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.
$\therefore\ \begin{bmatrix}5&\text{x}\\\text{y}&0\end{bmatrix}=\begin{bmatrix}5&\text{y}\\\text{x}&0\end{bmatrix}$
$\Rightarrow\text{x}=\text{y}$
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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.
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