix}$is:
- A
$9x^2(x + y)$
- B
$9y^2(x + y)$
- C
$3y^2(x + y)$
- ✓
$7x^2(x + y)$
AnswerCorrect option: D. $7x^2(x + y)$
We have, $\begin{vmatrix}\text{x}&\text{x}+\text{y}&\text{x}+2\text{y}\\\text{x}+2\text{y}&\text{x}&\text{x}+\text{y}\\\text{x}+\text{y}&\text{x}+2\text{y}&\text{x}\end{vmatrix}$
$=\begin{vmatrix}3(\text{x}+\text{y})&\text{x}+\text{y}&\text{y}\\3(\text{x}+\text{y})&\text{x}&\text{y}\\3(\text{x}+\text{y})&\text{x}+2\text{y}&-2\text{y}\end{vmatrix}\big[\because\ \text{C}_1\rightarrow\text{C}_1+\text{C}_2+\text{C}_3$ and $C_3\rightarrow\text{C}_3 -\text{C}_2\big]$
$=3(\text{x}+\text{y})\begin{vmatrix}1&(\text{x}+\text{y})&\text{y}\\1&\text{x}&\text{y}\\1&(\text{x}+2\text{y})&-2\text{y}\end{vmatrix} [$Taking $3(x + y)$ common from first column$]$
$=3(\text{x}+\text{y})\begin{vmatrix}0&\text{y}&0\\1&\text{x}&\text{y}\\1&(\text{x}+2\text{y})&-2\text{y}\end{vmatrix}[\because\ \text{R}_1\rightarrow\text{R}_1-\text{R}_2]$
Expanding along $R_1$,
$=3(\text{x}+\text{y})\big[-\text{y}(-2\text{y})-\text{y}\big]$
$=3\text{y}^2.3(\text{x}+\text{y})=9\text{y}^2(\text{x}+\text{y})$
View full question & answer→MCQ 1651 Mark
If $\text{A}=\frac{1}{3}\begin{bmatrix} 1 & 1 & 2 \\ 2 & 1 & -2 \\ \text{x} & 2 & \text{y} \end{bmatrix}$ is prthogonal, than x + y =
Answer$\text{A}=\frac{1}{3}\begin{bmatrix} 1 & 1 & 2 \\ 2 & 1 & -2 \\ \text{x} & 2 & \text{y} \end{bmatrix}$
$\text{A}^\text{T}\text{A}=\text{I}$
$\frac{1}{3}\begin{bmatrix} 1 & 2 & \text{x} \\ 1 & 1 & 2 \\ 2 & -2 & \text{y} \end{bmatrix}\frac{1}{3}\begin{bmatrix} 1 & 1 & 2 \\ 2 & 1 & -2 \\ \text{x} & 2 &\text{y} \end{bmatrix}=\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$
$\frac{1}{9}\begin{bmatrix} 1+4+\text{x}^2 & 1+2+2\text{x} & \text{xy}-2 \\ 1+2+2\text{x} & 1+1+4 & 2-2+2\text{y} \\ 2-4+\text{xy} & 2-2+2\text{y} & 4+4+\text{y}^2 \end{bmatrix}$
$=\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$
$\begin{bmatrix} 5+\text{x}^2 & 3+2\text{x} & \text{xy}-2 \\ 3+2\text{x} & 6 & 2\text{y} \\ -6+\text{xy} & 2\text{y} & 8+\text{y}^2 \end{bmatrix}$
$=\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$
Equality of two matrices does not hold Matrix A is not orthogonal.
Hence, the given question is incorrect.
View full question & answer→MCQ 1661 Mark
The matrix $\begin{bmatrix}1 &\text{amp; } 0 &\text{amp; 1} \\ 2 &\text{amp; 1}&\text{amp; 0} \\ 3 &\text{amp; 1} &\text{amp; 1}\end{bmatrix}$ is:
AnswerGiven $\text{A}=\begin{bmatrix}1 &\text{amp; } 0 &\text{amp; 1} \\ 2 &\text{amp; 1}&\text{amp; 0} \\ 3 &\text{amp; 1} &\text{amp; 1}\end{bmatrix}$
$\text{|A|}=\begin{bmatrix}1 &\text{amp; } 0 &\text{amp; 1} \\ 2 &\text{amp; 1}&\text{amp; 0} \\ 3 &\text{amp; 1} &\text{amp; 1}\end{bmatrix}$
$\Rightarrow∣\text{A}∣=1−1=0$
Hence, A is singular.
View full question & answer→MCQ 1671 Mark
If $\text{A}=\begin{bmatrix}2&5&9\\6&1&3\\4&8&2\end{bmatrix},$ find $|\text{A}|$
Answer$\Rightarrow|\text{A}|=\begin{bmatrix}2&5&9\\6&1&3\\4&8&2\end{bmatrix}$Evaluating along the first row, we get
$\triangle=2\begin{bmatrix}1&3\\8&2\end{bmatrix}-5\begin{bmatrix}6&3\\4&2\end{bmatrix}+9\begin{bmatrix}6&1\\4&8\end{bmatrix}$
$\triangle=2(2-24)-5(12-12)+9(48-4)$
$\triangle=2(-22)-0+9(44) $
$\triangle=-44+9(44)=44(-1+9)=352 $
View full question & answer→MCQ 1681 Mark
If $\text{S}=\begin{bmatrix}\text{a} & \text{b} \\ \text{c} & \text{d} \end{bmatrix},$ then adj A is:
- A
$\begin{bmatrix} -\text{d} & -\text{b} \\ -\text{c} & \text{a} \end{bmatrix}$
- ✓
$\begin{bmatrix} \text{d} & -\text{b} \\ -\text{c} & \text{a} \end{bmatrix}$
- C
$\begin{bmatrix} \text{d} & \text{b} \\ \text{c} & \text{a} \end{bmatrix}$
- D
$\begin{bmatrix} \text{d} & \text{c} \\ \text{b} & \text{a} \end{bmatrix}$
AnswerCorrect option: B. $\begin{bmatrix} \text{d} & -\text{b} \\ -\text{c} & \text{a} \end{bmatrix}$
Adjoint of a square matrix of order 2 is obtained by interchancing the diagoinal elements and changing the signs of off-diagonal elements.
Here,
$\text{A}=\begin{bmatrix}\text{a} & \text{bc} & \text{d} \end{bmatrix}$
$\Rightarrow\text{adj A}=\begin{bmatrix}\text{d} & -\text{b}-\text{c} & \text{a} \end{bmatrix}$
View full question & answer→MCQ 1691 Mark
A set of points which do not lie on the same line are called as
AnswerA set of points which do not lie on the same line are called as non collinear points
View full question & answer→MCQ 1701 Mark
If $\text{A} = \begin{bmatrix}1&\text{amp; } \log_{\text{b}}\text{a}\\ \log_\text{a}\text{b}&\text{amp; } 1\end{bmatrix}$then $ |\text{A}|$ is equal to:
- ✓
$0$
- B
$\log_\text{a}\text{b}$
- C
$-1$
- D
$\log_\text{b}\text{a}$
AnswerOn solving the given matrix,
$|\text{A}|=1-\log_\text{a}\text{b}.\log_\text{b} \text{a}=1-1=0$
View full question & answer→