MCQ 11 Mark
If $A$ is a square matrix of order $3$ and $|A| = 5$, then the value of $|2A\ '|$ is:
Answer$\left|2 A^{\prime}\right|$
$=2^3\left|A^{\prime}\right|$
$=8|4|$
$=8 \times 5$
$=40$
View full question & answer→MCQ 21 Mark
If $A$ is a square matrix such that $A ^2= A,$ then $(I-A)^3+ A$ is equal to:
Answer$A^2=A$
$(I+A)^3+A$
$\Rightarrow 1^3-A^3-3 I^2 A+31 A^2+A$
$\Rightarrow 1-A^3-3 A+3 A+A\left( \therefore A^2=A\right)$
$\Rightarrow 1-A \cdot A^2+A$
$\Rightarrow 1-A \cdot A+A$
$\Rightarrow 1-A+A$
$=1$
View full question & answer→MCQ 31 Mark
If $\text{A}=\begin{bmatrix}\text{n}&0&0\\0&\text{n}&0\\0&0&\text{n}\end{bmatrix}$ and $\text{B}=\begin{bmatrix}\text{a}_1&\text{a}_2&\text{a}_3\\\text{b}_1&\text{b}_2&\text{b}_3\\\text{c}_1&\text{c}_2&\text{c}_3\end{bmatrix},$ then $AB$ is equal to:
AnswerHere,
$\text{A}=\begin{bmatrix}\text{n}&0&0\\0&\text{n}&0\\0&0&\text{n}\end{bmatrix}$ and $\text{B}=\begin{bmatrix}\text{a}_1&\text{a}_2&\text{a}_3\\\text{b}_1&\text{b}_2&\text{b}_3\\\text{c}_1&\text{c}_2&\text{c}_3\end{bmatrix}$
$\therefore\ \text{AB}=\begin{bmatrix}\text{n}&0&0\\0&\text{n}&0\\0&0&\text{n}\end{bmatrix}\begin{bmatrix}\text{a}_1&\text{a}_2&\text{a}_3\\\text{b}_1&\text{b}_2&\text{b}_3\\\text{c}_1&\text{c}_2&\text{c}_3\end{bmatrix}$
$\Rightarrow\text{AB}=\begin{bmatrix}\text{na}_1&\text{na}_2&\text{na}_3\\\text{nb}_1&\text{nb}_2&\text{nb}_3\\\text{nc}_1&\text{nc}_2&\text{nc}_3\end{bmatrix}$
$\Rightarrow\text{AB}=\text{n}\begin{bmatrix}\text{a}_1&\text{a}_2&\text{a}_3\\\text{b}_1&\text{b}_2&\text{b}_3\\\text{c}_1&\text{c}_2&\text{c}_3\end{bmatrix}$
$\Rightarrow\text{AB}=\text{n}\text{B}$
View full question & answer→MCQ 41 Mark
If $A$ is a square matrix, then $AA$ is $a:$
- A
Skew$-$symmetric matrix.
- B
- C
- ✓
AnswerGiven$:\ A$ is a square matrix.
Let $\text{A}=\begin{bmatrix}1&2\\1&0\end{bmatrix}$
$\Rightarrow\text{AA}=\begin{bmatrix}1&2\\1&0\end{bmatrix}\begin{bmatrix}1&2\\1&0\end{bmatrix}=\begin{bmatrix}3&2\\1&2\end{bmatrix}$
View full question & answer→MCQ 51 Mark
The restriction on $n, k$ and $p$ so that $PY + WY$ will be defined are:
- ✓
$k = 3, p = n$
- B
$k$ is arbitary$, p = 2$
- C
$p$ is arbitary$, k = 3$
- D
AnswerCorrect option: A. $k = 3, p = n$
In this, order of $P = p \times k$ Order of $W = n \times 3$ Order of $Y = 3 \times k$
Thus, order of $PY = p \times k,$ when $k = 3$
And the order of $WY = p \times k,$ where $p = n$
View full question & answer→MCQ 61 Mark
A matrix has $16$ elements Which of the following can be the order of the matrix:
- A
$1 \times 16$
- B
$2 \times 8$
- C
$4 \times 4$
- ✓
AnswerA matrix of $mm$ rows and nn columns has $m \times nm\times n$ elements.
On multiplying the rows and columns in the given options, we notice.
that all $1 \times 16 = 16, 2 \times 8 = 16, 4 \times 4 = 16$
View full question & answer→MCQ 71 Mark
Suppose $A$ and $B$ are two square matrices of same order.If $A, B$ are symmetric matrices and $AB = BA$ then $AB$ is:
Answer$(\text{AB})^\text{T}=\text{B}^\text{T}\text{A}^\text{T}=\text{BA}=\text{AB}$
$\Rightarrow(\text{AB})^\text{T}=\text{AB}$
So that means the product of the matrices $\text{AB}$ is a symmetric matrix.
View full question & answer→MCQ 81 Mark
If $\text{A} = \begin{bmatrix} 2 &\text{amp; } 3\\ 6 &\text{amp; x} \end{bmatrix}, \text{B} = \begin{bmatrix} 2 &\text{amp; 3}\\ \text{p} &\text{amp; }2 \end{bmatrix}$ and $\text{A} = \text{B}, $ then $\text{p}$ and $ \text{x} $ are:
- A
$p = 6, x = 4$
- B
$p = 3, x = 4$
- C
$p = 4, x = 3$
- ✓
AnswerWeve, two matrices will be same, if the given two matrices have same number of rows and columns and each elements of that two matrices are same.
Now equating the given two matrices we get$, 6 = p$ and $x = 2.$
View full question & answer→MCQ 91 Mark
If $\text{A}=\begin{bmatrix}2&-1&3\\-4&5&1\end{bmatrix}$ and $\text{B}=\begin{bmatrix}2&3\\4&-2\\1&5\end{bmatrix},$ then:
AnswerCorrect option: C. $AB$ and $BA$ both are defined.
Given$:\ \text{A} = \begin{bmatrix}2& -1 &3\\-4 & 5 & 1 \end{bmatrix} \ \text{B} = \begin{bmatrix}2& 3\\4 & -2 \\1 & 5 \end{bmatrix}$
$\text{AB} = \begin{bmatrix}2& -1 &3\\-4 & 5 & 1 \end{bmatrix} \begin{bmatrix}2& 3 \\4 & -2 \\1 & 5 \end{bmatrix}$
$ \begin{bmatrix}3 & 23\\13 & -17 \end{bmatrix}$
So$, AB$ is defined as of columns in $A$ is equal to number of rows in $B.$
$\text{BA} = \begin{bmatrix}2&3\\4 &-2\\1 & 5 \end{bmatrix} \begin{bmatrix}2& -1 &3\\-4 & 5 & 1 \end{bmatrix}$
$ = \begin{bmatrix}-8& 13 &9\\16 & -14 & 10\\-18 & 24 & 8 \end{bmatrix}$
So$, BA$ is also defined of columns in $B$ is equal to number of rows in $A.$
View full question & answer→MCQ 101 Mark
The possible dimension of a matrix consisting $27$ elements is $4.$Reason: The number of ways of expressing $27$ as a product of two positive integers is $4.$
- A
Both Assertion $\&$ Reason are individually correct $\&$ Reason is correct explanation of Assertion,
- B
Both Assertion $\&$ Reason are individually true but Reason is Not the correct explanation of Assertion.
- ✓
Assertion is correct but Reason is incorrect.
- D
Assertion is incorrect but Reason is correct.
AnswerCorrect option: C. Assertion is correct but Reason is incorrect.
$27 = 1 \times 27$ and $3 \times 9$
Thus the number of ways of expressing $27$ as a product of two numbers is only $2$
Thus the Reason is false.
Also the possible dimensions of a matrix having $27$ elements are $27, 27 \times 1, 3 \times 9, 9 \times 3$
View full question & answer→MCQ 111 Mark
The number of all possible matrices of order $3 \times 3$ with each entry $0$ or $1$ is:
AnswerThe given matrix of the order $3\times3$ has $9$ elements and each of these elements can be either $0$ or $1.$
Now, each of the $9$ elements can be filled in two possible ways.
Therefore, by the multiplication principle, the required number of possible matrices is $2^9=512$
View full question & answer→MCQ 121 Mark
If $S =\left[ S _{ ij }\right]$ is a scalar matrix such that $S _{ ij }= k$ and $A$ is a square matrix of the same order, then $AS = SA = ?$
AnswerHere,
$\text{S}=\big[\text{S}_{\text{ij}}\big]$
$\Rightarrow\text{S}=\begin{bmatrix}\text{k}&0\\0&\text{k}\end{bmatrix}\big[\because\ \text{S}_\text{ij} = \text{k}\big]$
Let $\text{A}=\begin{bmatrix}\text{a}_{11}&\text{a}_{12}\\\text{a}_{21}&\text{a}_{22}\end{bmatrix}\big[\because A$ is square matrix$\big]$
Now,
$\text{AS}=\begin{bmatrix}\text{a}_{11}&\text{a}_{12}\\\text{a}_{21}&\text{a}_{22}\end{bmatrix}\begin{bmatrix}\text{k}&0\\0&\text{k}\end{bmatrix}=\begin{bmatrix}\text{ka}_{11}&\text{ka}_{12}\\\text{ka}_{21}&\text{ka}_{22}\end{bmatrix}$$=\text{k}\begin{bmatrix}\text{a}_{11}&\text{a}_{12}\\\text{a}_{21}&\text{a}_{22}\end{bmatrix}=\text{kA}$
$\text{SA}=\begin{bmatrix}\text{k}&0\\0&\text{k}\end{bmatrix}\begin{bmatrix}\text{a}_{11}&\text{a}_{12}\\\text{a}_{21}&\text{a}_{22}\end{bmatrix}=\begin{bmatrix}\text{ka}_{11}&\text{ka}_{12}\\\text{ka}_{21}&\text{ka}_{22}\end{bmatrix}$$=\text{k}\begin{bmatrix}\text{a}_{11}&\text{a}_{12}\\\text{a}_{21}&\text{a}_{22}\end{bmatrix}=\text{kA}$
$\therefore\ \text{AS}=\text{SA}=\text{kA}$
View full question & answer→MCQ 131 Mark
If a matrix $A$ is both symmetric and skew-symmetric, then:
AnswerCorrect option: B. $A$ is a zero matrix.
$A$ is symmetric $\Rightarrow a _{ ij }= a _{ ji } \rightarrow$ (1)$A$ is skew$-$symmetric
$\Rightarrow a_{i j}=-a_{i j} \rightarrow(2)$ and
$a_{i j}=-a_{i j}$
$\Rightarrow a_{i j}=0$ means the diagonal entries are zero.
From $(1)$ and $(2)$ we can write $a_{i j}=a_{i j}=0$ which means all the off diagonal entries are zero.
So, $A$ is a null matrix.
View full question & answer→MCQ 141 Mark
Which one of the following statements is not true:
- A
A scalar matrix is a square matrix
- B
A diagonal matrix is a square matrix
- C
A scalar matrix is a diagonal matrix
- ✓
A diagonal matrix is a scalar matrix
AnswerCorrect option: D. A diagonal matrix is a scalar matrix
Option $A$ and Option $C$ and option $B -$ true $A$ scalar matrix is a diagonal matrix and every diagonal matrix is a square matrix Hence every scalar matrix is also square matrix Option $D -$ not trueEvery diagonal matrix is square matrix but not vice versa
View full question & answer→MCQ 151 Mark
If the matrix $\begin{bmatrix}1 &\text{amp;3}& \text{amp}\lambda+2 \\2& \text{amp;4}&\text{amp;8} \\3&\text{amp;5}&\text{amp;}10 \end{bmatrix}$ is singular, then $\lambda=$
AnswerA matrix is singular if and only if it has a determinant of $0.$
$\begin{bmatrix}1 &\text{amp;3}& \text{amp}\lambda+2 \\2& \text{amp;4}&\text{amp;8} \\3&\text{amp;5}&\text{amp;}10 \end{bmatrix}$
$(40-40)-2(20-24)+(\lambda+2)(10-12)=0$
$2\lambda=4$
$\Rightarrow\lambda=2$
View full question & answer→MCQ 161 Mark
If $\begin{bmatrix}\cos\frac{2\pi}{7}&-\sin\frac{2\pi}{7}\\\sin\frac{2\pi}{7}&\cos\frac{2\pi}{7}\end{bmatrix}^\text{k}=\begin{bmatrix}1&0\\0&1\end{bmatrix},$ then the least positive integral value of $k$ is:
Answer$\text{A}=\begin{bmatrix}\cos\frac{2\pi}{7}&-\sin\frac{2\pi}{7}\\\sin\frac{2\pi}{7}&\cos\frac{2\pi}{7}\end{bmatrix}$
$\Rightarrow\text{A}^2=\text{A}\times\text{A}$
$\Rightarrow\text{A}^2=\begin{bmatrix}\cos\frac{2\pi}{7}&-\sin\frac{2\pi}{7}\\\sin\frac{2\pi}{7}&\cos\frac{2\pi}{7}\end{bmatrix}\begin{bmatrix}\cos\frac{2\pi}{7}&-\sin\frac{2\pi}{7}\\\sin\frac{2\pi}{7}&\cos\frac{2\pi}{7}\end{bmatrix}$
$\Rightarrow\text{A}^2=\begin{bmatrix}\cos^2\frac{2\pi}{7}-\sin^2\frac{2\pi}{7}&\Big(-2\cos\frac{2\pi}{7}-\sin\frac{2\pi}{7}\Big)\\2\cos\frac{2\pi}{7}\sin\frac{2\pi}{7}&\cos\frac{2\pi}{7}-\sin^2\frac{2\pi}{7}\end{bmatrix}$
$\Rightarrow\text{A}^2=\begin{bmatrix}\cos\frac{4\pi}{7}&-\sin\frac{4\pi}{7}\\\sin\frac{4\pi}{7}&\cos\frac{4\pi}{7}\end{bmatrix}$$\begin{bmatrix}\because\ \cos^2\theta-\sin^2\theta=\cos2\theta\\2\sin\theta\cos\theta=\sin2\theta\end{bmatrix}$
$\Rightarrow\text{A}^3=\text{A}^2\times\text{A}$
$\Rightarrow\text{A}^3=\begin{bmatrix}\cos\frac{4\pi}{7}&-\sin\frac{4\pi}{7}\\\sin\frac{4\pi}{7}&\cos\frac{4\pi}{7}\end{bmatrix}\begin{bmatrix}\cos\frac{2\pi}{7}&-\sin\frac{2\pi}{7}\\\sin\frac{2\pi}{7}&\cos\frac{2\pi}{7}\end{bmatrix}$
$\Rightarrow\text{A}^3=\begin{bmatrix}\Big(\cos\frac{4\pi}{7}\cos\frac{2\pi}{7}-\sin\frac{4\pi}{7}\sin\frac{2\pi}{7}\Big)&\Big(-\cos\frac{4\pi}{7}\sin\frac{2\pi}{7}-\sin\frac{4\pi}{7}\cos\frac{2\pi}{7}\Big)\\\Big(\sin\frac{4\pi}{7}\cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}\sin\frac{2\pi}{7}\Big)&\Big(-\sin\frac{2\pi}{7}\sin\frac{4\pi}{7}+\cos\frac{4\pi}{7}\cos\frac{2\pi}{7}\Big)\end{bmatrix}$
$\Rightarrow\text{A}^3=\begin{bmatrix}\cos\frac{6\pi}{7}&-\sin\frac{6\pi}{7}\\\sin\frac{6\pi}{7}&\cos\frac{6\pi}{7}\end{bmatrix}$$\begin{bmatrix}\because\ \cos\text{(A+B)}=\cos\text{A}\cos\text{B}-\sin\text{A}\sin\text{B}\\\sin\text{(A+B)}=\sin\text{A}\cos\text{B}+\cos\text{A}\sin\text{B}\end{bmatrix}$
Now we check if the pattern is same for $k = 6.$
Here,
$\text{A}^6=\text{A}^3.\text{A}^3$
$\Rightarrow\text{A}^6=\begin{bmatrix}\cos\frac{6\pi}{7}&-\sin\frac{6\pi}{7}\\\sin\frac{6\pi}{7}&\cos\frac{6\pi}{7}\end{bmatrix}\begin{bmatrix}\cos\frac{6\pi}{7}&-\sin\frac{6\pi}{7}\\\sin\frac{6\pi}{7}&\cos\frac{6\pi}{7}\end{bmatrix}$
$\Rightarrow\text{A}^6=\begin{bmatrix}\cos\frac{12\pi}{7}&-\sin\frac{12\pi}{7}\\\sin\frac{12\pi}{7}&\cos\frac{12\pi}{7}\end{bmatrix}$
Now, we check if the pattern is same for $k = 7.$
Here,
$\text{A}^7=\text{A}^6\times\text{A}$
$\Rightarrow\text{A}^7=\begin{bmatrix}\cos\frac{6\pi}{7}&-\sin\frac{6\pi}{7}\\\sin\frac{6\pi}{7}&\cos\frac{6\pi}{7}\end{bmatrix}\begin{bmatrix}\cos\frac{2\pi}{7}&-\sin\frac{2\pi}{7}\\\sin\frac{2\pi}{7}&\cos\frac{2\pi}{7}\end{bmatrix}$
$\Rightarrow\text{A}^7=\begin{bmatrix}\cos\frac{14\pi}{7}&-\sin\frac{14\pi}{7}\\\sin\frac{14\pi}{7}&\cos\frac{14\pi}{7}\end{bmatrix}$
$\Rightarrow\text{A}^7=\begin{bmatrix}\cos2\pi&-\sin2\pi\\\sin2\pi&\cos2\pi\end{bmatrix}$$\begin{bmatrix}\because\ \frac{14\pi}{7}=2\pi\end{bmatrix}$
$=\begin{bmatrix}1&0\\0&1\end{bmatrix}$
So, the least positive integral value of $k$ is $7.$
View full question & answer→MCQ 171 Mark
If matrix $\text{A}=\big[\text{a}_{\text{ij}}\big]_{2\times2'}$ where $\text{a}_\text{ij}=\begin{cases}1,&\text{if }\text{i }\neq\text{j}\\0,&\text{if }\text{i }=\text{j}\end{cases},$ then $A^2$ is equal to:
Answer$\text{A}=\begin{bmatrix}0 &1\\1&0\end{bmatrix}$
$\text{A}^2=\begin{bmatrix}0&1\\1&0\end{bmatrix}\begin{bmatrix}0&1\\1&0\end{bmatrix}$
$=\begin{bmatrix}1&0\\0&1\end{bmatrix}=1$
View full question & answer→MCQ 181 Mark
If $A=\left[a_{i j}\right]$ is a square matrix of even order such that $a_{i j}=i^2-j^2,$ then
- A
$A$ is a skew$-$symmetric matrix and $|A| = 0$
- B
$A$ is symmetric matrix and $|A|$ is a square
- C
$A$ is symmetric matrix and $|A| = 0$
- ✓
AnswerLet $\text{A}=\begin{bmatrix}0&-3\\3&0\end{bmatrix}$$\big[\because\ \text{a}_\text{ij} = \text{i}^2 -\text{j}^2\big]$
$|\text{A}|=0-(-9)=9\neq0$
View full question & answer→MCQ 191 Mark
Which of the given value of $x$ and $y$ make the following pair of matrices equal
$\begin{bmatrix}3\text{x}+7&5\\ \text{y}+1&2-3\text{x}\end {bmatrix},\ \begin{bmatrix}0& \text {y}-2\\8&4 \end{bmatrix}$
- A
$x = \frac{-1}{3}, y = 7$
- ✓
- C
$y = 7, x = \frac{-2}{3}$
- D
$x = \frac{-1}{3}, y = \frac{-2}{3}$
AnswerWe are given that
$\begin{bmatrix}3\text{x}+7&5\\ \text{y}+1&2-3\text{x}\end {bmatrix},\ \begin{bmatrix}0& \text {y}-2\\8&4 \end{bmatrix}$
By defination of equality of matrices.
$3x + 7 = 0, 5 = y - 2, y + 1 = 8, 2 - 3x = 4$
$\therefore \text x = \frac{7}{3},\ \text y = 7,\ \text x = -\frac {2}{3}$
$\therefore (b)$ is correct answer.
View full question & answer→MCQ 201 Mark
If $ \displaystyle \begin{vmatrix}\text{a} &\text{amp; }\text{b} &\text{amp; 0}\\ 0 &\text{amp; a} &\text{amp; b}\\\text{b}&\text{amp; a}&\text{amp; 0}\end{vmatrix}=0,$ then the order is:
- ✓
$3 \times 3$
- B
$2 \times 3$
- C
$2 \times 2$
- D
AnswerCorrect option: A. $3 \times 3$
There are $3$ rows and $3$ columns.Therefore the order of the matrix is $3 \times 3.$
View full question & answer→MCQ 211 Mark
If $\text{A}=\frac{1}{\pi}\begin{bmatrix}\sin^{-1}(\pi\text{x})&\tan^{-1}(\frac{\text{x}}{\pi})\\\sin^{-1}(\frac{\text{x}}{\pi})&\cot^{-1}(\pi\text{x})\end{bmatrix}, \text{B}=\frac{1}{\pi}\begin{bmatrix}-\cos^{-1}(\pi\text{x})&\tan^{-1}(\frac{\text{x}}{\pi})\\\sin^{-1}(\frac{\text{x}}{\pi})&-\tan^{-1}(\pi\text{x})\end{bmatrix},$ then $A - B$ is equal to :
- ✓
$I$
- B
$0$
- C
$2I$
- D
$\frac{1}{2}\text{I}$
AnswerGiven $\text{A}=\frac{1}{\pi}\begin{bmatrix}\sin^{-1}(\pi\text{x})&\tan^{-1}(\frac{\text{x}}{\pi})\\\sin^{-1}(\frac{\text{x}}{\pi})&\cot^{-1}(\pi\text{x})\end{bmatrix}$
$\text{B}=\frac{1}{\pi}\begin{bmatrix}-\cos^{-1}(\pi\text{x})&\tan^{-1}(\frac{\text{x}}{\pi})\\\sin^{-1}(\frac{\text{x}}{\pi})&-\tan^{-1}(\pi\text{x})\end{bmatrix}$
$\text{A}-\text{B}=\frac{1}{\pi}\begin{bmatrix}\sin^{-1}(\text{x}\pi)+\cos^{-1}(\pi\text{x})&0\\0&\cot^{-1}(\pi\text{x})+\tan^{-1}(\pi\text{x})\end{bmatrix}$
$=\frac{1}{\pi}\begin{bmatrix}\frac{\pi}{2}&0\\0&\frac{\pi}{2}\end{bmatrix}=\frac{1}{2}\begin{bmatrix}1&0\\0&1\end{bmatrix}=\frac{1}{2}\text{I}$
View full question & answer→MCQ 221 Mark
Out of the given matrices, choose that matrix which is a scalar matrix :
- ✓
$\begin{bmatrix}0&0\\0&0\end{bmatrix}$
- B
$\begin{bmatrix}0&0&0\\0&0&0\end{bmatrix}$
- C
$\begin{bmatrix}0&0\\0&0\\0&0\end{bmatrix}$
- D
$\begin{bmatrix}0\\0\\0\end{bmatrix}$
AnswerCorrect option: A. $\begin{bmatrix}0&0\\0&0\end{bmatrix}$
A diagonal matrix with all diagonal elements are equal is a scalar matrix.
View full question & answer→MCQ 231 Mark
If $\text{A} = \begin{bmatrix}\cos\alpha&-\sin\alpha\\ \sin\alpha&\cos\alpha\end{bmatrix},\text{then}\ \text{A + A}'=\text{I}$, if the value of a is:
- ✓
$\frac{\pi}{6}$
- B
$\frac{\pi}{3}$
- C
$\text{n}$
- D
$\frac{3\pi}{2}$
AnswerCorrect option: A. $\frac{\pi}{6}$
The correct answer is B.
$\text{A}=\begin{bmatrix}\cos\alpha&-\sin\alpha\\ \sin\alpha&\cos\alpha \end{bmatrix}$
$\Rightarrow\ \text{A}'=\begin{bmatrix}\cos\alpha&\sin\alpha\\ -\sin\alpha&\cos\alpha \end{bmatrix}$
Now, $\text{A + A}'=\text{I}$
$\therefore\ \begin{bmatrix}\cos\alpha&-\sin\alpha\\\sin\alpha&\cos\alpha\end{bmatrix}+\begin{bmatrix}\cos\alpha&\sin\alpha\\-\sin\alpha&\cos\alpha\end{bmatrix}=\begin{bmatrix}1&0\\0&1\end{bmatrix}$
$\Rightarrow\ \begin{bmatrix}2\cos\alpha&0\\0&2\cos\alpha\end{bmatrix}=\begin{bmatrix}1&0\\0&1\end{bmatrix}$
Comparing the corresponding elements of the two matrices, we have:
$2\cos\alpha=1$
$\Rightarrow\ \cos\alpha=\frac{1}{2}=\cos\frac{\pi}{3}$
$\therefore\ \alpha=\frac{\pi}{3}$
View full question & answer→MCQ 241 Mark
If the matrix $AB$ is zero, then :
AnswerCorrect option: A. It is not necessary that either $A = 0$ or, $B = 0$
Let $\text{A}=\begin{bmatrix}0&2\\0&0\end{bmatrix}$ and $\text{B}=\begin{bmatrix}1&0\\0&0\end{bmatrix}$
$\therefore \text{AB}=\begin{bmatrix}0&2\\0&0\end{bmatrix}\begin{bmatrix}1&0\\0&0\end{bmatrix}=\begin{bmatrix}0&0\\0&0\end{bmatrix}$
View full question & answer→MCQ 251 Mark
If $A$ and $B$ are two matrices such that $A B=B$ and $B A=A, A^2+B^2$ is equal to :
- A
$2AB$
- B
$2BA$
- ✓
$A + B$
- D
$AB$
AnswerCorrect option: C. $A + B$
Given $AB = A$ and $BA = B,$ then
$\Rightarrow \text{BAB}=B^2$ and $\text{ABA}=A^2$
$\Rightarrow B A=B^2$ and $A B=A^2$
$\Rightarrow B=B^2$ and $A=A^2$
$\Rightarrow A^2+B^2=A+B$
View full question & answer→MCQ 261 Mark
Let $\text{A}=\begin{bmatrix}\text{a}&0&0\\0&\text{a}&0\\0&0&\text{a}\end{bmatrix},$ then $A^n$ is equal to :
- A
$\begin{bmatrix}\text{a}^\text{n}&0&0\\0&\text{a}^\text{n}&0\\0&0&\text{a}\end{bmatrix}$
- B
$\begin{bmatrix}\text{a}^\text{n}&0&0\\0&\text{a}&0\\0&0&\text{a}\end{bmatrix}$
- ✓
$\begin{bmatrix}\text{a}^\text{n}&0&0\\0&\text{a}^\text{n}&0\\0&0&\text{a}^\text{n}\end{bmatrix}$
- D
$\begin{bmatrix}\text{na}&0&0\\0&\text{na}&0\\0&0&\text{na}\end{bmatrix}$
AnswerCorrect option: C. $\begin{bmatrix}\text{a}^\text{n}&0&0\\0&\text{a}^\text{n}&0\\0&0&\text{a}^\text{n}\end{bmatrix}$
$\text{A}=\begin{bmatrix}\text{a}&0&0\\0&\text{a}&0\\0&0&\text{a}\end{bmatrix}=\text{a}\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}$
$\text{A}^\text{n}=\text{a}^\text{n}\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}=\begin{bmatrix}\text{a}^\text{n}&0&0\\0&\text{a}^\text{n}&0\\0&0&\text{a}^\text{n}\end{bmatrix}$
View full question & answer→MCQ 271 Mark
If $\text{A}=\begin{bmatrix}\alpha&\beta\\\gamma&-\alpha\end{bmatrix}$ is such that $A^2 = I$, then :
- A
$1+\alpha^2+\beta\gamma=0$
- B
$1-\alpha^2+\beta\gamma=0$
- ✓
$1-\alpha^2-\beta\gamma=0$
- D
$1+\alpha^2-\beta\gamma=0$
AnswerCorrect option: C. $1-\alpha^2-\beta\gamma=0$
Given $\text{A}=\begin{bmatrix}\alpha&\beta\\\gamma&-\alpha\end{bmatrix}$ and $A^2 = I,$ then
$\text{A}^2=\text{I}$
$\Rightarrow\begin{bmatrix}\alpha&\beta\\\gamma&-\alpha\end{bmatrix}\begin{bmatrix}\alpha&\beta\\\gamma&-\alpha\end{bmatrix}=\begin{bmatrix}1&0\\0&1\end{bmatrix}$
$\Rightarrow\begin{bmatrix}\alpha^2+\beta\gamma&0\\0&\alpha^2+\beta\gamma\end{bmatrix}=\begin{bmatrix}1&0\\0&1\end{bmatrix}$
$\Rightarrow\alpha^2+\beta\gamma=1$
$\Rightarrow1-\alpha^2-\beta\gamma=0$
View full question & answer→MCQ 281 Mark
$\text{A}^2=\text{I} \Rightarrow$
- A
$|\text{A}|=0$
- B
$|\text{A}|=1$
- C
$|\text{A}|=-1$
- ✓
$|\text{A}|=\pm1$
AnswerCorrect option: D. $|\text{A}|=\pm1$
Given, $\text{A}^2=\text{I}$
Take determinant both sides,
$|\text{A}^2|=|\text{I}|$
$\Rightarrow|\text{A}^2|=1$
$\Rightarrow|\text{A}|=\pm1$
View full question & answer→MCQ 291 Mark
Choose the correct answer from the given four options.If matrix $A=\left[a_{i j}\right] 2 \times 2$, where $a_{i j}=1,$ if $i \neq j$ and 0 if $i=j$ then $A^2$ equal to :
AnswerWe have, $A=\left[a_{i j}\right] 2 \times 2$, where $a_{i j}=1$, if $i \neq j$ and 0 if $i=j$
$\therefore\ \text{A}=\begin{bmatrix}0&1\\1&0\end{bmatrix}$
And $\text{A}^2=\begin{bmatrix}0&1\\1&0\end{bmatrix}\begin{bmatrix}0&1\\1&0\end{bmatrix}$
$=\begin{bmatrix}1&0\\0&1\end{bmatrix}=\text{I}$
View full question & answer→MCQ 301 Mark
Choose the correct answer from the given four options.If $A$ and $B$ are two matrices of the order $3 \times m$ and $3 \times n,$ respectively and $m = n$, then order of matrix $(5A – 2B)$ is:
- A
$m \times 3$
- B
$3 \times 3$
- C
$m \times n$
- ✓
$3 \times n$
AnswerCorrect option: D. $3 \times n$
We are given that, the order of the matrices $A$ and $B$ are $3 \times m$ and $3 \times n$ respectively. Now, If $m = n$, then $A$ and $B$ have same orders as $3 \times n$ each, so the order of $(5A – 2B)$ should be same as $3 \times n$.
View full question & answer→MCQ 311 Mark
If matrix $A$ is of order $p \times q$ and matrix $B$ is of order $r \times s$ then $A − B$ will exist if :
- A
$p = q$
- ✓
$p = r, q = s$
- C
$p = q, r = s$
- D
AnswerCorrect option: B. $p = r, q = s$
If matrix $A$ is of order $p \times q$ and matrix $B$ is of order $r \times s$ then $A − B$ will exist if order of $A$ and $B$ is same.
Therefore $, p = r, q = s$
View full question & answer→MCQ 321 Mark
The matrix $\text{A}=\begin{bmatrix}0&-5&8\\5&0&12\\-8&-12&0\end{bmatrix}$ is a :
AnswerCorrect option: C. Skew $-$ symmetric matrix.
Given $\text{A}=\begin{bmatrix}0&-5&8\\5&0&12\\-8&-12&0\end{bmatrix}$
$\text{A}^\text{T}=\begin{bmatrix}0&5&-8\\-5&0&-12\\8&12&0\end{bmatrix}$
$\Rightarrow\text{A}=-\text{A}^\text{T}$
So, $A$ is skew $-$ symmetric matrix.
View full question & answer→MCQ 331 Mark
If $\text{A} = \displaystyle \left[ \begin{matrix} 1 &\text{amp ; 2} \\ 3&\text{amp; 4} \end{matrix} \right],$ then number of elements in $A$ are :
AnswerSince, given matrix $A$ is of order $2\times2 = 4$
$\therefore$ Number of elements in $\text{A} = 4$
View full question & answer→MCQ 341 Mark
The possible number of different orders that a matrix can have when it has $24$ elements, is :
AnswerPossible order of matrices $24 \times 1, 1 \times 24, 2 \times 12, 12 \times 2, 3 \times 8, 8 \times 3, 4 \times 6, 6 \times 4$
So, the number of possible matrices with $24$ elements is $8.$
View full question & answer→MCQ 351 Mark
If $A = [1$ amp; $2], B = [3$ amp; $4]$ then $A + B =$
- A
$[1$ amp; $4]$
- B
$[4$ amp; $4]$
- ✓
$[4$ amp; $6]$
- D
AnswerCorrect option: C. $[4$ amp; $6]$
Given, $A = [1$ amp; $2], B = [3$ amp; $4]$ then $A + B =$
$[1 + 3$ amp; $2 + 4] A + B = [4$ amp; $6]$
View full question & answer→MCQ 361 Mark
If $A = [1] ,$ then $A$ is :
- A
- B
- ✓
Non $-$ singular matrix
- D
AnswerCorrect option: C. Non $-$ singular matrix
$\text{A} = \big[1\big] $ is an identity matrix with order $1\times1.|\text{A}|\neq0$
So $A$ is nonsingular.
View full question & answer→MCQ 371 Mark
If $A =\left[a_{ ij }\right]$ is a scalar matrix of order $n \times n$ such that $a_{i j}=k$, for all $i^\text{th,}$
Answer$\text{Trace}=\sum\limits^\text{n}_{\text{i}-1}\text{a}_{\text{ij}}=\text{nk}$
View full question & answer→MCQ 381 Mark
If $A$ and $B$ are non $-$ zero square matrices of the same order such that $AB = 0,$ then
AnswerCorrect option: C. $|A| = 0$ or $|B| = 0$
From the properties of the matrices, if $A, B$ are non $-$ zero square matrices of same order such that $AB = 0$ then the either of the matrices must be singular matrix.
$A$ singular matrix is a matrix whose determinant is zero.
$\therefore |A| = 0$ or $|B| = 0$
View full question & answer→MCQ 391 Mark
If $\displaystyle \begin{vmatrix}\text{x} &\text{amp; } 1 \\ \text{y} &\text{amp; } 2 \end{vmatrix}-\displaystyle \begin{vmatrix}\text{y} &\text{amp; } 1 \\ 8&\text{amp; } 0\end{vmatrix}=\displaystyle \begin{vmatrix}2 &\text{amp; } 0 \\ \text{-x}&\text{amp; } 2\end{vmatrix}$ then the values of $x$ and $y$ respectively are :
- A
$5$ and $1$
- ✓
$5$ and $3$
- C
$5$ and $2$
- D
$3$ and $4$
AnswerCorrect option: B. $5$ and $3$
$x - y = 2$
$y - 8 = -x$
Solving we get $x = 5$ and $y = 3$
View full question & answer→MCQ 401 Mark
If $A$ is $3 \times 4$ matrix and $B$ is a matrix such that $A\ 'B$ and $BA\ '$ are both defined. Then $,B$ is of the type :
- ✓
$3\times 4$
- B
$3\times 3$
- C
$4\times 4$
- D
$4\times 3$
AnswerCorrect option: A. $3\times 4$
The order of $A$ is $3\times 4$.
So, the order of $A\ '$ is $4\times 3.$
Now, both $A\ 'B$ and $BA\ '$ are defined.
So, the number of columns in $A\ '$ should be equal to the number of rows in $B$ for $A\ 'B.$
Also, the number of columns in $B$ should be equal to number of rows in $A\ '$ for $BA\ '.$
Hence, the order of matrix $B$ is $3\times 4.$
View full question & answer→MCQ 411 Mark
If $\text{A}$ is a square of order $3,$ then $|\text{Adj}(\text{Adj}\text{A}^2)|=$
- A
$|\text{A}|^2$
- B
$|\text{A}|^4$
- ✓
$|\text{A}|^8$
- D
$|\text{A}|^{16}$
AnswerCorrect option: C. $|\text{A}|^8$
$\text{KEY} : 3$
$|\text{Adj}(\text{Adj}\text{A}^2)|$
$\text{Q}=|\text{A}^2|^{(3-1)^2}=|\text{A}^2|^4=|\text{A}|^8$
View full question & answer→MCQ 421 Mark
The matrix $\text{A}=\begin{bmatrix}0&\text{amp};0&\text{amp};0\\0&\text{amp};4&\text{amp};0\\4&\text{amp};0&\text{amp};0\end{bmatrix}$ is $a$ :
AnswerGiven, $\text{A}=\begin{bmatrix}0&\text{amp};0&\text{amp};4\\0&\text{amp};4&\text{amp};0\\4&\text{amp};0&\text{amp};0\end{bmatrix}$
The matrix is a square as it has same no. of rows and columns,
But it is not a diagonal matrix as there are elements other than diagonal ones which are not zero.
View full question & answer→MCQ 431 Mark
The element in the second row and third column of the matrix $\begin{bmatrix}4&\text{amp; }5&\text{amp; }6 \\3 &\text{amp;}-4&\text{amp; }3\\2 &\text{amp; }1&\text{amp; }0 \end{bmatrix}$ is :
AnswerThe element in the second row, third column is represented by $a_{23}=3$
View full question & answer→MCQ 441 Mark
If $\text{A}=\begin{bmatrix}1&\text{a}\\0&1\end{bmatrix},$ then $A^n\ ($where $n \in N$) equals :
- ✓
$\begin{bmatrix}1&\text{na}\\0&1\end{bmatrix}$
- B
$\begin{bmatrix}1&\text{n}^2\text{a}\\0&1\end{bmatrix}$
- C
$\begin{bmatrix}1&\text{n}\text{a}\\0&0\end{bmatrix}$
- D
$\begin{bmatrix}\text{n}&\text{n}\text{a}\\0&\text{n}\end{bmatrix}$
AnswerCorrect option: A. $\begin{bmatrix}1&\text{na}\\0&1\end{bmatrix}$
Given $\text{A}=\begin{bmatrix}1&\text{a}\\0&1\end{bmatrix}$
$\text{A}^2=\begin{bmatrix}1&\text{a}\\0&1\end{bmatrix}\begin{bmatrix}1&\text{a}\\0&1\end{bmatrix}$
$=\begin{bmatrix}1&2\text{a}\\0&1\end{bmatrix}$
$\text{A}^3=\text{A}^2\text{A}$
$=\begin{bmatrix}1&2\text{a}\\0&1\end{bmatrix}\begin{bmatrix}1&\text{a}\\0&1\end{bmatrix}$
$=\begin{bmatrix}1&3\text{a}\\0&1\end{bmatrix}$
On genaralising we get
$\text{A}^\text{n}=\begin{bmatrix}1&\text{na}\\0&1\end{bmatrix}$
View full question & answer→MCQ 451 Mark
If a matrix has mm rows and nn columns then its order is :
- A
$\text{m}+\text{n}$
- B
$\text{n}\times\text{n}$
- C
$\text{m}\times\text{m}$
- ✓
$\text{m}\times\text{n}$
AnswerCorrect option: D. $\text{m}\times\text{n}$
A matrix has mm rows and $n$ columns then its order is $\text{m}\times\text{n}$
View full question & answer→MCQ 461 Mark
If $A$ and $B$ are square matrices such that $AB = I$ and $BA = I,$ then $B$ is :
AnswerCorrect option: C. Multiplicative inverse matrix of $A$
$\text{AB}=\begin{bmatrix}\text{I}&\text{amp; }\end{bmatrix}\text{BA}=\text{I}$ is the multiplicative inverse of $A$.
Hence, the answer is multiplicative inverse matrix of $A.$
View full question & answer→MCQ 471 Mark
If $\text{A}=\begin{bmatrix}0&2\\3&-4\end{bmatrix}$ and $\text{kA}=\begin{bmatrix}0&3\text{a}\\2\text{b}&24\end{bmatrix},$ then the values of $\text{k, a, b},$ are respectively
- A
$-6, -12, -18$
- B
$-6, 4, 9$
- ✓
$-6, -4, -9$
- D
$-6, 12, 18$
AnswerCorrect option: C. $-6, -4, -9$
$\text{A}=\begin{bmatrix}0&2\\3&-4\end{bmatrix}$
$\text{kA}=\begin{bmatrix}0&3\text{a}\\2\text{b}&24\end{bmatrix}$
$\Rightarrow\begin{bmatrix}0&2\text{k}\\3\text{k}&-4\text{k}\end{bmatrix}=\begin{bmatrix}0&3\text{a}\\2\text{b}&24\end{bmatrix}$
$\Rightarrow-4\text{k}=24$
$\Rightarrow\text{k}=-6$
$2\text{k}=3\text {a}$
$\Rightarrow\text{a}=-4$
$3\text{k}=2\text{b}$
$\Rightarrow\text{b}=-9$
View full question & answer→MCQ 481 Mark
If $\text{A}=\begin{bmatrix}\text{i}&0\\0&\text{i}\end{bmatrix},\text{n}\in\text{N},$ then $A^{4 n}$ equals :
- A
$\begin{bmatrix}0&\text{i}\\\text{i}&0\end{bmatrix}$
- B
$\begin{bmatrix}0&0\\0&0\end{bmatrix}$
- ✓
$\begin{bmatrix}1&0\\0&1\end{bmatrix}$
- D
$\begin{bmatrix}0&\text{i}\\\text{i}&0\end{bmatrix}$
AnswerCorrect option: C. $\begin{bmatrix}1&0\\0&1\end{bmatrix}$
Given $\text{A}=\begin{bmatrix}\text{i}&0\\0&\text{i}\end{bmatrix}=\text{i}\begin{bmatrix}1&0\\0&1\end{bmatrix}$
$\text{A}^4=\text{i}^4\begin{bmatrix}1&0\\0&1\end{bmatrix}$
$=\begin{bmatrix}1&0\\0&1\end{bmatrix}$
$\text{A}^{4\text{n}}=\begin{bmatrix}1&0\\0&1\end{bmatrix}$
View full question & answer→MCQ 491 Mark
If $A$ is $3 \times 4$ matrix and $B$ is matrix such that $AB$ and $BA$ are both defined, then $B$ is of the type :
- ✓
$3 \times 4$
- B
$3 \times 3$
- C
$4 \times 4$
- D
$4 \times 3$
AnswerCorrect option: A. $3 \times 4$
Given that matrix $A$ is $3 \times 4.$
Let the $B$ matrix be $P \times Q.$
$\therefore A$ is $4 \times 3.$
Since $AB$ is defined, so number of columns of $A$ must be equal to number of rows of $B,$ therefore, $P = 3.$
Also, $BA$ is defined, so the number of columns of $B$ must be equal to number of rows of $A,$ then $Q = 4.$
Therefore, matrix $B$ is $3 \times 4.$
View full question & answer→MCQ 501 Mark
Choose the correct answer from the given four options. If $A$ is a square matrix such that $A^2=1$, then $(A-1)^3+(A+1)^3 - 7$ A is equal to :
- ✓
$A$
- B
$I - A$
- C
$I + A$
- D
$3A$
AnswerWe have, $A ^2=1$
Now, $(A-I)^3+(A+I)^3-7 A=[(A-I)+(A+I)]$
$\left[(A-I)^2+(A+I)^2-(A-I)(A+I)\right]-7 A$
${\left[\because a^3+b^3=(a+b)\left(a^2+b^2-a b\right)\right]}$
$=\left[(2 A)\left\{A^2+I^2-2 A I+A^2+I^2+2 A I-\left(A^2-I^2\right)\right\}\right]-7 A$
$=\left[(2 A)\left\{A I+I^2-2 A I+A I+I^2+2 A I-A I+I^2\right\}\right]-7 A\left[\because A^2=A I\right]$
$=2 A\left[I+I^2+I+I^2-I+I^2\right]-7 A$
$=2 A[5 I-I]-7 A$
$=8 A I-7 A I\ [\because A=A I]$
$=A I$
$=A$
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