MCQ 11 Mark
If $A$ is a square matrix of order $3$ and $|A| = 5,$ then the value of $|2A'|$ is:
Answer$|2A'|$
$= 2^3 |A'|$
$= 8 |4|$
$= 8 \times 5$
$= 40$
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$A^2 = A$
$(I + A)^3 +A$
$\Rightarrow I^3 - A^3 - 3I^2A + 3IA^2 + A$
$\Rightarrow I - A^3 - 3A+ 3A + A [ \therefore A^2 = A]$
$\Rightarrow I - A.A^2 + A$
$\Rightarrow I - A.A + A$
$\Rightarrow I - A + A$
$= I$
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:
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:
AnswerIn this, order of P = p × k Order of W = n × 3 Order of Y = 3 × k
Thus, order of PY = p × k, when k = 3
And the order of WY = p × 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:
AnswerA matrix of mm rows and nn columns has m \times nm×n elements.
On multiplying the rows and columns in the given options, we notice.
that all 1 × 16 = 16, 2 × 8 = 16, 4 × 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:
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:
- A
- B
- ✓
AB and BA both are defined.
- D
AB and BA both are not defined.
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 × 27 and 3 × 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 × 1, 3 × 9, 9 × 3
View full question & answer→MCQ 111 Mark
The number of all possible matrices of order 3 × 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 $S = [S_{ij}]$ 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 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
$\Rightarrow a_{ij} = - a_{ij} \rightarrow (2)$ 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 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 matrixHence every scalar matrix is also square matrixOption 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}$$\begin{bmatrix}\because\ \frac{14\pi}{7}=2\pi\end{bmatrix}$
$=\begin{bmatrix}1&0\\0&1\end{bmatrix}$
So, the least positive integral value of k is 7.
View full question & answer→MCQ 171 Mark
If matrix $\text{A}=\big[\text{a}_{\text{ij}}\big]_{2\times2'}$ where $\text{a}_\text{ij}=\begin{cases}1,&\text{if }\text{i }\neq\text{j}\\0,&\text{if }\text{i }=\text{j}\end{cases},$ then $A^2$ is equal to:
Answer$\text{A}=\begin{bmatrix}0 &1\\1&0\end{bmatrix}$
$\text{A}^2=\begin{bmatrix}0&1\\1&0\end{bmatrix}\begin{bmatrix}0&1\\1&0\end{bmatrix}$
$=\begin{bmatrix}1&0\\0&1\end{bmatrix}$
$=1$
View full question & answer→MCQ 181 Mark
If $A = [a_{ij}]$ is a square matrix of even order such that $a_{ij} = i^2 - j^2,$ then
- A
$A$ is a skew$-$symmetric matrix and $|A| = 0$
- B
$A$ is symmetric matrix and $|A|$ is a square
- C
$A$ is symmetric matrix and $|A| = 0$
- ✓
AnswerLet $\text{A}=\begin{bmatrix}0&-3\\3&0\end{bmatrix}$$\big[\because\ \text{a}_\text{ij} = \text{i}^2 -\text{j}^2\big]$
$|\text{A}|=0-(-9)=9\neq0$
View full question & answer→MCQ 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:
AnswerThere are 3 rows and 3 columns.Therefore the order of the matrix is 3 × 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:
- ✓
- B
- C
- 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:
- A
$\frac{\pi}{6}$
- ✓
$\frac{\pi}{3}$
- C
$\text{n}$
- D
$\frac{3\pi}{2}$
AnswerCorrect option: B. $\frac{\pi}{3}$
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:
- ✓
It is not necessary that either A = 0 or, B = 0
- B
- C
- D
All the above statements are wrong
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 $AB = B$ 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 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 $A = [a_{ij}]_{2\times 2},$ where $a_{ij} = 1,$ if $\text{i}\neq\text{j}$ and $0$ if $i = j$ then $A^2$ equal to:
AnswerWe have, $A = [a_{ij}]_{2\times 2}$, where $a_{ij} = 1,$ if $\text{i}\neq\text{j}$ 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 × m and 3 × n, respectively and m = n, then order of matrix (5A – 2B) is:
AnswerWe are given that, the order of the matrices A and B are 3 × m and 3 × n respectively. Now, If m = n, then A and B have same orders as 3 × n each, so the order of (5A – 2B) should be same as 3 × n.
View full question & answer→MCQ 311 Mark
If matrix A is of order p × q and matrix B is of order r × s then A − B will exist if:
AnswerIf matrix A is of order p × q and matrix B is of order r × 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:
AnswerGiven $\text{A}=\begin{bmatrix}0&-5&8\\5&0&12\\-8&-12&0\end{bmatrix}$
$\text{A}^\text{T}=\begin{bmatrix}0&5&-8\\-5&0&-12\\8&12&0\end{bmatrix}$
$\Rightarrow\text{A}=-\text{A}^\text{T}$
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 × 1, 1 × 24, 2 × 12, 12 × 2, 3 × 8, 8 × 3, 4 × 6, 6 × 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 =
AnswerGiven, 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
Answer$\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 $A = [a_{ij}]$ 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 381 Mark
If A and B are non - zero square matrices of the same order such that AB = 0, then
Answer
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:
Answerx - y = 2
y - 8 = -x
Solving we get x = 5 and y = 3
View full question & answer→MCQ 401 Mark
If A is 3×4 matrix and B is a matrix such that A'B and BA' are both defined. Then, B is of the type:
AnswerThe order of A is 3×4. So, the order of A' is 4×3.
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 3×4.
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. $|\text{A}|^8$
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:
- A
- B
- ✓
Multiplicative inverse matrix of A
- D
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 k, a, b, are respectively
Answer$\text{A}=\begin{bmatrix}0&2\\3&-4\end{bmatrix}$
$\text{kA}=\begin{bmatrix}0&3\text{a}\\2\text{b}&24\end{bmatrix}$
$\Rightarrow\begin{bmatrix}0&2\text{k}\\3\text{k}&-4\text{k}\end{bmatrix}=\begin{bmatrix}0&3\text{a}\\2\text{b}&24\end{bmatrix}$
$\Rightarrow-4\text{k}=24$
$\Rightarrow\text{k}=-6$
$2\text{k}=3\text {a}$
$\Rightarrow\text{a}=-4$
$3\text{k}=2\text{b}$
$\Rightarrow\text{b}=-9$
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 $A^{4n}$ 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. $\begin{bmatrix}1&0\\0&1\end{bmatrix}$
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 × 4 matrix and B is matrix such that AB and BA are both defined, then B is of the type:
AnswerGiven that matrix A is 3 x 4.
Let the B matrix be P x Q.
$\therefore$ A is 4 x 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 x 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 $A^2 = I,$ then $(A - I)^3 + (A + I)^3 - 7A$ is equal to:
- ✓
$A$
- B
$I - A$
- C
$I + A$
- D
$3A$
AnswerWe have, $A^2 = I$
Now, $(A - I)^3 + (A + I)^3 - 7A $
$= [(A - I) + (A + I)][(A - I)^2 + (A + I)^2 - (A - I)(A + I)] - 7A$
$[\because$ $a^3 + b^3 = (a + b)(a^2 + b^2 - ab)]$
$= [(2A){A^2 + I^2 - 2AI + A^2 + I^2 + 2AI - (A^2 - I^2)}] - 7A$
$= [(2A){AI + I^2 - 2AI + AI + I^2 + 2AI - AI +I^2}] - 7A [\because A^2 = AI ]$
$= 2A[I + I^2 + I + I^2 - I + I^2] - 7A$
$= 2A[5I - I] - 7A$
$= 8AI - 7AI [\because A = AI ]$
$= AI$
$= A$
View full question & answer→MCQ 511 Mark
Choose the correct answer from the given four options.Total number of possible matrices of order $3 ×3 $ with each entry $2$ or $0$ is:
AnswerTotal number of possible matrices of order $3 \times 3$ with each entry $2$ or $0$ is $2^9$ i.e., $512.$
View full question & answer→MCQ 521 Mark
The order of a matrix $\begin{bmatrix}2&\text{amp};5 &\text{amp};7 \end{bmatrix}$ is:
AnswerSince, Order of a matrix is represented by m × n, where mm is the number of rows and nn is the number of columns.
Given, $\begin{bmatrix}2&\text{amp};5 &\text{amp};7 \end{bmatrix}$ is a matrix in which number of row is 1 and number of columns are 3.
$\therefore\begin{bmatrix}2&\text{amp};5 &\text{amp};7 \end{bmatrix}$ is a matrix of order 1 × 3
View full question & answer→MCQ 531 Mark
Which of the following is correct:
- A
Determinant is a square matrix
- B
Determinant is a number associated to a matrix
- ✓
Determinant is a number associated to a square matrix
- D
AnswerCorrect option: C. Determinant is a number associated to a square matrix
Determinant is defined only for a square matrix.and its denotes the value of that square matrix.
View full question & answer→MCQ 541 Mark
Choose the correct answer from the given four options.For any two matrices A and B, we have:
- A
$\text{AB}=\text{BA}$
- B
$\text{AB}\neq\text{BA}$
- C
$\text{AB}=\text{O}$
- ✓
AnswerFor any two matrices A and S, we may have AB = BA = I, $\text{AB}\neq\text{BA}$ and AB = O but it is not always true.
View full question & answer→MCQ 551 Mark
Two matrices A and B are added if:
- A
- ✓
- C
No of columns of A is equal to columns of B
- D
No of rows of A is equal to no of columns of B
AnswerWhile adding two matrices we add the numbers which belong to some row and column of each matrixo two matrices can be added.
If there are equal number of rows and columns in both.Both matrices should have same order therefore.
View full question & answer→MCQ 561 Mark
Matrix $\text{A} = [\text{a}_\text{ij}]_{\text{m} \times \text{n}}$ is a square matrix if:
AnswerMatrix $\text{A} = [\text{a}_\text{ij}]_{\text{m} \times \text{n}}$ is a square matrix Number of columns = number of rows = m
View full question & answer→MCQ 571 Mark
If A and B are symmetric matrices, then ABA is:
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 581 Mark
The matrix $\text{A}=\begin{bmatrix}1&0&0\\0&2&0\\0&0&4\end{bmatrix}$ is:
AnswerA matrix is called Diagonal matrix if all the elements, except those in the leading diagonal, are zero.
View full question & answer→MCQ 591 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 601 Mark
If $A$ and $B$ are matrices of order $3 \times 2$ and $C$ is of order $2 \times 3,$ then which of the following matrices is not defined:
- ✓
$A^T+ B$
- B
$A^T+ B^T$
- C
$A^T+ C$
- D
$B + C^T$
AnswerCorrect option: A. $A^T+ B$
Given order of $A$ is $3 \times 2$
$\Rightarrow$ order of $A^T$ is $2 \times 3$
Also, given order of $B$ is $3 \times 2$
$\Rightarrow$ order of $B^T$ is $2 \times 3$
Order of $C\ 2 \times 3$
Since,$ A^T, B^T, C$ have same order, so addition of any $2$ or all three matrices are defined.
$A^T+ B$ is not defined as their orders are different.
View full question & answer→MCQ 611 Mark
- A
$\begin{bmatrix} -1 & 3 \\ 2 & 4 \end{bmatrix}$
- B
$\begin{bmatrix} 0 & 3 \\ 2 & 0 \end{bmatrix}$
- ✓
$\begin{bmatrix} 4 & 0 \\ 0 & 4 \end{bmatrix}$
- D
$\text{None of these}$
AnswerCorrect option: C. $\begin{bmatrix} 4 & 0 \\ 0 & 4 \end{bmatrix}$
A diagonalmatrixwith all its main diagonal entries equal is ascalar matrix, that is, ascalarmultiple of the identitymatrix
$\therefore \begin{bmatrix} 4 &\text{amp; 0} \\ 0 &\text{amp; } 4 \end{bmatrix}$ is a scalar matrix.
View full question & answer→MCQ 621 Mark
If the matrices has 13 elements , then the possible dimension (order) it can have are:
AnswerAs we know the number of elements in a matrix = (no.of rows) × No.of columns.
Therefore for 13 elements the rows and columns could only be (13 × 1) or (1 × 13)
View full question & answer→MCQ 631 Mark
If $\text{A}=\displaystyle \begin{vmatrix} 1 &\text{amp; } 0 \\ 1 &\text{amp; } 0 \end{vmatrix}$ And $\text{B}=\displaystyle \begin{vmatrix} 1 &\text{amp; } 0 \\ 0 &\text{amp; } 1 \end{vmatrix}$ then $\text{A+B}=$
AnswerCorrect option: C. $\displaystyle \begin{vmatrix}2&0 \\ 1 &1 \end{vmatrix}$
$\text{A+B}=\displaystyle \begin{vmatrix} 2 &\text{amp; } 0 \\ 1 &\text{amp; } 1 \end{vmatrix}$
View full question & answer→MCQ 641 Mark
A square matrix A has 9 elements. What is the possible order of A?
AnswerThe factors of 9 are 1, 3 and 9.So, the possible orders of a matrix containing 9 elements is 1 × 9, 9 × 1, 3 × 3.
In a square matrix, the number of rows is equal to the number of columns.So, the required order is 3 × 3.
View full question & answer→MCQ 651 Mark
If A, B are symmetric matrices of same order, then AB - BA is a
AnswerThe correct answer is A.
A and B are symmetric matrices, therefore, we have:
A' = A and B' = B ...(i)
Consider $ (\text{AB} - \text{BA})' = (\text{AB})' - (\text{BA})' \ \ \big[(\text{A} - \text{B})' = \text{A}' - \text{B}'\big]$
$= \text{B}'\text{A}' - \text{A}'\text{B}'\ \ \big[(\text{AB}) = \text{B}'\text{A}'\big]$
$= \text{BA} -\text{AB}\ \ \big[\text{by}(1)\big]$
$=-(\text{A}\text{B}- \text{B}\text{A})$
$\therefore\ (\text{AB}- \text{BA})' = -(\text{AB} - \text{BA})$
Thus, (AB - BA) is a skew-symmetric matrix.
View full question & answer→MCQ 661 Mark
If $A$ and $B$ are matrices of the same order, then $AB^T - BA^T$ is :
- ✓
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 671 Mark
Choose the correct answer from the given four options.
If A and B are matrices of same order, then (AB′ – BA′) is a:
AnswerWe have matrices A and B of same order.
Let P = (AB' - BA')
Then, P' = (AB' - BA')'
= (AB')' - (BA')'
= (B')'(A)' - (A')'B' = BA' - AB' = -(AB' - BA') = -P
Hence, (AB' - BA') is a Skew symmetric matrix.
View full question & answer→MCQ 681 Mark
If $A$ is square matrix of order $3$, then $∣Adj(Adj\ A^2)∣ =$
- A
$|A|^2$
- B
$|A|^4$
- ✓
$|A|^8$
- D
$|A|^{16}$
AnswerCorrect option: C. $|A|^8$
$∣\text{adj}(\text{adj}\text{A}^2)∣=\text{Q}=\begin{vmatrix}\text{A}^2\end{vmatrix}^{(3-1)^2} =\begin{vmatrix}\text{ A}^2 \end{vmatrix} ^4 =\begin{vmatrix} \text{A}\end{vmatrix}^8$
View full question & answer→MCQ 691 Mark
If A and B are two matrices of order 3 × m and 3 × n respectively and m = n, then the order of 5A - 2B is:
AnswerA matrix of order 3 × mB matrix of order 3 × n
It is also given that m = nThen the order of the matrix will be sameSo order 5A - 2B is 3 × n.
View full question & answer→MCQ 701 Mark
If $\text{A}=\displaystyle \begin{vmatrix} 5 &\text{amp; x} \\ \text{y} &\text{amp; 6} \end{vmatrix}\text{B}=\displaystyle \begin{vmatrix} -4 &\text{amp; y} \\-4 &\text{amp; 5} \end{vmatrix}$ and $\text{A}+\text{B}=1$ then the values of x and y respectively are:
Answer$\text{A+B =1},\text{ i.e.,} \displaystyle \begin{vmatrix} 1&\text{amp; }\text{x+y} \\\text{y}-4 &\text{amp;} 1 \end{vmatrix}=\begin{vmatrix} 1&\text{amp; } 0 \\ 0 &\text{amp; } 1 \end{vmatrix}$
or $\text{x}=\text{y}=0,\text{ y}-4=0$
$\therefore\text{ x} = -4, \text{ y}=4$
View full question & answer→MCQ 711 Mark
A matrix has 18 elements.Find the number of possible orders of the matrix:
AnswerA matrix of mm rows and nn columns has m × n elements.
18 can be got by all combinations of 1 × 18,18 × 1, 2 × 9, 9 × 2, 3 ×6, 6 × 3
Hence, there are 6 possible matrices which have 18 elements.
View full question & answer→MCQ 721 Mark
If $\text{A}=\displaystyle \left[ \begin{matrix} 1 &\text{amp; 2} \\ 3&\text{amp; 4} \end{matrix} \right],$ then which of the following is not an element of A ?
Answer0 is not present in given matrix.
View full question & answer→MCQ 731 Mark
If m[-3 amp; 4] + n[-3 amp; 4] = [10 amp;-11], then 3m + 7n =
AnswerGiven m[-3 + 4] + n[4 - 3] = [10 - 11]
⇒ -3m + 4n = 10 and 4m - 3n = -11
by solving we get m= -2 and n=1
$\therefore$ 3m + 7n = -6 + 7 = 1
View full question & answer→MCQ 741 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 751 Mark
If n = p, then order of matrix 7X - 5Z is:
AnswerHere n = p (given), the order of matrices X and Z are equal.
$\therefore$ 7X – 5Z is well defined and the order of 7X – 5Z is same as the order of X and Z.
$\therefore$ The order of 7X – 5Z is either equal to 2 × n or 2 × p
But it is given that n = p
Therefore, the option (B) is correct.
View full question & answer→MCQ 761 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
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 771 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 781 Mark
The number of different possible orders of matrices having 18 identical elements is:
AnswerLet the order of the matrix is $ (\text{a}\times\text{b})$ There are 18 elements in the matrix.
So, $\text{a}\times\text{b} = 18$
Possible orders can be $ (1\times18),( 18\times1),( 2\times9),( 9\times2),( 3\times6),( 6\times3)$
There are 6 possible orders.
View full question & answer→MCQ 791 Mark
Choose the correct answer from the given four options.If $\text{A}=\begin{bmatrix}1&0\\0&1\end{bmatrix},$ then $A^2$ is equal to :
- A
$\begin{bmatrix}0&1\\1&0\end{bmatrix}$
- B
$\begin{bmatrix}1&0\\1&0\end{bmatrix}$
- C
$\begin{bmatrix}0&1\\0&1\end{bmatrix}$
- ✓
$\begin{bmatrix}1&0\\0&1\end{bmatrix}$
AnswerCorrect option: D. $\begin{bmatrix}1&0\\0&1\end{bmatrix}$
$\because\ \text{A}^2=\text{A}.\text{A}$
$=\begin{bmatrix}0&1\\1&0\end{bmatrix}.\begin{bmatrix}0&1\\1&0\end{bmatrix}=\begin{bmatrix}1&0\\0&1\end{bmatrix}$
View full question & answer→MCQ 801 Mark
If $A$ and $B$ are symmetric matrices of the same order, then:
- A
$\text{AB}$ is a symmetric matrix.
- B
$\text{A - B}$ is askew$-$symmetric matrix.
- ✓
$\text{AB + BA}$ is a symmetric matrix.
- D
$\text{AB - BA}$ is a symmetric matrix.
AnswerCorrect option: C. $\text{AB + BA}$ is a symmetric matrix.
View full question & answer→MCQ 811 Mark
If the order of matrices A and B are 3 × 2 and 2 × 1 respectively, then find the order of matrix (if possible) AB:
AnswerOrder of A : 3 × 2 Order of B : 2 × 1 Multiplication of matrices is possible if and only.
if the number of columns of first matrix is equal to the number of rows of second matrix In AB
No.of columns in A is No.of rows in B is 2 $\therefore$ AB exists.
Order of AB is (number of rows of A x number of columns of B)
$\therefore$ Order of AB is (3 × 1)
View full question & answer→MCQ 821 Mark
The order of the matrix $\begin{bmatrix}1\\3\\4<\text{br}> \end{bmatrix}$ is:
AnswerOrder of matrix with mm rows and nn columns is given as $\text{m} \times\text{n}$ Let $\text{A}=\begin{bmatrix}-1\\3\\4 \end{bmatrix}$
In the given matrix, there are $3 $ rows and $1$ column.
Hence, the order of A is $3\times 1$.
View full question & answer→MCQ 831 Mark
$B = A + A^2+ A^3 + A^4$ If order of $A$ is $3$ then order of $B$ is:
AnswerThe order of matrix doesnt change when the operation are done on it.
So The order of $B$ remains same as the order of $A$
View full question & answer→MCQ 841 Mark
If A, B are square matrices of order 3, A is non-singular and AB = 0, then B is a:
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 851 Mark
If the sum of the matrices $\begin{bmatrix}\text{x}\\\text{x}\\\text{y}\end{bmatrix},\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}$ and $\begin{bmatrix}\text{z}\\0\\0\end{bmatrix}$ is the matrix $\begin{bmatrix}10\\5\\5\end{bmatrix},$ then what is the value of y ?
Answer$\begin{bmatrix}\text{x}\\\text{x}\\\text{y}\end{bmatrix}+\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}+\begin{bmatrix}\text{z}\\0\\0\end{bmatrix}=\begin{bmatrix}10\\5\\5\end{bmatrix}\therefore\text{x}+\text{y}+\text{z}=10,\text{ x}+\text{y}=5$
$\text{y}+\text{z}=5$ Replacing $\text{x}+\text{y}=5$ in $\text{x}+\text{y}=\text{z}=10$
We have, $\text{z}=5$
Also, $\text{y}+\text{z}=5$
$\therefore\text{y}=5-\text{z}=0$
View full question & answer→MCQ 861 Mark
If $\text{A}= \begin{bmatrix} 1 &\text{amp; } 2 &\text{amp;} 3\end{bmatrix},$ then order is:
AnswerAn $\text{m}\times\text{n} $ matrix has m row and n columns.
The given matrix $\text{A}= \begin{bmatrix} 1 &\text{amp; } 2 &\text{amp;} 3\end{bmatrix},$ has 1 row and 3 columns.
Thus, order of A is $ 1\times3.$
View full question & answer→MCQ 871 Mark
If $ \text{A}+\displaystyle \begin{vmatrix} 4 &\text{amp; } 2 \\ 1 &\text{amp; } 3 \end{vmatrix}=\displaystyle \begin{vmatrix} 6 &\text{amp; } 9 \\ 1 &\text{amp; } 4\end{vmatrix} $ then $\text{A}=$
- ✓
$\displaystyle \begin{vmatrix} 2 & 7 \\ 0 & 1\end{vmatrix} $
- B
$\displaystyle \begin{vmatrix} 0 & 1 \\ 2 & 7\end{vmatrix} $
- C
$\displaystyle \begin{vmatrix} 1 & 0 \\ 2 & 7\end{vmatrix} $
- D
$\displaystyle \begin{vmatrix} 2 & 1 \\ 0 & 7\end{vmatrix} $
AnswerCorrect option: A. $\displaystyle \begin{vmatrix} 2 & 7 \\ 0 & 1\end{vmatrix} $
$ \text{A}=\displaystyle \begin{vmatrix} 6 &\text{amp; } 9 \\ 1 &\text{amp; } 4\end{vmatrix}-\displaystyle \begin{vmatrix} 4 &\text{amp; } 2 \\ 1 &\text{amp; } 3 \end{vmatrix}=\displaystyle \begin{vmatrix} 2 &\text{amp; } 7 \\ 0 &\text{amp; } 1 \end{vmatrix}$
View full question & answer→MCQ 881 Mark
If $A$ and $B$ are two matrices of same order, then $A + B$ is equal to :
- ✓
$B + A$
- B
$BA$
- C
$(A + B)^T$
- D
$A - B$
AnswerCorrect option: A. $B + A$
Yes, matrices are commutative.
We can see it as follows, Let element of $A$ matrix be denoted by $\text{a}_\text{ij}$ and $B$ matrix be denoted by $\text{b}_\text{ij},$
Then corresponding elements of$\text{ A + B}$ matrix will be $(\text{a}_\text{ij} +\text{b}_\text{ij}) $ and corresponding
elements of $\text{B + A}$ matrix will be $(\text{b}_\text{ij} +\text{ a}_\text{ij}) $ But since addition is commutative, corresponding elements
of$\text{ A + B}$ and $\text{B + A}$ matrices are the same, So they are equal.
View full question & answer→MCQ 891 Mark
If $ \displaystyle \begin{vmatrix} \text{x} &\text{amp;}\text{ y} \\ 1 &\text{amp; } 6 \end{vmatrix}= \displaystyle \begin{vmatrix} 1&\text{amp; }8 \\ 1 &\text{amp; } 6 \end{vmatrix}$ then $\text{x}+2\text{y}=$
Answerx = 1.y = 8
$\therefore$ x + 2y = 17
View full question & answer→MCQ 901 Mark
If $A$ is a square matrix, then $' A – A\ ’$ is a:
- A
- ✓
Skew$-$symmetric matrix.
- C
- D
AnswerCorrect option: B. Skew$-$symmetric matrix.
View full question & answer→MCQ 911 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 921 Mark
If A is a matrix of order m × n and B is a matrix such that AB′ and B′A are both defined, the order of the matrix B is:
AnswerGiven that order of matrix A is m\times nm×nNow if AB′ is defined then
number of column of A should be same as number of rows of B′, which is
n Also since B′A is defined,so number column of B′ should be same as number of rows of A
which is m Thus order of B′ is n × m.Hence, order of matrix B is m × n.
Note: Product of two matrix A and B, AB is defined only if number of columns of A is same as number of rows of B.
And B′ represents transpose of matrix B.
View full question & answer→MCQ 931 Mark
If $\displaystyle \text{a}_{\text{ij}}=0\left (\text{i}\neq \text{j} \right )$ and $\displaystyle\text{a}_{\text{ij}}=2\left (\text{i= j} \right )$ then the matrix $\text{A}=\displaystyle \left [ \text{a}_{\text{ij}} \right ]_{\text{n}\times\text{n}}$ is a _______ matrix ?
AnswerGiven A is a square matrix as the number of rows and columns are same as n
The elements $\text{a}_\text{ij}$ where $\text{i} = \text{j} $ lie along the diagonal.
and the elements $\text{a}_\text{ij}$ where $\text{i}\neq\text{j}$ do not lie along the diagonal.
Given, diagonal elements = 2 and the rest of the elements = 0
Such a diagonal matrix where all diagonal elements are equal, is called a scalar matrix.
View full question & answer→MCQ 941 Mark
If the matrix is a square matrix and it contains 36 elements, then the order of the matrix is:
AnswerThe number of elements in a matrix are equal to the product of number of rows and columns.
As the matrix is a square matrix so that if matrix has 36 elements then the order of the matrix is 6 × 6.
Hence, the answer is 6 × 6.
View full question & answer→MCQ 951 Mark
$A = [a_{ij}]_{m \times n}$ is a square matrix, if:
- A
$m < n$
- B
$m > n$
- ✓
$m = n$
- D
AnswerCorrect option: C. $m = n$
For $A = [a_{ij}]_{m \times n}$ to be square matrix.
number of row $s =$ number of columns
$\therefore m = n$
$\therefore (c)$ is correct.
View full question & answer→MCQ 961 Mark
If a matrix has 13 elements, then the possible dimensions (orders) of the matrix are:
- ✓
$1\times13$ or $13\times1$
- B
$1\times26$ or $26\times1$
- C
$2\times13$ or $13\times2$
- D
$13\times13$
AnswerCorrect option: A. $1\times13$ or $13\times1$
If order of matrix $\text{A}=\text{a}\times\text{b}$
Then number of element in $\text{A}=\text{ab}$
Given $\text{ab}=13$
So, $\text{a}=1,\text{b}=13$
or $\text{b}=1,\text{a}=13$
So, $1\times13$ or $13\times1$ are possible order of $\text{A}$
View full question & answer→MCQ 971 Mark
Choose the correct answer from the given four options. On using elementary column operations $C_2 → C_2 – 2C_1$ in the following matrix equation $\begin{bmatrix}1&-3\\2&4\end{bmatrix}=\begin{bmatrix}1&-1\\0&1\end{bmatrix}\begin{bmatrix}3&1\\2&4\end{bmatrix},$ we have:
- A
$\begin{bmatrix}1&-5\\0&4\end{bmatrix}=\begin{bmatrix}1&-1\\-2&2\end{bmatrix}\begin{bmatrix}3&-5\\2&0\end{bmatrix}$
- B
$\begin{bmatrix}1&-5\\0&4\end{bmatrix}=\begin{bmatrix}1&-1\\0&1\end{bmatrix}\begin{bmatrix}3&-5\\-0&2\end{bmatrix}$
- C
$\begin{bmatrix}1&-5\\2&0\end{bmatrix}=\begin{bmatrix}1&-3\\0&1\end{bmatrix}\begin{bmatrix}3&1\\-2&4\end{bmatrix}$
- ✓
$\begin{bmatrix}1&-5\\2&0\end{bmatrix}=\begin{bmatrix}1&-1\\0&1\end{bmatrix}\begin{bmatrix}3&-5\\2&0\end{bmatrix}$
AnswerCorrect option: D. $\begin{bmatrix}1&-5\\2&0\end{bmatrix}=\begin{bmatrix}1&-1\\0&1\end{bmatrix}\begin{bmatrix}3&-5\\2&0\end{bmatrix}$
Given that, $\begin{bmatrix}1&-3\\2&4\end{bmatrix}=\begin{bmatrix}1&-1\\0&1\end{bmatrix}\begin{bmatrix}3&1\\2&4\end{bmatrix}$
On using $C_2 → C_2 - 2C_1;$ $\begin{bmatrix}1&-5\\2&0\end{bmatrix}=\begin{bmatrix}1&-1\\0&1\end{bmatrix}\begin{bmatrix}3&-5\\2&0\end{bmatrix}$
Since, on using elementary column operation on $X = AB,$ we apply these operations simultaneously on $X$ and on the second matrix $B$ of the product $AB$ on $\text{RHS}.$
View full question & answer→MCQ 981 Mark
If $A$ and $B$ are two matrices such that $AB = A$ and $BA = B,$ then $B^2$ is equal to:
AnswerHere, $AB = A ...(1)$
$BA = B ...(2)$
$\Rightarrow \text{BAB} = BB [$Multiplying both sides by $B]$
$\Rightarrow BA = B^2 [$From eq.$ (1)]$
$\Rightarrow B = B^2 [$From eq. $(2)]$
View full question & answer→MCQ 991 Mark
The matrix $\begin{bmatrix}0&5&-7\\-5&0&11\\7&-11&0\end{bmatrix}$ is:
- ✓
- B
- C
- D
An uppertriangular matrix.
AnswerHere,
$\text{A}=\begin{bmatrix}0&5&-7\\-5&0&11\\7&-11&0\end{bmatrix}$
$\Rightarrow\text{A}^\text{T}=\begin{bmatrix}0&-5&7\\5&0&-11\\-7&11&0\end{bmatrix}$
$\Rightarrow\text{A}^\text{T}=-\begin{bmatrix}0&5&-7\\-5&0&11\\7&-11&0\end{bmatrix}$
$\Rightarrow\text{A}^\text{T}=-\text{A}$
Thus, A is a skew-symmetric matrix.
View full question & answer→MCQ 1001 Mark
If every row of a matrix A contains p elements and its column contains q elements, then the order of A is:
Answer$\begin{bmatrix}\text{a}_{11} &\text{amp;}\text{ a}_{12} \\\text{a}_{21}& \text{amp;}\text{ a}_{22}\\\text{a}_{31}&\text{amp; }\text{a}_{32} \end{bmatrix}$
Hence order of $\text{A}$ is $3\times2$
Row contains pp elements
So number of columns $=\text{P}$
Each column contains $\text{q}:$ element
So number of rows $=\text{q}$
Therefore, order $=\text{q}\times\text{p}$
View full question & answer→MCQ 1011 Mark
The order of $\begin{bmatrix}\text{x}&\text{amp;}\text{ y}&\text{amp; }\text{z}\end{bmatrix}$ $\begin{bmatrix}\text{x} &\text{amp;}\text{ h}&\text{amp;}\text{ g} \\\text{h} &\text{amp;}\text{ b}&\text{amp; }\text{f}\\\text{g} &\text{amp;}\text{ f}&\text{amp; }\text{c} \end{bmatrix}\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}$ is:
- A
$3\times1$
- ✓
$1\times1$
- C
$1\times3$
- D
$3\times3$
AnswerCorrect option: B. $1\times1$
Let $\text{ABC}=\begin{bmatrix}\text{x}&\text{amp;}\text{ y}&\text{amp; }\text{z}\end{bmatrix}\begin{bmatrix}\text{x} &\text{amp;}\text{ h}&\text{amp;}\text{ g} \\\text{h} &\text{amp;}\text{ b}&\text{amp; }\text{f}\\\text{g} &\text{amp;}\text{ f}&\text{amp; }\text{c} \end{bmatrix}\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}$
Here, the order of $\text{A}$ is $1\times3$
Order of $\text{B}$ is $3\times3$
Since, matrix multiplication satisfies associative property
$\text{i}.\text{e}. (\text{AB})\text{C} = \text{A}(\text{BC})$
Hence, the order of $\text{AB}$ is $1\times3$
Hence, the order of $\text{ABC}$ is $1\times1$
View full question & answer→MCQ 1021 Mark
If $A$ is a square matrix such that $A^2 = A,$ then $(I + A)^3– 7A$ is equal to:
AnswerGiven: $A^2 = A...(i)$
Multiplying both sides by $A, A^3 = A^2= A [$From eq. $(i)]...(ii)$
Also given $(I + A)^3 – 7A = I^3+ A^3 + 3I^2A + 3IA^2 – 7A$
Putting $A^2 = A [$from eq. $(i)]$ and $A^3 = A [$From eq. $(ii)],$
$= I + A + 3IA + 3IA – 7A = I + A + 3A + 3A – 7A [\because IA = A]$
$= I + 7A – 7A = I$
Therefore, option $(C)$ is correct.
View full question & answer→MCQ 1031 Mark
Matrices A and B will be inverse of each other only if:
AnswerBy definition of inverse of square matrix,
Option (a) is correct.
View full question & answer→MCQ 1041 Mark
Choose the correct answer from the given four options. On using elementary row operation $R_1 \rightarrow R_1 – 3R_2$ in the following matrix equation $\begin{bmatrix}4&2\\3&3\end{bmatrix}=\begin{bmatrix}1&2\\0&3\end{bmatrix}\begin{bmatrix}2&0\\1&1\end{bmatrix},$ we have:
- ✓
$\begin{bmatrix}-5&-7\\3&3\end{bmatrix}=\begin{bmatrix}1&-7\\0&3\end{bmatrix}\begin{bmatrix}2&0\\1&1\end{bmatrix}$
- B
$\begin{bmatrix}-5&-7\\3&3\end{bmatrix}=\begin{bmatrix}1&2\\0&3\end{bmatrix}\begin{bmatrix}-1&-3\\1&1\end{bmatrix}$
- C
$\begin{bmatrix}-5&-7\\3&3\end{bmatrix}=\begin{bmatrix}1&2\\1&-7\end{bmatrix}\begin{bmatrix}2&0\\1&1\end{bmatrix}$
- D
$\begin{bmatrix}4&2\\-5&-7\end{bmatrix}=\begin{bmatrix}1&2\\-3&-3\end{bmatrix}\begin{bmatrix}2&0\\1&1\end{bmatrix}$
AnswerCorrect option: A. $\begin{bmatrix}-5&-7\\3&3\end{bmatrix}=\begin{bmatrix}1&-7\\0&3\end{bmatrix}\begin{bmatrix}2&0\\1&1\end{bmatrix}$
We have, $\begin{bmatrix}4&2\\3&3\end{bmatrix}=\begin{bmatrix}1&2\\0&3\end{bmatrix}\begin{bmatrix}2&0\\1&1\end{bmatrix}$
Using elementary row operation $R_1 → R_1 - 3R_2$
$\begin{bmatrix}-5&-7\\3&3\end{bmatrix}=\begin{bmatrix}1&-7\\0&3\end{bmatrix}\begin{bmatrix}2&0\\1&1\end{bmatrix}$
Since, on using elementary row operation on $X = AB,$ we apply these operation simultaneously on $X$ and on the first matrix $A$ of the product $AB$ on $\text{RHS}.$
View full question & answer→MCQ 1051 Mark
Choose the correct answer from the given four options. If $A$ is matrix of order $m \times n$ and $B$ is a matrix such that $AB\ '$ and $B\ 'A$ are both defined, then order of matrix $B$ is:
- A
$m \times m$
- B
$n \times n$
- C
$n \times m$
- ✓
$m \times n$
AnswerCorrect option: D. $m \times n$
Let $A = [a_{ij}]_{m\times n}$ and $B = [b_{ij}]_{p\times q}$
$B\ ' = [b_{ji}]_{q\times p}$
Now, $AB\ ’$ is defined, so $n = q$
and $B\ ’A$ is also defined, so $p = m$
$\therefore$ Order of $B\ ' = [b_{ji}]_{n\times m}$
And order of $B = B = [b_{ij}]m_{\times n}$
View full question & answer→MCQ 1061 Mark
The order of the matrix $ \displaystyle \left[ \begin{matrix} 1 &\text{amp; }2 \\ 3&\text{amp; } 4 \end{matrix} \right]$ is:
AnswerIf a matrix has mm rows and n columns then its order is m × n Clearly in the given matrix, number of rows and columns are each 2Hence its order is 2 × 2.
View full question & answer→MCQ 1071 Mark
A matrix consisting of a single column of m elements is know as:
MCQ 1081 Mark
If m[-3, 4] + n[4, -3] = [10, -11] then 3m + 7n = 3m + 7n =
Answerm[-3 amp; 4] + n[10 amp; -11] = [10 amp; -11]
[-3m + 4n amp; 4m − 3n] = [10 amp; -11]
−3m + 4n = 10 ⟹ 12m − 16n = −40 ........(1)
4m − 3n = −11 ⟹ 12m − 9n = -33 ........(2)
Solving equation 1 and 2, we
get, n = 1 and m= -2m = −2 Therefore, 3m + 7n = 3(-2) + 7 = -6 + 7 = 1
View full question & answer→MCQ 1091 Mark
The number of possible orders of a matrix containing $24$ elements are:
View full question & answer→MCQ 1101 Mark
If A = [1, 2, 3], then the set of elements of A is:
AnswerSince, $ \text{A}=\begin{bmatrix} 1 &\text{amp; } 2 &\text{amp; }3 \end{bmatrix}$ represents a matrix with three,
elements 1, 2, 3 $\therefore$ Elements of A = (1, 2, 3)
View full question & answer→MCQ 1111 Mark
Choose the correct answer from the given four options.
The matrix $\begin{bmatrix}1&0&0\\0&2&0\\0&0&4\end{bmatrix}$ is a:
AnswerWe have $\text{A}=\begin{bmatrix}1&0&0\\0&2&0\\0&0&4\end{bmatrix}$
$\therefore\ \text{A}'=\text{A}$
So, the given matrix is a symmetric matrix.
View full question & answer→MCQ 1121 Mark
Choose the correct answer from the given four options.
The matrix $\begin{bmatrix}0&-5&8\\5&0&12\\-8&-12&0\end{bmatrix}$ is a:
AnswerWe have $\text{B}=\begin{bmatrix}0&-5&8\\5&0&12\\-8&-12&0\end{bmatrix}$
$\Rightarrow\ \text{B}'=\begin{bmatrix}0&5&8\\-5&0&-12\\8&12&0\end{bmatrix}$
$=-\begin{bmatrix}0&-5&8\\5&0&12\\-8&-12&0\end{bmatrix}$
$=-\text{B}$
Since, B' = -B,
Thus, B is a skew-symmetric matrix.
View full question & answer→MCQ 1131 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:
Answer$\begin{bmatrix}2\text{x}+\text{y}&4\text{x}\\5\text{x}-7&4\text{x}\end{bmatrix}=\begin{bmatrix}7&7\text{y}-13\\\text{y}&\text{x}+6\end{bmatrix},$
Equating the terms, we get
4x = x + 6
⇒ x = 2
And
2x + y = 7
⇒ y = 3
View full question & answer→MCQ 1141 Mark
If $\text{AB}=\text{A}$ and $\text{BA = B}$ then $\text{B}^2 $ is equal to:
- ✓
$\text{B}$
- B
$\text{A}$
- C
$\text{-B}$
- D
$\text{B}^2$
AnswerCorrect option: A. $\text{B}$
We have, $\text{AB}=\text{A}$and $\text{BA = B}$
Since, $\text{B}^2=\text{B.B}$
$\text{B}^2=\text{(BA)}.\text{B}$
$\text{B}^2=\text{B}.\text{(AB)}$
$\text{B}^2=\text{B.A}$
$\text{B}^2=\text{B}$
Hence, this is the answer.
View full question & answer→MCQ 1151 Mark
What is the order of the product- $\begin{bmatrix}\text{x}&\text{amp;}\text{ y}&\text{amp;}\text{ z}\end{bmatrix}\begin{bmatrix}\text{a} &\text{amp;}\text{ h}&\text{amp;}\text{ g} \\\text{h} &\text{amp;}\text{ b}&\text{amp; }\text{f}\\\text{g} &\text{amp;}\text{ f}&\text{amp; }\text{c} \end{bmatrix}\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}$ is:
- A
$3\times1$
- ✓
$1\times1$
- C
$1\times3$
- D
$3\times3$
AnswerCorrect option: B. $1\times1$
If two matrix of order $\text{x}\times\text{m}$ and $\text{m}\times\text{n}$ are multiplied then the order of resultant matrix is $\text{x}\times\text{n}$
Now in the given equation going from left, matrices of order $1\times3$ and $3\times3$ are multiplied.So, the order of resultant matrix is $1\times3 $ And now this is multiplied by matrix of order $3\times1.$
This will give resultant matrix of order $1\times1$
View full question & answer→MCQ 1161 Mark
The matrix $\begin{bmatrix}0&\text{amp; }1\\1&\text{amp; }0\end{bmatrix}$ is the matrix reflection in the line:
AnswerWe know that the reflection matrix through a line $\text{y}=\text{mx}$ making an $\angle \theta$ with x - axis is given as.
$\begin{bmatrix} \cos { 2\theta }&\text{amp;}\sin { 2\theta } \\ \sin { 2\theta } &\text{amp;} -\cos { 2\theta }\end{bmatrix}$
Given transformation matrix is $\begin{bmatrix}0&\text{amp; }1\\1&\text{amp; }0\end{bmatrix}$
$\Rightarrow\cos2\theta=0\sin2\theta=1$
$\Rightarrow 2\theta ={90}^{0}$
$\Rightarrow \theta={45}^{0}$
$\Rightarrow \tan \theta=1$
Hence, the line of reflection is $\text{y}=\text{x}$
View full question & answer→MCQ 1171 Mark
The transpose of a square matrix is a?
AnswerThe transpose of square matrix is a new square matrix whose rows are.
the columns of original. this makes the columns the new square matrix row of the original. Answer is square matrix.
View full question & answer→MCQ 1181 Mark
If the order of a matrix is 20 × 5 then the number of elements in the matrix is _____?
AnswerAs the matrix has 20 rows and 5 columns, the number of elements in the matrix is 20 × 5 = 100
View full question & answer→MCQ 1191 Mark
If $\text{A} =\displaystyle \begin{bmatrix} -1 &\text{amp; } 0 &\text{amp; }0 \\ 0 &\text{amp; }\text{x} &\text{amp; } 0 \\ 0 &\text{amp; } 0 &\text{amp; }\text{m} \end{bmatrix}$is a scalar matrix then $\text{x}+\text{m}=$
AnswerA scalar matrix has all the elements of the diagonals same.For example:$ \begin{bmatrix} 3 &\text{amp; } 0\\ 0&\text{amp; } 3 \end{bmatrix}$
In our case A is given to be a scalar matrix hence all the diagonal elements must be same.
So, $\text{x} = \text{m} = -1$
And $\text{x}+\text{m} = -1 -1 = -2$
View full question & answer→MCQ 1201 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 1211 Mark
If $\text{A}=\begin{bmatrix}\text{a}&\text{b}\\\text{b}&\text{a}\end{bmatrix}$ and $\text{A}^2=\begin{bmatrix}\alpha&\beta\\\beta&\alpha\end{bmatrix},$ then:
- A
$\alpha=\text{a}^2+\text{b}^2,\beta=\text{ab}$
- ✓
$\alpha=\text{a}^2+\text{b}^2,\beta=2\text{ab}$
- C
$\alpha=\text{a}^2+\text{b}^2,\beta=\text{a}^2-\text{b}^2$
- D
$\alpha=2\text{ab},\beta=\text{a}^2+\text{b}^2$
AnswerCorrect option: B. $\alpha=\text{a}^2+\text{b}^2,\beta=2\text{ab}$
View full question & answer→MCQ 1221 Mark
If A and B are two matrices such that A + B and AB are both defined, then
- A
A and B can be any matrices
- B
A, B are square matrices not necessarily of the same order
- ✓
A, B are square matrices of the same order
- D
Number of columns of A = Number of rows of B
AnswerCorrect option: C. A, B are square matrices of the same order
Let A and B both have a matrices of order m × n
Also AB is defined, it means m = n
Hence A and B are square matrices of same order
View full question & answer→MCQ 1231 Mark
If $A = \begin{bmatrix}\alpha&\beta\\ \gamma&-\alpha\end{bmatrix}$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: $\text{A}=\begin{bmatrix}\alpha&\beta\\ \gamma&-\alpha\end{bmatrix} \text{and}\ \text{A}^{2}=\text{I}$
$\Rightarrow\ \begin{bmatrix}\alpha&\beta\\ \gamma&-\alpha\end{bmatrix}\begin{bmatrix}\alpha&\beta\\ \gamma&-\alpha\end{bmatrix}=\begin{bmatrix}1&0\\0&1\end{bmatrix}$
$\Rightarrow\ \begin{bmatrix}\alpha^{2}+\beta\gamma&\alpha\beta-\alpha\beta\\ \alpha\gamma-\gamma\alpha&\beta\gamma+\alpha^{2}\end{bmatrix}=\begin{bmatrix}1&0\\0&1\end{bmatrix}$
$\Rightarrow\ \begin{bmatrix}\alpha^{2}+\beta\gamma&0\\0&\beta\gamma+\alpha^{2}\end{bmatrix}=\begin{bmatrix}1&0\\0&1\end{bmatrix}$
Equating corresponding entries, we have
$\alpha^{2}+\beta\gamma=1$
$\Rightarrow 1-\alpha^{2}-\beta\gamma=0$
Therefore, option $(C)$ is correct.
View full question & answer→MCQ 1241 Mark
The matrix$ \displaystyle \begin{bmatrix}-12\\10 \\13 \\4 \end{bmatrix}$ is a:
AnswerMatrix $ \displaystyle \begin{bmatrix}-12\\10 \\13 \\4 \end{bmatrix}$ is a column matrix.Hence, the answer is column matrix.
View full question & answer→MCQ 1251 Mark
If $\text{A}=\begin{bmatrix}3&\text{x}-1\\2\text{x}+3&\text{x}+2\end{bmatrix}$ is a symmetric matrix, then $x =$
View full question & answer→MCQ 1261 Mark
Choose the correct answer from the given four options.The matrix $\text{P}=\begin{bmatrix}0&0&4\\0&4&0\\4&0&0\end{bmatrix}$ is a:
AnswerWe know that, in a square matrix number of rows are equal to the number of columns.
So, the matrix $\text{P}=\begin{bmatrix}0&0&4\\0&4&0\\4&0&0\end{bmatrix}$ is a square matrix.
View full question & answer→MCQ 1271 Mark
The order of any matrix is 3 × 2 then no.of element in the matrix:
AnswerOrder of matrix is 3 × 2, then number of elements in the matrix is 6.
Hence, the answer is 6.
View full question & answer→MCQ 1281 Mark
The transpose of a row matrix is:
AnswerTranspose of row matrix Let $ \text{A}=\begin{bmatrix}\text{x} &\text{amp; y} &\text{amp; z} \end{bmatrix}$ be a row
matrix $\text{A}^\text{T}=\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}$Clearly $\text{A}^\text{T}$ is a column matrix $\therefore$ Transpose of row.
View full question & answer→MCQ 1291 Mark
If order of A + B is n × n, then the order of AB is:
AnswerIf order of $\text{A}+\text{B}$ is $\text{n}\times\text{n},$ then the order of both$\begin{bmatrix}\text{A}&\text{amp;}\text{ B}\end{bmatrix}$ is $\text{n}\times\text{n}$
Therefore, order of $\text {AB }$is $\text{n}\times\text{n}$
View full question & answer→MCQ 1301 Mark
The possibility for the formation of rectangular matrices in the matrix algebra are?
- ✓
rows greater than columns
- B
- C
rows greater than column by 2 times
- D
AnswerCorrect option: A. rows greater than columns
The possibilities of formation of rectangular matrix are the following:
(1) Rows are greater then columns.
(2) Columns are greater then rows.
(3) Rows greater then column by 2 times.
View full question & answer→MCQ 1311 Mark
If $\displaystyle \text{a}_{\text{ij}}=0\left (\text{i}\neq \text{j} \right )$ and $\displaystyle \text{a}_{\text{ij}}=1\left (\text{i}= \text{j} \right )$ then the matrix $\text{A}=\displaystyle \left [\text{a}_{\text{ij}} \right ]_{\text{n}\times\text{n}}$ is a _____ matrix:
AnswerThe elements $\text{a}_\text{ij}$ of a matrix where i = j lie along its diagonal and
the elements $\text{a}_\text{ij}$ of a matrix where $\text{i}\neq\text{j}$ are not along the diagonal.
As the diagonal elements are 11 and the rest of the elements are 0, the matrix A is an identity matrix.
View full question & answer→MCQ 1321 Mark
If $\text{A}=\displaystyle \begin{vmatrix} 1 \\ 3 \end{vmatrix}\text{B}=\displaystyle \begin{vmatrix} -1 \\ 4 \end{vmatrix}$ then $ 2\text{A}+\text{B} =$
- A
$\displaystyle \begin{vmatrix} 10 \\ 9 \end{vmatrix}$
- B
$\displaystyle \begin{vmatrix} 10 \\ 1 \end{vmatrix}$
- ✓
$\displaystyle \begin{vmatrix} 1 \\ 10 \end{vmatrix}$
- D
$\displaystyle \begin{vmatrix} 1 \\ 9 \end{vmatrix}$
AnswerCorrect option: C. $\displaystyle \begin{vmatrix} 1 \\ 10 \end{vmatrix}$
$2\text{A+B}=|26|$
View full question & answer→MCQ 1331 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,
$AB = A ...(1)$
$BA = B ...(2)$
$\Rightarrow \text{ABA} = AA \ [$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 1341 Mark
The number of possible matrices of order $3×3$ 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 1351 Mark
If $\text{A} = \begin{bmatrix}1\end{bmatrix}$ then the order of the matrix is:
AnswerSince, given matrix contain a single element means it contain one row and one column.
View full question & answer→MCQ 1361 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:
- A
$m \times m$
- B
$n \times n$
- C
$n \times m$
- ✓
$m \times n$
AnswerCorrect option: D. $m \times n$
Given $A$ is a matrix of order $m \times m\times n$ and $B$ is a matrix such that $AB^T$ and $B^TA$ are both defined.
Since $AB^T$ is defined then number of columns of $A$ must be equal to number of rows of $B^T.$
So number of rows in $B^T$ is $n$.
This gives number of columns in $B$ is $n$.
Again since $ AB^TA$ is defined then number of columns of $B^T$ is equal to the number of rows of $A$.
So number of columns of $B^T$is m this gives the number of rows of $B$ is m.
So order of $B$ is $m \times n.$
View full question & answer→MCQ 1371 Mark
If $\text{A}=\displaystyle \begin{vmatrix} 2 &\text{amp;}-3 \\ 3 &\text{amp; 2} \end{vmatrix}$ and $\text{B}=\displaystyle \begin{vmatrix} 3 &\text{amp;}-2 \\ 2 &\text{amp; 3} \end{vmatrix}$ then $2\text{ A-B}=$
- A
$\displaystyle \begin{vmatrix} 1 &4 \\ 4 &1\end{vmatrix}$
- B
$\displaystyle \begin{vmatrix} 1 &4 \\ 1 &4\end{vmatrix}$
- ✓
$\displaystyle \begin{vmatrix} 1 &-4 \\ 4 &1\end{vmatrix}$
- D
$\displaystyle \begin{vmatrix} 4 &1\\ 1 &4\end{vmatrix}$
AnswerCorrect option: C. $\displaystyle \begin{vmatrix} 1 &-4 \\ 4 &1\end{vmatrix}$
$2\text{ A-B}=\displaystyle \begin{vmatrix} 4 &\text{amp;}-6 \\ 6 &\text{amp; 4} \end{vmatrix}-\displaystyle \begin{vmatrix} 3 &\text{amp;}-2 \\ 2 &\text{amp; 3} \end{vmatrix}=\displaystyle \begin{vmatrix} 1 &\text{amp;}-4\\ 4 &\text{amp; 1} \end{vmatrix}$
View full question & answer→MCQ 1381 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 1391 Mark
The number of all possible matrices of order $3 \times 3$ with each entry $0$ or $1$ is:
AnswerA matrix of order $3 \times 3$ has $9$ elements.
Now each elements can be $0$ or $1$ .
$\therefore 9$ places can be filled up in $2^9$ ways
required number of matrices $=2^9$
$=2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 $
$=512$
$\therefore(d)$ is correct answer.
View full question & answer→MCQ 1401 Mark
If $A$ is any square matrix, then which of the following is skew $-$ symmetric?
- A
$A + A^T$
- ✓
$A - A^T$
- C
$AA^T$
- D
$A^TA$
AnswerCorrect option: B. $A - A^T$
$A - A^T$
View full question & answer→MCQ 1411 Mark
If the matrix A is both symmetric and skew symmetric, then:
AnswerSince, A is symmetric, therefore, A’ = A ...(i)
And A is skew-symmetric, therefore, A’ = –A ⇒ A = –A [From eq. (i)]
⇒ A + A = 0 ⇒ 2A = 0 ⇒ A = 0
Therefore, A is zero matrix.
Therefore, option (B) is correct.
View full question & answer→MCQ 1421 Mark
Let $A$ is a square matrix of order $n$ and a being a scalar then $∣aA∣ =$
- A
$a∣A∣$
- B
$∣a∣∣A∣$
- ✓
$a^n∣A∣$
- D
AnswerCorrect option: C. $a^n∣A∣$
Given, $A$ is a square matrix of order $n$ and $a$ being a scalar.
Now $A$ is the matrix in which each elements of $A$ is multiplied by a.
So when we take determinant of $A$ then form each row or column a will be common.
Then $∣aA∣ = a^n∣A∣.$
View full question & answer→MCQ 1431 Mark
If $A$ and $B$ are square matrices of order $n \times n$ such that, $A^2− B^2= (A − B)(A + B),$ then of the following will always be true?
AnswerCorrect option: B. $AB = BA$
$A^2− B^2= (A − B)(A + B) \rightarrow A^2− B^2$
$= A^2− BA + AB − B^2\rightarrow BA = AB$
View full question & answer→MCQ 1441 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 1451 Mark
If $\begin{bmatrix}2&\text{amp; }3\\4&\text{amp; }4\end{bmatrix}+\begin{bmatrix}\text{x}&\text{amp; }3\\\text{y}&\text{amp; }1\end{bmatrix}=\begin{bmatrix}10&\text{amp; }6\\8&\text{amp; }5\end{bmatrix}$ then $(\text{x, y})=$
Answer2 + x = 10 or x = 8
4 + y = 8 or y = 4
View full question & answer→MCQ 1461 Mark
The restriction on n, k and p so that PY + WY will be define are:
AnswerGiven: $ \text{x}_{2\times n}, \text{y}_{3\times k},\text{ z}_{2 \times p}, \text{w}_{n \times 3}, \text{p}_{p\times k}$
Now, $\text{py + wy} = \text{p}_{p \times k}\times \text{y}_{3\times k}+\text{w}_{n\times3}\times\text{y}_{3 \times k}$
On comparing, k = 3 and p = n
Therefore, option (A) is correct.
View full question & answer→MCQ 1471 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 1481 Mark
If a matrix P has 8 elements then how many different values the order of the matrix can take?
AnswerA matrix of mm rows and n columns has m × n elements.
8 can be got by all combinations of 1 × 8, 8 × 1, 2 × 4, 4 × 2
Hence, there are 4 possible matrices which have 8 elements.
View full question & answer→MCQ 1491 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 1501 Mark
If A and B are two matrices of order 3×m and 3×n respectively and m = n, then the order of 5A - 2B is:
AnswerIn scalar multiplicaion and in addition or substraction of matrics the order doesn't change.
View full question & answer→MCQ 1511 Mark
If $\text{A}=\begin{bmatrix}a^2 &\text{amp; ab}&\text{amp; ac} \\\text{ab}&\text{amp; }\text{b}^2&\text{amp;}\text{ bc}\\\text{ac}&\text{amp;}\text{bc}&\text{amp;}\text{c}^2 \end{bmatrix}$and $\text{a}^2+\text{b}^2+\text{c}^3=1$ then $\text{A}^2=$
- A
$2\text{A}$
- ✓
$\text{A}$
- C
$3\text{A}$
- D
$\frac{1}{2}\text{A}$
AnswerCorrect option: B. $\text{A}$
$\text{A}^2=\begin{bmatrix}\text{a}^2 &\text{amp; ab}&\text{amp; ac} \\\text{ab}&\text{amp; }\text{b}^2&\text{amp;}\text{ bc}\\\text{ac}&\text{amp;}\text{bc}&\text{amp;}\text{c}^2 \end{bmatrix}\begin{bmatrix}\text{a}^2 &\text{amp; ab}&\text{amp; ac} \\\text{ab}&\text{amp; }\text{b}^2&\text{amp;}\text{ bc}\\\text{ac}&\text{amp;}\text{bc}&\text{amp;}\text{c}^2 \end{bmatrix}=\text{A}$
View full question & answer→MCQ 1521 Mark
The order the matrix is $\begin{bmatrix}2&\text{amp; }3&\text{amp; }4\\9&\text{amp; }8&\text{amp; }7\end{bmatrix}$ is :
- A
$4 \times 3$
- B
$3 \times 2$
- ✓
$2 \times 3$
- D
$3 \times 1$
AnswerCorrect option: C. $2 \times 3$
If $A$ is a matrix with mm rows and $n$ columns.
Then the order of a matrix is nothing but a size of a matrix, which is given by $m \times n.$
Since, in the given matrix, there are $2$ rows and $3$ columns.
So, order of given matrix will be $2 \times 3$.
View full question & answer→MCQ 1531 Mark
If $\begin{bmatrix}\text{r}+4&\text{amp; 6}\\3&\text{amp; 3}\end{bmatrix}=\begin{bmatrix}{5}&\text{amp;}\text{ r}+5\\\text{r+2}&\text{amp; 4}\end{bmatrix}$ then $\text{r}=$
AnswerWe know that two matrices are equal iff their corresponding elements are equal.
Thus comparing corresponding elements we get, for the first entry of.
the given matrices r + 4 = 5 and r is satisfying other equations which are involving r ⇒ r = 1
View full question & answer→MCQ 1541 Mark
If a matrix is of order $2 \times 3,$ then the number of elements in the matrix is:
AnswerGiven a matrix $2\times3$
$\Rightarrow \begin{bmatrix} { \text{a} }_{11} &\text{amp; } {\text{a} }_{12} &\text{amp; } { \text{a} }_{13} \\ { \text{a} }_{21} &\text{amp; } {\text{a} }_{22} &\text{amp; } {\text{a} }_{23} \end{bmatrix}$
Clearly there are $6 $ elements.
View full question & answer→MCQ 1551 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 1561 Mark
Choose the correct answer from the given four options.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 + y is:
AnswerWe have, $\begin{bmatrix}2\text{x}+\text{y}&4\text{x}\\5\text{x}-7&4\text{x}\end{bmatrix}=\begin{bmatrix}7&7\text{y}-13\\\text{y}&\text{x}+6\end{bmatrix}$
⇒ 4x = x + 6 ⇒ x = 2
and 4x = 7y - 13
⇒ 8 = 7y - 13
⇒ y = 3
$\therefore$ x + y = 2 + 3 = 5
View full question & answer→MCQ 1571 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.
$\therefore\ \text{y}+\text{x}=0$
View full question & answer→MCQ 1581 Mark
Choose the correct answer from the given four options.If $\text{A}=\frac{1}{\pi}\begin{bmatrix}\sin^{-1}(\text{x}\pi)&\tan^{-1}\Big(\frac{\text{x}}{\pi}\Big)\\\sin^{-1}\Big(\frac{\text{x}}{\pi}\Big)&\cot^{-1}(\pi\text{x})\end{bmatrix}$ and $\text{B}=\frac{1}{\pi}\begin{bmatrix}-\cos^{-1}(\text{x}\pi)&\tan^{-1}\Big(\frac{\text{x}}{\pi}\Big)\\\sin^{-1}\Big(\frac{\text{x}}{\pi}\Big)&\tan^{-1}(\pi\text{x})\end{bmatrix}$ then A - B is:
- A
$\text{I}$
- B
$0$
- C
$2\text{I}$
- ✓
$\frac{1}{2}\text{I}$
AnswerCorrect option: D. $\frac{1}{2}\text{I}$
We have, $\text{B}=\begin{bmatrix}-\frac{1}{\pi}\cos^{-1}\text{x}\pi&\frac{1}{\pi}\tan^{-1}\frac{\text{x}}{\pi}\\\frac{1}{\pi}\sin^{-1}\frac{\text{x}}{\pi}&-\frac{1}{\pi}\tan^{-1}\pi\text{x}\end{bmatrix}$
and $\text{A}=\begin{bmatrix}\frac{1}{\pi}\sin^{-1}\text{x}\pi&\frac{1}{\pi}\tan^{-1}\frac{\text{x}}{\pi}\\\frac{1}{\pi}\sin^{-1}\frac{\text{x}}{\pi}&\frac{1}{\pi}\cot^{-1}\pi\text{x}\end{bmatrix}$
$\therefore\ \text{A}-\text{B}=\begin{bmatrix}\frac{1}{\pi}(\sin^{-1}\text{x}\pi+\cos^{-1}\text{x}\pi)&0\\0&\frac{1}{\pi}\big(\cot^{-1}\text{x}\pi+\tan^{-1}\pi\text{x}\big)\end{bmatrix}$
$=\begin{bmatrix}\frac{1}{\pi}\Big(\frac{\pi}{2}\Big)&0\\0&\frac{1}{\pi}\Big(\frac{\pi}{2}\Big)\end{bmatrix}$
$=\frac{1}{2}\begin{bmatrix}1&0\\0&1\end{bmatrix}$
$=\frac{1}{2}\text{I}$
View full question & answer→MCQ 1591 Mark
The total number of matrices formed with the help of 6 different numbers are:
AnswerNo.of numbers in Matrix is 6
The possible orientations of Matrix is.
1 × 6, 2 × 3, 3 × 2, 6 × 1
The numbers in each orientation can be arranged in 6! ways.
$\implies$The total possibilities are 4(6!).
View full question & answer→MCQ 1601 Mark
If order of a matrix is 3 × 3, then it is a?
AnswerSince, order of given matrix is 3 × 3.
$\therefore$ No of rows = No. of columns
So, given matrix is a square matrix.
View full question & answer→MCQ 1611 Mark
A matrix having mm rows and nn columns with m = n is said to be a?
AnswerA matrix having mm rows and nn columns with m = n, means number of rows are equal to number of columns.
$\therefore$ given matrix is square matrix.
View full question & answer→