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
$\Rightarrow\begin{bmatrix}\text{x} & 0 & 0\\ 0 & \text{y} & 0 \\ 0 & 0 & \text{z}\end{bmatrix}=\begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}\text{A}$
$\Rightarrow\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}=\begin{bmatrix}\text{x}^1 & 0 & 0\\ 0 & \text{y}^1 & 0 \\ 0 & 0 & \text{z}^1\end{bmatrix}\text{A}$
$\Big[\text{Applying R}_1=\frac{1}{\text{x}}\text{R}_1,\text{R}_2=\frac{1}{\text{y}}\text{R}_2\text{ and R}_3=\frac{1}{\text{z}}\text{R}_3\Big]$
$\Rightarrow\text{A}^{-1}=\begin{bmatrix}\text{x}^{-1} & 0 & 0\\ 0 & \text{y}^{-1} & 0 \\ 0 & 0 & \text{z}^{-1}\end{bmatrix}$
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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.
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