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 $(1+A)^3-7 A$ is equal to:
AnswerGiven: $A^2=A \ldots (i)$
Multiplying both sides by $A, A^3=A^2=A[$ From eq. (i) $] \ldots (ii)$
Also given $(I+A)^3-7 A=I^3+A^3+3 I^2 A+3 I A^2-7 A$
Putting $A^2=A\left[\right.$ from eq. (i)] and $A^3=A[$ from eq. $(ii)],$
$=I+A+3 I A+3 I A-7 A=I+A+3 A+3 A-7 A[\because I A=A]$
$=I+7 A-7 A=I$
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 $R_1 \rightarrow R_1-3 R_2$ 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 $R_1 \rightarrow R_1-3 R_2$
$\begin{bmatrix}-5&-7\\3&3\end{bmatrix}=\begin{bmatrix}1&-7\\0&3\end{bmatrix}\begin{bmatrix}2&0\\1&1\end{bmatrix}$
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 $A=\left[a_{i j}\right]_{m \times n}$ and $B=\left[b_{i j}\right]_{p \times q}$
$B^{\prime}=\left[b_{ ji }\right]_{q \times p}$
Now, $A B^{\prime}$ is defined, so $n=q$
and $B^{\prime} A$ is also defined, so $p=m$
$\therefore$ Order of $B^{\prime}=\left[b_{j j}\right]_{n \times m}$
And order of $B=B=\left[b_{i j}\right] m_{\times n}$
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:
- ✓
$2 \times 2$
- B
$4 \times 1$
- C
$1 \times 4$
- D
AnswerCorrect option: A. $2 \times 2$
If a matrix has mm rows and n columns then its order is $m \times n$ Clearly in the given matrix, number of rows and columns are each $2.$
Hence its order is $2 \times 2.$
View full question & answer→MCQ 1071 Mark
A matrix consisting of a single column of $m$ elements is know as:
View full question & answer→MCQ 1081 Mark
If $m[-3, 4] + n[4, -3] = [10, -11]$ then $3m + 7n = 3m + 7n =$
Answer$m[-3\ \text{amp} ; 4] + n[10\ \text{amp}; -11] = [10 \ \text{amp} ; -11]$
$[-3m + 4n \ \text{amp} ; 4m − 3n] = [10 \ \text{amp} ; -11]$
$−3m + 4n = 10 $
$\Rightarrow 12m − 16n = −40 ........(1)$
$4m − 3n = −11 $
$\Rightarrow12m − 9n = -33 ........(2)$
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 :
- ✓
$[1, 2, 3]$
- B
$[2, 0]$
- C
Only $2$
- D
AnswerCorrect option: A. $[1, 2, 3]$
Since, $ \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:
- A
- ✓
- C
Skew$-$symmetric matrix.
- D
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:
- A
- B
- ✓
Skew$-$symmetric matrix.
- D
AnswerCorrect option: C. Skew$-$symmetric matrix.
We 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 :
- 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 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 :
- A
$x = 1$
- B
$x + y = 1$
- C
$y = 1$
- ✓
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 $\begin{bmatrix}0&\text{amp; }1\\1&\text{amp; }0\end{bmatrix}$
$\Rightarrow\cos2\theta=0\sin2\theta=1$
$\Rightarrow 2\theta ={90}^\circ$
$\Rightarrow \theta={45}^\circ$
$\Rightarrow \tan \theta=1$
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 $20 \times 5$ 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 $20 \times 5 = 100$
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 $m \times n$
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 $A ^2=1$ 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 $3 \times 2$ then no.of element in the matrix:
AnswerOrder of matrix is $3 \times 2,$ 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 $n \times n,$ then the order of $AB$ is :
- ✓
$n \times n$
- B
$n \times m$
- C
$m \times n$
- D
AnswerCorrect option: A. $n \times n$
If 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?
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 :
- ✓
$B^2=B$ and $A^2=A$
- B
$B^2 \neq B$ and $A^2=A$
- C
$A^2 \neq A, B^2=B$
- D
$A^2 \neq A, B^2 \neq B$
AnswerCorrect option: A. $B^2=B$ and $A^2=A$
Here,
$A B=A \cdot .(1)$
$B A=B \ldots(2)$
$\Rightarrow A B A=A A\ [$ Multiplying both sides by $A]$
$B A B=B B\ [$ Multiplying both sides by $ A]$
$\Rightarrow A B=A^2\ [$ From eq. $(2)]$
$B A=B^2\ [$ 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 $3 \times 3$ 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 1351 Mark
If $\text{A} = \begin{bmatrix}1\end{bmatrix}$ then the order of the matrix is:
- ✓
$1 \times 1$
- B
$2 \times 1$
- C
$1 \times 2$
- D
AnswerCorrect option: A. $1 \times 1$
Since, 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 $A B^{\top}$ and $B^{\top} A$ 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 $A B^{\top}$ and $B^{\top} A$ are both defined.
Since $A B^{\top}$ is defined then number of columns of $A$ must be equal to number of rows of $B^{\top}$.
So number of rows in $B^{\top}$ is $n$.
This gives number of columns in $B$ is $n$.
Again since $A B^{\top} A$ is defined then number of columns of $B^{\top}$ is equal to the number of rows of $A$.
So number of columns of $B^{\top}$ 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 :
Answer"margin $-$ left : $40px\ " > A$ 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
$A+A^{\top}$
- ✓
$A-A^{\top}$
- C
$A A^{\top}$
- D
$A^{\top} A$
AnswerCorrect option: B. $A-A^{\top}$
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
$∣a∣∣A∣$
- ✓
$a^{\text {n }}| A |$
- D
AnswerCorrect option: C. $a^{\text {n }}| A |$
Given, $A$ is a square matrix of order $n$ and $a$ being a scalar.
Now $aA$ is the matrix in which each elements of $A$ is multiplied by $a$.
So when we take determinant of $aA$ then form each row or column a will be common.
Then $ ∣aA∣ = a^{\text {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. $AB = BA$
$A^2-B^2=(A-B)(A+B) \rightarrow A^2-B^2$
$=A^2-B A+A B-B^2 \rightarrow B A=A B$
View full question & answer→MCQ 1441 Mark
If $\text{A}=\begin{bmatrix}5&\text{x}\\\text{y}&0\end{bmatrix}$ and $A = A ^{\top}$, 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 $(\text{x, y})=$
- A
$(4, 8)$
- ✓
$(8, 4)$
- C
$(1, 2)$
- D
AnswerCorrect option: B. $(8, 4)$
$2 + x = 10$ 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 $m \times n$ elements.
$8$ can be got by all combinations of $1 \times 8, 8 \times 1, 2 \times 4, 4 \times 2$
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 $A^2=1$, then $(A-1)^3+(A+1)^3-7 A$ is equal to :
- ✓
$A$
- B
$I - A$
- C
$I + A$
- D
$3A$
Answer$(A-I)^3+(A+I)^3-7 A$
$=A^3-I^3-3 A^2 I+3 A I^2+A^3+I^3+3 A^2 I+3 A I^2-7 A$
$=2 A^3+6 A I^2-7 A$
$=2 A \cdot A^2+6 A-7 A$
$=2 A \cdot I-A\left(\because A^2=1\right)$
$=2 A-A$
$=A$
Hence, the correct option is $(a).$
View full question & answer→MCQ 1501 Mark
If $A$ and $B$ are two matrices of order $3\times m$ 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→