MCQ 11 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 21 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 31 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 41 Mark
If $A$ is a square matrix such that $A^2 = A$, then $(I + A)^3 - 7A$ is equal to:
AnswerHere,
$A^2 = A ...(1)$
$A^3 = A^2A$
$= A^2 [$From $eq. (1)]$
$= A$
$\therefore A^3 = A ...(2)$
We know that $(I + A)^3 = I^3 + 3(I)^2 A + 3(I) A^2 + A^3$
$\Rightarrow (I + A)^3 = I + 3A + 3A + A [$From $eqs. (1)$ and $(2)]$
$\Rightarrow (I + A)^3 = I + 7A$
$\Rightarrow (I + A)^3 - 7A $
$= I$
View full question & answer→MCQ 51 Mark
If $S = [S_{ij}]$ 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}$$\big[\because\ \text{A is square matrix}\big]$
Now,
$\text{AS}=\begin{bmatrix}\text{a}_{11}&\text{a}_{12}\\\text{a}_{21}&\text{a}_{22}\end{bmatrix}\begin{bmatrix}\text{k}&0\\0&\text{k}\end{bmatrix}=\begin{bmatrix}\text{ka}_{11}&\text{ka}_{12}\\\text{ka}_{21}&\text{ka}_{22}\end{bmatrix}$$=\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}$$=\text{k}\begin{bmatrix}\text{a}_{11}&\text{a}_{12}\\\text{a}_{21}&\text{a}_{22}\end{bmatrix}=\text{kA}$
$\therefore\ \text{AS}=\text{SA}=\text{kA}$
View full question & answer→MCQ 61 Mark
If a matrix $A$ is both symmetric and skew$-$symmetric, then:
AnswerCorrect option: B. $A$ is a zero matrix.
$A$ is symmetric
$\Rightarrow a_{ij} = a_{ji} \rightarrow (1)$
$A$ is skew$-$symmetric
$\Rightarrow a_{ij} = - a_{ij} \rightarrow (2)$ and
$a_{ij} = - a_{ij}$
$\Rightarrow a_{ij} = 0$ means the diagonal entries are zero.
From $(1)$ and $(2)$ we can write
$a_{ij} = a_{ij} = 0$ which means all the off diagonal entries are zero.
So, $A$ is a null matrix.
View full question & answer→MCQ 71 Mark
If $\begin{bmatrix}\cos\frac{2\pi}{7}&-\sin\frac{2\pi}{7}\\\sin\frac{2\pi}{7}&\cos\frac{2\pi}{7}\end{bmatrix}^\text{k}=\begin{bmatrix}1&0\\0&1\end{bmatrix},$ then the least positive integral value of $k$ is:
Answer$\text{A}=\begin{bmatrix}\cos\frac{2\pi}{7}&-\sin\frac{2\pi}{7}\\\sin\frac{2\pi}{7}&\cos\frac{2\pi}{7}\end{bmatrix}$
$\Rightarrow\text{A}^2=\text{A}\times\text{A}$
$\Rightarrow\text{A}^2=\begin{bmatrix}\cos\frac{2\pi}{7}&-\sin\frac{2\pi}{7}\\\sin\frac{2\pi}{7}&\cos\frac{2\pi}{7}\end{bmatrix}\begin{bmatrix}\cos\frac{2\pi}{7}&-\sin\frac{2\pi}{7}\\\sin\frac{2\pi}{7}&\cos\frac{2\pi}{7}\end{bmatrix}$
$\Rightarrow\text{A}^2=\begin{bmatrix}\cos^2\frac{2\pi}{7}-\sin^2\frac{2\pi}{7}&\Big(-2\cos\frac{2\pi}{7}-\sin\frac{2\pi}{7}\Big)\\2\cos\frac{2\pi}{7}\sin\frac{2\pi}{7}&\cos\frac{2\pi}{7}-\sin^2\frac{2\pi}{7}\end{bmatrix}$
$\Rightarrow\text{A}^2=\begin{bmatrix}\cos\frac{4\pi}{7}&-\sin\frac{4\pi}{7}\\\sin\frac{4\pi}{7}&\cos\frac{4\pi}{7}\end{bmatrix}$$\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 81 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$\text{A}=\begin{bmatrix}0 &1\\1&0\end{bmatrix}$
$\text{A}^2=\begin{bmatrix}0&1\\1&0\end{bmatrix}\begin{bmatrix}0&1\\1&0\end{bmatrix}$
$=\begin{bmatrix}1&0\\0&1\end{bmatrix}=I$
View full question & answer→MCQ 91 Mark
If $A = [a_{ij}]$ is a square matrix of even order such that $a_{ij} = i^2 - j^2$, 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 101 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:
- A
$I$
- B
$0$
- C
$2I$
- ✓
$\frac{1}{2}\text{I}$
AnswerCorrect option: D. $\frac{1}{2}\text{I}$
Given $\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 111 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 121 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 131 Mark
If $A$ and $B$ are two matrices such that $AB = B$ and $BA = 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 BAB = B^2$ and $ABA = A^2$
$\Rightarrow BA = B^2$ and $AB = A^2$
$\Rightarrow B = B^2$ and $A = A^2$
$\Rightarrow A^2 + B^2 = A + B$
View full question & answer→MCQ 141 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}$
$\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 151 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 161 Mark
The matrix $\text{A}=\begin{bmatrix}0&-5&8\\5&0&12\\-8&-12&0\end{bmatrix}$ is a:
- A
- B
- ✓
Skew$-$symmetric matrix.
- D
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 171 Mark
If $A = [a_{ij}]$ is a scalar matrix of order $n \times n$ such that $a_{ij} = k$, for all $i$, then trace of $A$ is equal to:
Answer$\text{Trace}=\sum\limits^\text{n}_{\text{i}-1}\text{a}_{\text{ij}}=\text{nk}$
View full question & answer→MCQ 181 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 191 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}$
$\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 201 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 $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 211 Mark
If $\text{A}=\begin{bmatrix}\text{i}&0\\0&\text{i}\end{bmatrix},\text{n}\in\text{N},$ then $A^{4n}$ 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}$
$\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 221 Mark
If $A$ and $B$ are symmetric matrices, then $\text{ABA}$ is:
- ✓
- B
Skew$-$symmetric matrix.
- C
- D
AnswerLet $\text{A}=\begin{bmatrix}1&2\\2&1\end{bmatrix}$ and $\text{B}=\begin{bmatrix}3&2\\2&3\end{bmatrix}$
$\text{AB}=\begin{bmatrix}1&2\\2&1\end{bmatrix}\begin{bmatrix}3&2\\2&3\end{bmatrix}=\begin{bmatrix}7&8\\8&7\end{bmatrix}$
$\text{ABA}=\begin{bmatrix}7&8\\8&7\end{bmatrix}\begin{bmatrix}1&2\\2&1\end{bmatrix}=\begin{bmatrix}23&22\\22&23\end{bmatrix}$
View full question & answer→MCQ 231 Mark
The matrix $\text{A}=\begin{bmatrix}1&0&0\\0&2&0\\0&0&4\end{bmatrix}$ is:
- A
- B
- C
Skew$-$symmetric matrix.
- ✓
AnswerA matrix is called Diagonal matrix if all the elements, except those in the leading diagonal, are zero.
View full question & answer→MCQ 241 Mark
Which of the given values of $X$ and $Y$ make the following pairs 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
$\text{x}=-\frac{1}{3},\text{y}=7$
- B
$\text{y}=7,\text{x}=-\frac{2}{3}$
- C
$\text{x}=-\frac{1}{3},\text{y}=-\frac{2}{5}$
- ✓
$\text{Not possible to find}$
AnswerCorrect option: D. $\text{Not possible to find}$
$\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}$
$\Rightarrow3\text{x}+7=0$
$\Rightarrow\text{x}=\frac{-7}{3}$
$5=\text{y}-2$
$\Rightarrow\text{y}=7$
$\text{y}+1=8$
$\Rightarrow\text{y}=7$
$2-3\text{x}=4$
$\Rightarrow\text{x}=\frac{-2}{3}$
We are getting two values of $x.$
So, it is not possible to find.
View full question & answer→MCQ 251 Mark
If $A$ and $B$ are matrices of the same order, then $AB^T - BA^T$ is a:
- ✓
Skew$-$symmetric matrix.
- B
- C
- D
AnswerCorrect option: A. Skew$-$symmetric matrix.
$(AB^T - BA^T)^T $
$= (AB^T)^T - (BA^T)^T$
$= BA^T - AB^T$
$= -(AB^T - BA^T)$
Therefore, $AB^T - BA^T$ is a skew$-$symmetric matrix.
Hence, the correct option is $(a).$
View full question & answer→MCQ 261 Mark
The matrix $\text{A}=\begin{bmatrix}0&0&4\\0&4&0\\4&0&0\end{bmatrix}$ is a:
AnswerGiven: $\text{A}=\begin{bmatrix}0&0&4\\0&4&0\\4&0&0\end{bmatrix}$
Since, number of rows is equal to number of columns.
Therefore, $A$ is a square matrix.
Hence, the correct option is $(a).$
View full question & answer→MCQ 271 Mark
If $\text{A}=\begin{bmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{bmatrix},$ then $A^T + A = I_2$, if:
- A
$\theta=\text{n}\pi,\text{n}\in\text{Z}$
- B
$\theta=(2\text{n}+1)\frac{\pi}{2},\text{n}\in\text{Z}$
- ✓
$\theta=2\text{n}\pi+\frac{\pi}{3},\text{n}\in\text{Z}$
- D
$\text{None of these}$
AnswerCorrect option: C. $\theta=2\text{n}\pi+\frac{\pi}{3},\text{n}\in\text{Z}$
Here,
$\text{A}=\begin{bmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{bmatrix}$
$\Rightarrow\text{A}^\text{T}=\begin{bmatrix}\cos\theta&\sin\theta\\-\sin\theta&\cos\theta\end{bmatrix}$
Now,
$\text{A}^\text{T}+\text{A}=\text{I}_2$
$\Rightarrow\begin{bmatrix}\cos\theta&\sin\theta\\-\sin\theta&\cos\theta\end{bmatrix}+\begin{bmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{bmatrix}=\begin{bmatrix}1&0\\0&1\end{bmatrix}$
$\Rightarrow\begin{bmatrix}2\cos\theta&0\\0&2\cos\theta\end{bmatrix}=\begin{bmatrix}1&0\\0&1\end{bmatrix}$
$\Rightarrow2\cos\theta=1$
$\Rightarrow\cos\theta=\frac{1}{2}$
$\Rightarrow\cos\theta=\cos\frac{\pi}{3}$
$\Rightarrow\theta=2\text{n}\pi\pm\frac{\pi}{3}$ $(\text{n}\in\text{Z})$
View full question & answer→MCQ 281 Mark
If $\text{A}=\begin{bmatrix}1&0&0\\0&1&0\\\text{a}&\text{b}&-1\end{bmatrix},$ then $A^2$ is equal to:
- A
$A$ null matrix
- ✓
$A$ unit matrix
- C
$-A$
- D
$A$
AnswerCorrect option: B. $A$ unit matrix
$\text{A}^2=\text{AA}$
$\Rightarrow\text{A}^2=\begin{bmatrix}1&0&0\\0&1&0\\\text{a}&\text{b}&-1\end{bmatrix}\begin{bmatrix}1&0&0\\0&1&0\\\text{a}&\text{b}&-1\end{bmatrix}$
$\Rightarrow\text{A}^2=\begin{bmatrix}1+0+0&0+0+0&0+0-0\\0+0+0&0+1+0&0+0-0\\\text{a}+0-\text{a}&0+\text{b}-\text{b}&0+0+1\end{bmatrix}$
$\Rightarrow\text{A}^2=\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}$
View full question & answer→MCQ 291 Mark
If $A, B$ are square matrices of order $3, A$ is non$-$singular and $AB = 0,$ then $B$ is a:
- ✓
- B
- C
Unit$-$matrix.
- D
Non$-$singular matrix.
AnswerSince $A$ is non$-$singular matrix and the determinant of a non$-$singular matrix is non$-$zero, $B$ should be a null matrix.
View full question & answer→MCQ 301 Mark
If $A$ and $B$ are square matrices of the same order, then $(A + B)(A - B)$ is equal to:
- A
$A^2 - B^2$
- B
$A^2 - BA - AB - B^2$
- ✓
$A^2 - B^2 + BA - AB$
- D
$A^2 - BA + B^2+ AB$
AnswerCorrect option: C. $A^2 - B^2 + BA - AB$
$(A + B)(A - B) $
$= A^2 - AB + BA - B^2$
Hence, the correct option is $(c).$
View full question & answer→MCQ 311 Mark
If $A$ and $B$ are two matrices such that $AB = A$ and $BA = B$, then $B^2$ is equal to:
AnswerHere, $AB = A ...(1)$
$BA = B ...(2)$
$\Rightarrow BAB = BB [$Multiplying both sides by $B]$
$\Rightarrow BA = B^2 [$From $eq. (1)]$
$\Rightarrow B = B^2 [$From $eq. (2)]$
View full question & answer→MCQ 321 Mark
The matrix $\begin{bmatrix}0&5&-7\\-5&0&11\\7&-11&0\end{bmatrix}$ is:
AnswerCorrect option: A. A skew$-$symmetric matrix.
Here,
$\text{A}=\begin{bmatrix}0&5&-7\\-5&0&11\\7&-11&0\end{bmatrix}$
$\Rightarrow\text{A}^\text{T}=\begin{bmatrix}0&-5&7\\5&0&-11\\-7&11&0\end{bmatrix}$
$\Rightarrow\text{A}^\text{T}=-\begin{bmatrix}0&5&-7\\-5&0&11\\7&-11&0\end{bmatrix}$
$\Rightarrow\text{A}^\text{T}=-\text{A}$
Thus, $A$ is a skew$-$symmetric matrix.
View full question & answer→MCQ 331 Mark
The number of all possible matrices of order $3\times 3$ with each entry $0$ or $1$ is:
AnswerThere are $9$ elements in a $3\times 3$ matrix and one element can be filled in two ways, either with $0$ or $1.$
Thus,
Total possible matrices $= 2^9 = 512$
View full question & answer→MCQ 341 Mark
If $\begin{bmatrix}2\text{x}+\text{y}&4\text{x}\\5\text{x}-7&4\text{x}\end{bmatrix}=\begin{bmatrix}7&7\text{y}-13\\\text{y}&\text{x}+6\end{bmatrix},$ then the value of $x$ and $y$ is:
- A
$x = 3, y = 1$
- B
$x = 2, y = 3$
- ✓
$x = 2, y = 4$
- D
$x = 3, y = 3$
AnswerCorrect option: C. $x = 2, y = 4$
$\begin{bmatrix}2\text{x}+\text{y}&4\text{x}\\5\text{x}-7&4\text{x}\end{bmatrix}=\begin{bmatrix}7&7\text{y}-13\\\text{y}&\text{x}+6\end{bmatrix},$
Equating the terms, we get
$4x = x + 6$
$\Rightarrow x = 2$
And
$2x + y = 7$
$\Rightarrow y = 3$
View full question & answer→MCQ 351 Mark
If $\text{I}=\begin{bmatrix}1&0\\0&1\end{bmatrix},\text{J}=\begin{bmatrix}0&1\\-1&0\end{bmatrix}$ and $\text{B}=\begin{bmatrix}\cos\theta&\sin\theta\\-\sin\theta&\cos\theta\end{bmatrix},$ then $B$ equals:
- ✓
$\text{I}\cos\theta+\text{J}\sin\theta$
- B
$\text{I}\sin\theta+\text{J}\cos\theta$
- C
$\text{I}\cos\theta-\text{J}\sin\theta$
- D
$-\text{I}\cos\theta+\text{J}\sin\theta$
AnswerCorrect option: A. $\text{I}\cos\theta+\text{J}\sin\theta$
Here
$\text{I}\cos\theta+\text{J}\sin\theta$
$=\begin{bmatrix}1&0\\0&1\end{bmatrix}\cos\theta+\begin{bmatrix}0&1\\-1&0\end{bmatrix}\sin\theta$
$=\begin{bmatrix}\cos\theta&0\\0&\cos\theta\end{bmatrix}+\begin{bmatrix}0&\sin\theta\\-\sin\theta&0\end{bmatrix}$
$=\begin{bmatrix}\cos\theta&\sin\theta\\-\sin\theta&\cos\theta\end{bmatrix}=\text{B}$
View full question & answer→MCQ 361 Mark
If $AB = A$ and $BA = B$, where $A$ and $B$ are square matrices, then:
AnswerCorrect option: A. $B^2 = B$ and $A^2 = A$
Here,
$\text{AB = A} ...(1)$
$\text{BA = B} ...(2)$
$\Rightarrow \text{ABA = AA} [$Multiplying both sides by $A]$
$\text{BAB = BB} [$Multiplying both sides by $A]$
$\Rightarrow AB = A^2 [$From eq. $(2)]$
$BA = B^2 [$From eq. $(1)]$
$\Rightarrow A = A^2 [$From eq. $(1)]$
$B = B^2 [$From eq. $(2)]$
View full question & answer→MCQ 371 Mark
The number of possible matrices of order $3\times 3$ with each entry $2$ or $0$ is:
AnswerLet us consider a matrix $\begin{bmatrix}\text{a}&\text{b}&\text{c}\\\text{d}&\text{e}&\text{f}\\\text{g}&\text{h}&\text{i}\end{bmatrix}$
The element a can have two values $0$ or $2$ in two ways.
Similarly all other elements can also have two values $0$ or $2$ in two ways each.
So, the total number of combinations is $2^9$.
So, total no of matrices will be $2^9$.
View full question & answer→MCQ 381 Mark
If $\text{A}=\begin{bmatrix}1&-1\\2&-1\end{bmatrix},\text{B}=\begin{bmatrix}\text{a}&1\\\text{b}&-1\end {bmatrix}$ and $(A + B)^2 = A^2 + B^2$, values of $a$ and $b$ are:
- A
$a = 4, b = 1$
- ✓
$a = 1, b = 4$
- C
$a = 0, b = 4$
- D
$a = 2, b = 4$
AnswerCorrect option: B. $a = 1, b = 4$
Here,
$(\text{A+B})^2=\text{A}^2+\text{B}^2$
$\Rightarrow\text{A}^2+\text{AB}+\text{BA}+\text{B}^2=\text{A}^2+\text{B}^2$
$\Rightarrow\text{AB}+\text{BA}=0$
$\Rightarrow\text{AB}=-\text{BA}$
$\Rightarrow\begin{bmatrix}1&-1\\2&-1\end{bmatrix}\begin{bmatrix}\text{a}&1\\\text{b}&-1\end{bmatrix}=-\begin{bmatrix}\text{a}&1\\\text{b}&-1\end{bmatrix}\begin{bmatrix}1&-1\\2&-1\end{bmatrix}$
$\Rightarrow\begin{bmatrix}\text{a}-\text{b}&2\\2\text{a}-\text{b}&3\end{bmatrix}=-\begin{bmatrix}\text{a}+2&-\text{a}-1\\\text{b}-2&-\text{b}+1\end{bmatrix}$
$\Rightarrow\begin{bmatrix}\text{a}-\text{b}&2\\2\text{a}-\text{b}&3\end{bmatrix}=\begin{bmatrix}-\text{a}-2&\text{a}+1\\\text{b}+2&\text{b}-1\end{bmatrix}$
The corresponding elements of two equal matrices are equal.
$\Rightarrow\text{a}+1=2$ and $b-1=3$
$\therefore\ \text{a}=1$ and $b=4$
View full question & answer→MCQ 391 Mark
If $\text{A}=\begin{bmatrix}5&\text{x}\\\text{y}&0\end{bmatrix}$ and $A = A^T$, then:
- A
$x = 0, y = 5$
- B
$x + y = 5$
- ✓
$x = y$
- D
AnswerCorrect option: C. $x = y$
Here,
$\text{A}=\begin{bmatrix}5&\text{x}\\\text{y}&0\end{bmatrix}$
$\text{A}^\text{T}=\begin{bmatrix}5&\text{y}\\\text{x}&0\end{bmatrix}$
Now,
$\text{A}=\text{A}^\text{T}$
The corresponding elements of two equal matrices are equal.
$\therefore\ \begin{bmatrix}5&\text{x}\\\text{y}&0\end{bmatrix}=\begin{bmatrix}5&\text{y}\\\text{x}&0\end{bmatrix}$
$\Rightarrow\text{x}=\text{y}$
View full question & answer→MCQ 401 Mark
If $\text{A}=\begin{bmatrix}2&0&-3\\4&3&1\\-5&7&2\end{bmatrix}$ is expressed as the sum of a symmetric and skew$-$symmetric matrix, then the symmetric matrix is:
- ✓
$\begin{bmatrix}2&2&-4\\2&3&4\\-4&4&2\end{bmatrix}$
- B
$\begin{bmatrix}2&4&-5\\0&3&7\\-3&1&2\end{bmatrix}$
- C
$\begin{bmatrix}4&4&-8\\4&6&8\\-8&8&4\end{bmatrix}$
- D
$\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}$
AnswerCorrect option: A. $\begin{bmatrix}2&2&-4\\2&3&4\\-4&4&2\end{bmatrix}$
Here,
$\text{A}=\begin{bmatrix}2&0&-3\\4&3&1\\-5&7&2\end{bmatrix}$
$\Rightarrow\text{A}^\text{T}=\begin{bmatrix}2&4&-5\\0&3&7\\-3&1&2\end{bmatrix}$
Now,
$\text{A}+\text{A}^\text{T}=\begin{bmatrix}2&0&-3\\4&3&1\\-5&7&2\end{bmatrix}+\begin{bmatrix}2&4&-5\\0&3&7\\-3&1&2\end{bmatrix}$
$\Rightarrow\text{A}+\text{A}^\text{T}=\begin{bmatrix}2+2&0+4&-3-5\\4+0&3+3&1+7\\-5-3&7+1&2+2\end{bmatrix}$
$\Rightarrow\text{A}+\text{A}^\text{T}=\begin{bmatrix}4&4&-8\\4&6&8\\-8&8&4\end{bmatrix}$
$\text{A}-\text{A}^\text{T}=\begin{bmatrix}2&0&-3\\4&3&1\\-5&7&2\end{bmatrix}-\begin{bmatrix}2&4&-5\\0&3&7\\-3&1&2\end{bmatrix}$
$\Rightarrow\text{A}-\text{A}^\text{T}=\begin{bmatrix}2-2&0-4&-3+5\\4-0&3-3&1-7\\-5+3&7-1&2-2\end{bmatrix}$
$\Rightarrow\text{A}-\text{A}^\text{T}=\begin{bmatrix}0&-4&2\\4&0&-6\\-2&6&0\end{bmatrix}$
Let $\text{P}=\frac{1}{2}(\text{A}+\text{A}^\text{T})=\frac{1}{2}\begin{bmatrix}4&4&-8\\4&6&8\\-8&8&4\end{bmatrix}$ $=\begin{bmatrix}2&2&-4\\2&3&4\\-4&4&2\end{bmatrix}$
$\text{Q}=\frac{1}{2}(\text{A}-\text{A}^\text{T})=\frac{1}{2}\begin{bmatrix}0&-4&2\\4&0&-6\\-2&6&0\end{bmatrix}$$=\begin{bmatrix}0&-2&1\\2&0&-3\\-1&3&0\end{bmatrix}$
Now,
$\text{P}^\text{T}=\begin{bmatrix}2&2&-4\\2&3&4\\-4&4&2\end{bmatrix}^\text{T}=\begin{bmatrix}2&2&-4\\2&3&4\\-4&4&2\end{bmatrix}=\text{P}$
$\text{Q}^\text{T}=\begin{bmatrix}0&-2&1\\2&0&-3\\-1&3&0\end{bmatrix}=\begin{bmatrix}0&2&-1\\-2&0&3\\1&-3&0\end{bmatrix}$$=-\begin{bmatrix}0&-2&1\\2&0&-3\\-1&3&0\end{bmatrix}=-\text{Q}$
$\text{P+Q}=\begin{bmatrix}2&2&-4\\2&3&4\\-4&4&2\end{bmatrix}+\begin{bmatrix}0&-2&1\\2&0&-3\\-1&3&0\end{bmatrix}$
$=\begin{bmatrix}2+0&2-2&-4+1\\2+2&3+0&4-3\\-4-1&4+3&2+0\end{bmatrix}$
$=\begin{bmatrix}2&0&-3\\4&3&1\\-5&7&2\end{bmatrix}=\text{A}$
Thus, we have expressed $A$ is the sum of a symmetric and a skew$-$symmetric matrix.
Hence,the symmetric matrix is $\begin{bmatrix}2&2&-4\\2&3&4\\-4&4&2\end{bmatrix}.$
View full question & answer→MCQ 411 Mark
If $A$ is a square matrix such that $A^2 = I$, then $(A - I)^3 + (A + I)^3 - 7A$ is equal to:
- ✓
$A$
- B
$I - A$
- C
$I + A$
- D
$3A$
Answer$(A - I)^3 + (A + I)^3 - 7A$
$= A^3 - I^3 - 3A^2I + 3AI^2 + A^3 + I^3 + 3A^2I + 3AI^2 - 7A$
$= 2A^3 + 6AI^2 - 7A$
$= 2A.A^2 + 6A - 7A$
$= 2A.I - A (\because$ $A^2 = I)$
$= 2A - A$
$= A$
Hence, the correct option is $(a)$.
View full question & answer→MCQ 421 Mark
If $A$ and $B$ are two matrices of order $3\times m$ and $3\times n$ respectively and $m = n,$ then the order of $5A - 2B$ is:
- A
$m\times 3$
- B
$3\times 3$
- C
$m\times n$
- ✓
$3\times n$
AnswerCorrect option: D. $3\times n$
In scalar multiplicaion and in addition or substraction of matrics the order doesn't change.
View full question & answer→MCQ 431 Mark
If $A$ is a matrix of order $m\times n$ and $B$ is a matrix such that $AB^T$ and $B^TA$ are both defined, then the order of matrix $B$ is:
- ✓
$m\times n$
- B
$n\times n$
- C
$n\times m$
- D
$m\times n$
AnswerCorrect option: A. $m\times n$
$A$ is $m\times n$ matrix and $AB^T$ is defined then
number of columns in $A=$ number of rows in $B^T =n$
$B^TA$ is also defined then number of columns in $B^T=$ number of rows in $A = m$
Order of $B$ is $m\times n$
View full question & answer→MCQ 441 Mark
If $\text{A}=\begin{bmatrix}1&2&\text{x}\\0&1&0\\0&0&1\end{bmatrix}$ and $\text{B}=\begin{bmatrix}1&-2&\text{y}\\0&1&0\\0&0&1\end{bmatrix}$ and $AB = I_3$, then $x + y$ equals:
AnswerGiven: $AB = I_3$
$\Rightarrow\begin{bmatrix}1&2&\text{x}\\0&1&0\\0&0&1\end{bmatrix}\begin{bmatrix}1&-2&\text{y}\\0&1&0\\0&0&1\end{bmatrix}=\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}$
$\Rightarrow\begin{bmatrix}1&0&\text{y+x}\\0&1&0\\0&0&1\end{bmatrix}=\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}$
The corresponding elements of two equal matrices are equal.
$\therefore\ \text{y}+\text{x}=0$
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