Question 11 Mark
A determinant of second order is made with the elements 0 and 1. The number of determinants with non - negative values is:
- 3
- 10
- 11
- 13
Answer
- 13
Solution:
There 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→Question 21 Mark
If A is a singular matrix, then A (adj A) is a
- scalar matrix
- zero matrix
- identity matrix
- orthogonal matrix
Answer
- zero matrix
Solution
Given A is a singular matrix.
⇒ ∣A∣ = 0
A (adjA) = ∣A∣I = 0I = O
∴ A (adjA) is a zero matrix.
View full question & answer→Question 31 Mark
Evaluate $\begin{bmatrix}4&8&12\\6&12&18\\7&14&21\end{bmatrix}$ is:
- 168
- -1
- -168
- 0
Answer
- 0
Soluton:
$\triangle=\begin{bmatrix}4&8&12\\6&12&18\\7&14&21\end{bmatrix}$
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 41 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:
- ✓
$0$
- B
$1$
- C
$xyz$
- D
$x^2 yz^3$
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→Question 51 Mark
Evaluate $\begin{bmatrix}1&0&1\\0&0&1\\1&0&1\end{bmatrix}$ is:
- 2
- 0
- 1
- -1
Answer
- 0
Solution:
$\triangle=\begin{bmatrix}1&0&1\\0&0&1\\1&0&1\end{bmatrix}$
$\triangle=1\begin{bmatrix}0&1\\0&1\end{bmatrix}-0\begin{bmatrix}0&1\\1&1\end{bmatrix}+1\begin{bmatrix}0&0\\1&0\end{bmatrix}$
$\triangle=1(0-0)-0(0-1)+1(0-0)$
$\triangle=0-0+0=0.$
View full question & answer→MCQ 61 Mark
If $B$ is a non$-$singular matrix and $A$ is a square matrix, then det $(B^{-1} AB)$ is equal to:
- A
Det $(A^{-1})$
- B
Det $(B^{-1})$
- ✓
Det $(A)$
- D
Det $(B)$
AnswerCorrect option: C. 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 71 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
$a^3 + b^3 + c^3 - 3abc$
- D
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 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
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→Question 101 Mark
If $\begin{vmatrix}x&2\\18&x\end{vmatrix}=\begin{vmatrix}6&2\\18&6\end{vmatrix},$ then x is equal to:
- 6
- $\pm6$
- - 6
- 0
Answer$\begin{vmatrix}x&2\\18&x\end{vmatrix}=\begin{vmatrix}6&2\\18&6\end{vmatrix}$
$\Rightarrow x^2-36=36-36$
$\Rightarrow x^2-36=0$
$\Rightarrow x^2=36$
$\Rightarrow x=\pm6$
Hence, the correct answer is (b).
View full question & answer→Question 111 Mark
Choose the correct answer.
Let A be a square matrix of order 3 × 3, then | kA| is equal to:
- $k\left|\text{A}\right|$
- $k^2\left|\text{A}\right|$
- $k^3\left|\text{A}\right|$
- $3k\left|\text{A}\right|$
Answer$\text{Let A}=\begin{bmatrix}a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&a_{23}\\a_{31}&a_{32}&a_{33}\end{bmatrix}$ 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→Question 121 Mark
If area of triangle is 35 sq units with vertices (2, – 6), (5, 4) and (k, 4). Then k is
- 12
- -2
- -12, -2
- 12, -2
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 131 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 141 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 151 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 161 Mark
Let $A$ be a nonsingular square matrix of order $3 \times 3.$ Then $|adj. A|$ 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 $|adj. A| = |A|^{n-1}$
$\therefore$ Putting $n = 3, |adj. A| = |A|^2$
Therefore, option $(b)$ is correct.
View full question & answer→Question 171 Mark
For the system of equations:
x + 2y + 3z = 1
2x + y + 3z = 2
5x + 5y + 9z = 4
- There is only one solution.
- There exists infinitely many solution.
- There is no solution.
- None of these.
Answer
- There is only one solution.
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→Question 181 Mark
Which of the following matrices will not have a determinant?
- $\begin{bmatrix}4&2\\5&4\end{bmatrix}$
- $\begin{bmatrix}1&5&3\\3&6&2\\4&8&7\end{bmatrix}$
- $\begin{bmatrix}5&8&9\\3&4&6\end{bmatrix}$
- $\begin{bmatrix}1&2\\5&5\end{bmatrix}$
Answer
- $\begin{bmatrix}5&8&9\\3&4&6\end{bmatrix}$
Solution:
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→Question 191 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:
- $\text{x}=2,-\frac{1}{3}$
- $\text{x}=-1,-\frac{1}{3}$
- $\text{x}=-2,-\frac{1}{3}$
- $\text{x}=0,-\frac{1}{3}$
Answer
- $\text{x}=2,-\frac{1}{3}$
Solution:
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→Question 201 Mark
Find the determinant of the matrix $\text{A}=\begin{bmatrix}9&8\\7&6\end{bmatrix}$ is:
- -1
- 1
- 2
- -2
Answer
- -2
Solution:
Given 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 211 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→Question 221 Mark
What is the determinant of the matrix $\left [\begin{matrix} 3&\text{amp; } 6\\ -1 &\text{amp; } 2\end {matrix} \right]$ ?
- 0
- 12
- ∣0∣
- ∣6∣
Answer
- 12
Solution:
Given, $\left [\begin{matrix} 3&\text{amp; } 6\\ -1 &\text{amp; } 2\end {matrix} \right]$
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 231 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 241 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
$CBA^{-1}$
- ✓
$C^{-1} BA^{-1}$
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 251 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 261 Mark
Given that $A$ is a square matrix of order $3$ and $|A| = -4,$ then $|adj\ A|$ is equal to:
AnswerGiven that $A$ is a square matrix of order $3$ and $|A| = -4.$
We know that $|adj\ A| = |A|^{n−1},$ where n is the order of matrix $A.$
So, $|adj\ A| = (−4)^{3-1} = (-4)^2 = 16$
View full question & answer→Question 271 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:
- f(a) = 0
- f(b) = 0
- f(0) = 0
- f(1) = 0
Answer
- f(0) = 0
Solution:
Let $\text{f(x)}=\begin{vmatrix}0&\text{x}-\text{a}&\text{x}-\text{b}\\\text{x}+\text{a}&0&\text{x}-\text{c}\\\text{x}+\text{b}&\text{x}+\text{c}&0\end{vmatrix}$
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 281 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 :
- A
$\frac{1}{2}\begin{bmatrix} 2 & 4 \\ 3 & -5 \end{bmatrix}$
- B
$\frac{1}{2}\begin{bmatrix} -2 & 4 \\ 3 & 5 \end{bmatrix}$
- ✓
$\begin{bmatrix} 2 & 4 \\ 3 & -5 \end{bmatrix}$
- D
AnswerCorrect option: C. $\begin{bmatrix} 2 & 4 \\ 3 & -5 \end{bmatrix}$
$A = BX$
$B^{-1}A = B^{-1}BX$
$X = B^{-1}A$
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 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 $C_1 \rightarrow C_1+ C_2 + C_3]$
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→Question 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:
- 0
- -1
- 1
- None of these.
Answer
- 0
Solution:
We 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→Question 311 Mark
If A is any skew-symmetric matrix of odd order then ∣A∣ equals
- −1
- 0
- 1
- none of these
Answer
- 0
Solution:
if A is skew symmetric matrix
then A = -AT
Therefore, ∣A∣ = -∣AT∣ = -∣A∣
⇒ 2∣A∣ =0
⇒ ∣A∣ = 0
View full question & answer→MCQ 321 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 331 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→Question 341 Mark
Find the values of x, if: $\begin{vmatrix} 2 &\text{amp; } 4 \\ 5 &\text{amp; } 1 \end{vmatrix}= \begin{vmatrix} 2\text{x} &\text{amp; }4 \\ 6 &\text{amp; x} \end{vmatrix}$
- $\pm \sqrt{3}$
- $3$
- $-3$
- $\text{None of these}$
Answer
- $\pm \sqrt{3}$
Solution:
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→Question 351 Mark
If A and B are square matrices of order 2, then det (A + B) = 0 is possible only when:
- Det (A) = 0 or det (B) = 0
- Det (A) + det (B) = 0
- Det (A) = 0 and det (B) = 0
- A + B = 0
Answer
- A + B = 0
Solution:
Let $\text{A}=[\text{a}_{\text{ij}}]$ 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 361 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 same.
Also if $x = b$ we get $1^{st}$ and $2^{nd}$ column same.
View full question & answer→Question 371 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:
- 4
- 2
- a + b + c
- None of these.
Answer
- 4
Solution:
$\triangle=\begin{vmatrix}-2\text{a}&\text{a}+\text{b}&\text{a}+\text{c}\\\text{b}+\text{a}&-2\text{b}&\text{b}+\text{c}\\\text{c}+\text{a}&\text{c}+\text{b}&-2\text{c}\end{vmatrix}$
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 381 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 391 Mark
If $A$ and $B$ are invertible matrices, which of the following statement is not correct.
- A
$adj\ A = |A|A^{-1}$
- B
$det\ (A^{-1}) = (det\ A)^{-1}$
- ✓
$(A + B)^{-1} = A^{-1} + B^{-1}$
- D
$(AB)^{-1} = B^{-1} A^{-1}$
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 401 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
$\text{A}=\begin{bmatrix} -1 & 0 \\ 0 & 1 \end{bmatrix},\text{if n}\in\text{N}$
- 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→Question 411 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):
- $\frac{1}{2}$
- $\frac{\sqrt{3}}{2}$
- $\sqrt{2}$
- $\frac{2\sqrt{3}}{4}$
Answer
- $\frac{1}{2}$
Solution:
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→Question 421 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
- The value of determinant changes if any two rows or columns are interchanged
- The value of determinant is zero if any two rows and columns are identical
- The value of determinant gets multiplied by k, if each element of row or column is multiplied by k
Answer
- The value of determinant changes if all of its rows and columns are interchanged
Solution:
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→Question 431 Mark
Evaluate $\begin{bmatrix}3&-1&3\\6&-5&4\\3&-2&3\end{bmatrix}$ is:
- 100
- 223
- 240
- 230
Answer
- 240
Solution:
Expanding 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 441 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 :
- A
$\triangle_1+\triangle_2=0$
- ✓
$\triangle_1+2\triangle_2=0$
- C
$\triangle_1=\triangle_2$
- D
AnswerCorrect option: B. $\triangle_1+2\triangle_2=0$
$\triangle_2=\begin{vmatrix}1&\text{bc}&\text{a}\\1&\text{ca}&\text{b}\\1&\text{ab}&\text{c}\end{vmatrix}$
$=\frac{1}{\text{abc}}\begin{vmatrix}1&\text{abc}&\text{a}^2\\1&\text{bca}&\text{b}^2\\1&\text{cab}&\text{c}^2\end{vmatrix}$
$ [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 451 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 461 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, $|A^{-1}| = |A|^{-1}, (AT)^{-1} = (A^{-1})^T$ and $|\text{A}|\neq0$ all are the properties of the inverse of a matrix $A.$
View full question & answer→Question 471 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?
- $\text{adj A} = |\text{A}|.\text{A}^{-1}$
- $\text{det (A)}^{-1}=[\text{det(A)}]^{-1}$
- $(\text{AB})^{-1}=\text{B}^{-1}\text{A}^{-1}$
- $(\text{A}+\text{B})^{-1}=\text{B}^{-1}+\text{A}^{-1}$
Answer
- $(\text{A}+\text{B})^{-1}=\text{B}^{-1}+\text{A}^{-1}$
Solution:
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→Question 481 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:
- $\pm 3$
- $\pm 2$
- $\pm 5$
- $0$
Answer
- $\pm 3$
Solution:
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→Question 491 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?
- x = 3, y = 1
- x = 1, y = 3
- x = 0, y = 3
- x = 0, y = 0
Answer
- x = 0, y = 0
Solution:
$ \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→Question 501 Mark
If A is a skew symmetric matrix, then ∣A∣ is
- 1
- -1
- 0
- none
Answer
- 0
Solution:
SINCE 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→