MCQ 11 Mark
If $\left[\begin{array}{ll}2 & 0 \\ 0 & 4\end{array}\right]\left[\begin{array}{l}x \\ y\end{array}\right]=\left[\begin{array}{c}2 \\ -8\end{array}\right]$, the value of $x$ and $y$ respectively are:
AnswerCorrect option: A. $1,-2$
(a) $1,-2$
Explanation:
$\begin{array}{l}{\left[\begin{array}{ll}2 & 0 \\ 0 & 4\end{array}\right]\left[\begin{array}{l}x \\ y\end{array}\right]=\left[\begin{array}{r}2 \\ -8\end{array}\right]} \\ {\left[\begin{array}{l}2 x+0 y \\ 0 x+4 y\end{array}\right]=\left[\begin{array}{r}2 \\ -8\end{array}\right]} \\ {\left[\begin{array}{l}2 x \\ 4 y\end{array}\right]=\left[\begin{array}{r}2 \\ -8\end{array}\right]}\end{array}$
On equating the corresponding terms,
$2 x=2 \Rightarrow x=1$
and $4 y=-8 \Rightarrow y=-2$
$(1,-2) \text { Ans. }$
View full question & answer→MCQ 21 Mark
If matrix $A=\left[\begin{array}{ll}2 & 2 \\ 0 & 2\end{array}\right]$ and $A^2=\left[\begin{array}{ll}4 & x \\ 0 & 4\end{array}\right]$, then thevalue of $x$ is:
Answer(c) 8
Explanation:
Given, matrix $A =\left[\begin{array}{ll}2 & 2 \\ 0 & 2\end{array}\right]$.
Then $A ^2=\left[\begin{array}{lll}2 & 2 \\ 0 & 2\end{array}\right]\left[\begin{array}{ll}2 & 2 \\ 0 & 2\end{array}\right]-\left[\begin{array}{ll}4 & 8 \\ 0 & 4\end{array}\right]$
On comparing it with $A ^2=\left[\begin{array}{ll}4 & x \\ 0 & 4\end{array}\right]$, we get that
$x=8$.
View full question & answer→MCQ 31 Mark
Which of the following is/are correct?
Statement (A): If $A=\left[\begin{array}{rr}3 & 1 \\ -1 & 2\end{array}\right]$, then $A ^2=\left[\begin{array}{rr}8 & -5 \\ 5 & 3\end{array}\right]$
Statement (B): If $A=\left[\begin{array}{rr}3 & 1 \\ -1 & 2\end{array}\right], B=\left[\begin{array}{ll}5 & x \\ 1 & 0\end{array}\right]$ and $A^2=B$, then the value of $x$ is 5 .
Statement (C): For matrix multiplication, the number of columns in the first matrix must be equal to the number of rows in the second matrix.
- A
- ✓
- C
- D
All A, B and C are correct
Answer(b) Only B and C are correct
Explanation:
$\begin{array}{l}A=\left[\begin{array}{rr}3 & 1 \\ -1 & 2\end{array}\right]_{2 \times 2} \\ A^2=\left[\begin{array}{rr}3 & 1 \\ -1 & 2\end{array}\right]_{2 \times 2}\left[\begin{array}{rr}3 & 1 \\ -1 & 2\end{array}\right]_{2 \times 2}=\left[\begin{array}{rr}8 & 5 \\ -5 & 3\end{array}\right]\end{array}$
Statement A is incorrect
$\ldots A ^2= B$
$\left[\begin{array}{rr}8 & 5 \\ -5 & 3\end{array}\right]=\left[\begin{array}{ll}5 & x \\ 1 & 0\end{array}\right]$
x=5
Statement B and C are correct
View full question & answer→MCQ 41 Mark
Which of the following is/are correct?
Statement (A): A row matrix has only one row and multiple columns.
Statement (B): A column matrix has only one column and multiple rows.
Statement (C): The addition of a row matrix with a column matrix is always possible.
- ✓
- B
- C
- D
All A, B and C are correct
Answer(a) Only A and B are correct
Explanation:
Statement A and B are correct but statement C is incorrect because the addition of a row matrix and a column matrix is not possible as they have different dimensions and matrix addtion requires matrices of the same dimensions.
View full question & answer→MCQ 51 Mark
Statement (A): Adding the null matrix $\left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right]$ to any $2 \times 2$ matrix $\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]$ will result in the original matrix $\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]$.
Statement (B): Multiplying any $2 \times 2$ matrix by the identity matrix $\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$ will change thevalues of the elements in the original matrix.
Which of the statement is valid?
Answer(a) Only A
Explanation
For statement A,
If $\left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right]+\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]=\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]$
Hence, adding the null matrix to any $2 \times 2$ matrix will result in the original matrix.
Statement A is correct.
For statement B,
If, $\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]=\left[\begin{array}{ll}a+0 & b+0 \\ 0+c & 0+d\end{array}\right]=\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]$
Hence, Multiplying any $2 \times 2$ matrix by the identity matrix $\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$ will result in the original matrix.
View full question & answer→MCQ 61 Mark
Statement (A): The matrices $A=\left[\begin{array}{rr}2 & 5 \\ 7 & -4\end{array}\right]$ and $B=\left[\begin{array}{rr}2 & 5 \\ 7 & -4\end{array}\right]$ are equal.
Statement (B): The Matrices $A=\left[\begin{array}{rrr}2 & 3 & 0 \\ 7 & -6 & 5\end{array}\right]$ and $B=\left[\begin{array}{rrr}2 & 0 & 3 \\ 7 & -6 & 5\end{array}\right]$ are equal.
Which of the statement is valid?
Answer(a) Only A
Explanation
For statement A,
The matrices $A=\left[\begin{array}{rr}2 & 5 \\ 7 & -4\end{array}\right]$ and $B=\left[\begin{array}{rr}2 & 5 \\ 7 & -4\end{array}\right]$ are equal, because both are of the same order 2 $\times 2$ and their corresponding entries are equal.
Hence, Statement A is correct
For statement B,
The matrice $A=\left[\begin{array}{rrr}2 & 3 & 0 \\ 7 & -6 & 5\end{array}\right]$ and $B=\left[\begin{array}{rrr}2 & 0 & 3 \\ 7 & -6 & 5\end{array}\right]$ are not equal, because (1, 2)th entry of $A \neq$ $(1,2)$ th entry of $B$, even though both matrices $A$ and $B$ one of the same order $2 \times 3$.
Hence, statement B is incorrect.
View full question & answer→MCQ 71 Mark
If $\left[\begin{array}{cc}x & 0 \\ -3 & 1\end{array}\right]\left[\begin{array}{ll}1 & 1 \\ 0 & y\end{array}\right]=\left[\begin{array}{rr}2 & 2 \\ -3 & -2\end{array}\right]$ then value of $x$ and $y$ is:
- A
$x=1, y=2$
- B
$x=-1, y=2$
- ✓
$x=2, y=1$
- D
$x=2, y=-1$
AnswerCorrect option: C. $x=2, y=1$
(c) $x=2, y=1$
Explanation:
$\begin{array}{l}\text { Here, }\left[\begin{array}{cc}x & 0 \\ -3 & 1\end{array}\right]\left[\begin{array}{ll}1 & 1 \\ 0 & y\end{array}\right]=\left[\begin{array}{rr}2 & 2 \\ -3 & -2\end{array}\right] \\ \Rightarrow\left[\begin{array}{cc}x & x \\ -3 & -3+y\end{array}\right]=\left[\begin{array}{cc}2 & 2 \\ -3 & -2\end{array}\right] \\ \Rightarrow x=2 \text { and }-3+y=-2 \\ \Rightarrow x=2 \text { and } y=1\end{array}$
View full question & answer→MCQ 81 Mark
The value of $\left[\begin{array}{cc}\cot 45^{\circ} & \operatorname{cosec} 30^{\circ} \\ \sec 60^{\circ} & \sin 90^{\circ}\end{array}\right]\left[\begin{array}{cc}2 \cos 60^{\circ} & -2 \sin 30^{\circ} \\ \tan 45^{\circ} & \cos 0^{\circ}\end{array}\right]$ is:
- A
$\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$
- ✓
$\left[\begin{array}{cc}-1 & 0 \\ 0 & -1\end{array}\right]$
- C
$\left[\begin{array}{ll}0 & -1 \\ 0 & -1\end{array}\right]$
- D
$\left[\begin{array}{cc}0 & -1 \\ -1 & 0\end{array}\right]$
AnswerCorrect option: B. $\left[\begin{array}{cc}-1 & 0 \\ 0 & -1\end{array}\right]$
(b) $\left[\begin{array}{cc}-1 & 0 \\ 0 & -1\end{array}\right]$
Explanation:
$\begin{array}{l}{\left[\begin{array}{ll}1 & 2 \\ 2 & 1\end{array}\right]\left[\begin{array}{rr}1 & -1 \\ -1 & 1\end{array}\right]} \\ =\left[\begin{array}{ll}1-2 & -2+2 \\ 2-2 & -2+1\end{array}\right]=\left[\begin{array}{rr}-1 & 0 \\ 0 & -1\end{array}\right]\end{array}$
View full question & answer→MCQ 91 Mark
If I is the unit matrix of order $2 \times 2$, the value of matrix $M$, such that $M-2 I=3\left[\begin{array}{cc}-1 & 0 \\ 4 & 1\end{array}\right]$ is:
- A
$\left[\begin{array}{cc}1 & 0 \\ 12 & 5\end{array}\right]$
- ✓
$\left[\begin{array}{cc}-1 & 0 \\ 12 & 5\end{array}\right]$
- C
$\left[\begin{array}{cc}-1 & 0 \\ 4 & 3\end{array}\right]$
- D
$\left[\begin{array}{ll}-1 & 0 \\ 12 & 3\end{array}\right]$
AnswerCorrect option: B. $\left[\begin{array}{cc}-1 & 0 \\ 12 & 5\end{array}\right]$
(b) $\left[\begin{array}{ll}-1 & 0 \\ 12 & 5\end{array}\right]$
Explanation:
$\begin{array}{l} M -2 I =3\left[\begin{array}{cc}-1 & 0 \\ 4 & 1\end{array}\right] \\ M -2\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]=\left[\begin{array}{cc}-3 & 0 \\ 12 & 3\end{array}\right] \\ M =\left[\begin{array}{ll}-3 & 0 \\ 12 & 3\end{array}\right],\left[\begin{array}{ll}2 & 0 \\ 0 & 2\end{array}\right] \\ =\left[\begin{array}{cc}-3+2 & 0+0 \\ 12+0 & 3+2\end{array}\right] \\ =\left[\begin{array}{ll}-1 & 0 \\ 12 & 5\end{array}\right]\end{array}$
View full question & answer→MCQ 101 Mark
If $\left[\begin{array}{ll}1 & 2 \\ 0 & 0\end{array}\right]\left[\begin{array}{c}3 \\ -p\end{array}\right]=\left[\begin{array}{l}5 \\ 0\end{array}\right]$, then value of $p$ is:
Answer(a) -1
Explanation:
$\begin{array}{l}{\left[\begin{array}{cc}1 & 2 \\ 0 & 0\end{array}\right]\left[\begin{array}{c}3 \\ -p\end{array}\right]=\left[\begin{array}{l}5 \\ 0\end{array}\right]} \\ {\left[\begin{array}{c}3-2 p \\ 0-0\end{array}\right]=\left[\begin{array}{l}5 \\ 0\end{array}\right]} \\ \Rightarrow 3-2 p=5 \\ \Rightarrow-2 p=5-3 \\ \Rightarrow-2 p=2\end{array}$
$\Rightarrow p=-1$
View full question & answer→MCQ 111 Mark
Construct a $2 \times 2$ matrix whose elements ajj are given by $a j j=2 i+j$ :
- A
$\left[\begin{array}{ll}1 & 2 \\ 3 & 4\end{array}\right]$
- B
$\left[\begin{array}{ll}2 & 3 \\ 4 & 5\end{array}\right]$
- ✓
$\left[\begin{array}{ll}3 & 4 \\ 5 & 6\end{array}\right]$
- D
$\left[\begin{array}{ll}4 & 5 \\ 6 & 7\end{array}\right]$
AnswerCorrect option: C. $\left[\begin{array}{ll}3 & 4 \\ 5 & 6\end{array}\right]$
(c) $\left[\begin{array}{ll}3 & 4 \\ 5 & 6\end{array}\right]$
Explanation:
Here, $a j j=2 i+j$
$\begin{array}{l}a_{11}=2(1)+1=2+1=3 \\ a_{12}=2(1)+2=2+2=4 \\ a_{21}=2(2)+1=4+1=5 \\ a_{22}=2(2)+2=4+2=6\end{array}$
So, the matrix is $\left[\begin{array}{ll}3 & 4 \\ 5 & 6\end{array}\right]$
View full question & answer→MCQ 121 Mark
If $A=\left[\begin{array}{ll}-1 & 2\end{array}\right]$ and $B=\left[\begin{array}{cc}1 & -2 \\ 0 & 3\end{array}\right]$
Which of the following operation is possible?
Answer(c) $A B$
Explanation:
Two matrices A and B are conformable for the product AB if the number of columns in $A$ is same as the number of rows in $B$.
Number of columns in $A=$ Number of rows in $B$
2=2
View full question & answer→MCQ 131 Mark
If $\left[\begin{array}{ll}2 & x \\ 0 & 1\end{array}\right]+3\left[\begin{array}{ll}2 & 1 \\ 4 & 0\end{array}\right]=\left[\begin{array}{cc}8 & 8 \\ 12 & 1\end{array}\right]$
The value of $x$ is:
Answer(d) 5
Explanation:
Given,
$\begin{array}{l}\text { If }\left[\begin{array}{ll}2 & x \\ 0 & 1\end{array}\right]+3\left[\begin{array}{ll}2 & 1 \\ 4 & 0\end{array}\right]=\left[\begin{array}{cc}8 & 8 \\ 12 & 1\end{array}\right] \\ \Rightarrow\left[\begin{array}{ll}2 & x \\ 0 & 1\end{array}\right]+\left[\begin{array}{cc}6 & 3 \\ 12 & 0\end{array}\right]=\left[\begin{array}{cc}8 & 8 \\ 12 & 1\end{array}\right] \\ \Rightarrow\left[\begin{array}{cc}2+6 & x+3 \\ 0+12 & 1+0\end{array}\right]=\left[\begin{array}{cc}8 & 8 \\ 12 & 1\end{array}\right] \\ \Rightarrow\left[\begin{array}{cc}8 & x+3 \\ 12 & 1\end{array}\right]=\left[\begin{array}{cc}8 & 8 \\ 12 & 1\end{array}\right] \\ \Rightarrow x+3=8, x=8-3=5 \\ \Rightarrow x=5\end{array}$
View full question & answer→MCQ 141 Mark
Given $\left[\begin{array}{ll}a & b \\ c & d\end{array}\right] \times X =\left[\begin{array}{l}p \\ q\end{array}\right]$. The order of matrix X is:
- A
$2 \times 2$
- B
$1 \times 2$
- ✓
$2 \times 1$
- D
$1 \times 1$
AnswerCorrect option: C. $2 \times 1$
(c) $2 \times 1$
Explanation:
$\begin{array}{l} AX = B \\ A =\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]_{(2 \times 2)}, B =\left[\begin{array}{l}p \\ q\end{array}\right]_{(2 \times 1)}\end{array}$
Let the order of X be $(m \times n)$
... The order of matrix $X$ is $(2 \times 1)$.
View full question & answer→MCQ 151 Mark
If $A =\left[\begin{array}{cc}2 & 12 \\ 0 & 1\end{array}\right], B =\left[\begin{array}{ll}4 & x \\ 0 & 1\end{array}\right]$ and $A ^2= B$, then the value of $x$ is :
Answer(c) 36
Explanation :
$\begin{array}{l}A^2=B \\ \Rightarrow A \cdot A \cdot=B \\ \Rightarrow\left[\begin{array}{cc}2 & 12 \\ 0 & 1\end{array}\right]\left[\begin{array}{cc}2 & 12 \\ 0 & 1\end{array}\right]=\left[\begin{array}{ll}4 & x \\ 0 & 1\end{array}\right] \\ \Rightarrow\left[\begin{array}{cc}4 & 36 \\ 0 & 1\end{array}\right]=\left[\begin{array}{ll}4 & x \\ 0 & 1\end{array}\right] \\ \Rightarrow x=36\end{array}$
View full question & answer→MCQ 161 Mark
If $A=\left[\begin{array}{rr}3 & 1 \\ -1 & 2\end{array}\right]$, then $A^2+5 A$ is:
- A
$\left[\begin{array}{ll}23 & 10 \\ 10 & 13\end{array}\right]$
- B
$\left[\begin{array}{ll}10 & 23 \\ 13 & 10\end{array}\right]$
- C
$\left[\begin{array}{ll}23 & 10 \\ 10 & 23\end{array}\right]$
- ✓
$\left[\begin{array}{rr}23 & 10 \\ -10 & 13\end{array}\right]$
AnswerCorrect option: D. $\left[\begin{array}{rr}23 & 10 \\ -10 & 13\end{array}\right]$
(d) $\left[\begin{array}{rr}23 & 10 \\ -10 & 13\end{array}\right]$
Explanation:
$\begin{array}{l}A^2+5 A=A \cdot A+5 A \\ =\left[\begin{array}{rr}3 & 1 \\ -1 & 2\end{array}\right]\left[\begin{array}{rr}3 & 1 \\ -1 & 2\end{array}\right]+5\left[\begin{array}{rr}3 & 1 \\ -1 & 2\end{array}\right] \\ =\left[\begin{array}{rr}8 & 5 \\ -5 & 3\end{array}\right]+\left[\begin{array}{rr}15 & 5 \\ -5 & 10\end{array}\right] \\ =\left[\begin{array}{rr}23 & 10 \\ -10 & 13\end{array}\right]\end{array}$
View full question & answer→MCQ 171 Mark
The simplified form of $\left[\begin{array}{cc}4 \sin 30^{\circ} & 2 \cos 60^{\circ} \\ \sin 90^{\circ} & 2 \cos 0^{\circ}\end{array}\right]\left[\begin{array}{ll}4 & 5 \\ 5 & 4\end{array}\right]$ is:
- A
$\left[\begin{array}{ll}14 & 13 \\ 13 & 14\end{array}\right]$
- ✓
$\left[\begin{array}{ll}13 & 14 \\ 14 & 13\end{array}\right]$
- C
$\left[\begin{array}{ll}14 & 14 \\ 13 & 13\end{array}\right]$
- D
$\left[\begin{array}{ll}13 & 13 \\ 14 & 14\end{array}\right]$
AnswerCorrect option: B. $\left[\begin{array}{ll}13 & 14 \\ 14 & 13\end{array}\right]$
(b) $\left[\begin{array}{ll}13 & 14 \\ 14 & 13\end{array}\right]$
Explanation:
$\begin{array}{l}{\left[\begin{array}{cc}4 \sin 30^{\circ} & 2 \cos 60^{\circ} \\ \sin 90^{\circ} & 2 \cos 0^{\circ}\end{array}\right]\left[\begin{array}{ll}4 & 5 \\ 5 & 4\end{array}\right]} \\ =\left[\begin{array}{cc}4 \times \frac{1}{2} & 2 \times \frac{1}{2} \\ 1 & 2 \times 1\end{array}\right]\left[\begin{array}{ll}4 & 5 \\ 5 & 4\end{array}\right] \\ =\left[\begin{array}{ll}2 & 1 \\ 1 & 2\end{array}\right]\left[\begin{array}{ll}4 & 5 \\ 5 & 4\end{array}\right] \\ =\left[\begin{array}{ll}13 & 14 \\ 14 & 13\end{array}\right]\end{array}$
View full question & answer→MCQ 181 Mark
Given $\left[\begin{array}{rr}4 & 2 \\ -1 & 1\end{array}\right] M =7 I$. If I is a matrix of order $2 \times 2$, then the order of matrix M is:
- ✓
$2 \times 2$
- B
$2 \times 1$
- C
$2 \times 3$
- D
$3 \times 3$
AnswerCorrect option: A. $2 \times 2$
(a) $2 \times 2$
Explanation:
$\begin{array}{l}{\left[\begin{array}{rr}4 & 2 \\ -1 & 1\end{array}\right]_{2 \times 2}[ M ]_{p \times q}=[7I]_2 \times 2} \\ \therefore 2 \times q=2 \times 2 \text { and } 2=p \\ \Rightarrow q=2 \text { and } p=2 .\end{array}$
View full question & answer→MCQ 191 Mark
If $A=\left[\begin{array}{ll}7 & 3 \\ 5 & 2\end{array}\right]$ and $B=\left[\begin{array}{ll}2 & 5 \\ 4 & 5\end{array}\right]$, then the matrix $C$ such that $2 A+3 C=8 B$, is:
- A
$\left[\begin{array}{cc}2 & 24 \\ 22 & 36\end{array}\right]$
- ✓
$\left[\begin{array}{cc}\frac{2}{3} & \frac{34}{3} \\ \frac{22}{3} & \frac{36}{3}\end{array}\right]$
- C
$\left[\begin{array}{cc}\frac{2}{3} & \frac{22}{3} \\ \frac{36}{3} & \frac{34}{3}\end{array}\right]$
- D
$\left[\begin{array}{cc}1 & 17 \\ 11 & 18\end{array}\right]$
AnswerCorrect option: B. $\left[\begin{array}{cc}\frac{2}{3} & \frac{34}{3} \\ \frac{22}{3} & \frac{36}{3}\end{array}\right]$
(b) $\left[\begin{array}{cc}\frac{2}{3} & \frac{34}{3} \\ \frac{22}{3} & \frac{36}{3}\end{array}\right]$
Explanation:
We have,
$2 A+3 C=8 B$
$\begin{array}{l}\Rightarrow 3 C =8 B-2 A \\ =8\left[\begin{array}{ll}2 & 5 \\ 4 & 5\end{array}\right]-2\left[\begin{array}{ll}7 & 3 \\ 5 & 2\end{array}\right] \\ =\left[\begin{array}{ll}16 & 40 \\ 32 & 40\end{array}\right]-\left[\begin{array}{ll}14 & 6 \\ 10 & 4\end{array}\right] \\ =\left[\begin{array}{cc}2 & 34 \\ 22 & 36\end{array}\right] \\ \Rightarrow C=\frac{1}{3}\left[\begin{array}{cc}2 & 34\end{array}\right] \\ =\left[\begin{array}{cc}\frac{2}{3} & \frac{34}{3} \\ \frac{22}{3} & \frac{36}{3}\end{array}\right]\end{array}$
View full question & answer→MCQ 201 Mark
If $A=\left[\begin{array}{rr}5 & -1 \\ 6 & 7\end{array}\right]$ and $B=\left[\begin{array}{ll}2 & 1 \\ 3 & 4\end{array}\right]$, then which of the following is correct ?
AnswerCorrect option: D. $A B \neq B A$
(d) $A B \neq B A$
Explanation:
We have,
$AB =\left[\begin{array}{rr}5 & -1 \\ 6 & 7\end{array}\right]\left[\begin{array}{ll}2 & 1 \\ 3 & 4\end{array}\right]$
$\begin{array}{l}=\left[\begin{array}{cc}10-3 & 5-4 \\ 12-21 & 6-28\end{array}\right] \\ =\left[\begin{array}{rr}7 & 1 \\ -9 & -22\end{array}\right]\end{array}$
$\begin{array}{l} BA =\left[\begin{array}{ll}2 & 1 \\ 3 & 4\end{array}\right]\left[\begin{array}{rr}5 & -1 \\ 6 & 7\end{array}\right] \\ =\left[\begin{array}{rr}10+6 & -2+7 \\ 15+24 & -3+28\end{array}\right]=\left[\begin{array}{cc}16 & 5 \\ 39 & 25\end{array}\right]\end{array}$
So, $AB = BA$.
View full question & answer→MCQ 211 Mark
If $A=\left[\begin{array}{rr}4 & -1 \\ 2 & 1\end{array}\right]$, then $(A+2 I)(A-3 I)$ is equal to:
- ✓
$\left[\begin{array}{ll}4 & -4 \\ 8 & -8\end{array}\right]$
- B
$\left[\begin{array}{rr}4 & 8 \\ -4 & -8\end{array}\right]$
- C
$\left[\begin{array}{ll}-4 & 4 \\ -8 & 8\end{array}\right]$
- D
$\left[\begin{array}{rr}-4 & 4 \\ 8 & -8\end{array}\right]$
AnswerCorrect option: A. $\left[\begin{array}{ll}4 & -4 \\ 8 & -8\end{array}\right]$
(a) $\left[\begin{array}{ll}4 & -4 \\ 8 & -8\end{array}\right]$
Explanation:
(A+2 I)(A-3 I)
$=\left\{\left[\begin{array}{rr}4 & -1 \\ 2 & 1\end{array}\right]+2\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]\right\}$
$\left\{\left[\begin{array}{rr}4 & -1 \\ 2 & 1\end{array}\right]-3\left[\begin{array}{rr}1 & 0 \\ 0 & 1\end{array}\right]\right\}$
$\begin{array}{l}=\left[\begin{array}{rr}6 & -1 \\ 2 & 3\end{array}\right]\left[\begin{array}{rr}1 & -1 \\ 2 & -2\end{array}\right] \\ =\left[\begin{array}{rr}6-2 & -6+2 \\ 2+6 & -2-6\end{array}\right] \\ =\left[\begin{array}{rr}4 & -4 \\ 8 & -8\end{array}\right]\end{array}$
View full question & answer→MCQ 221 Mark
The order of the matrix $\left[\begin{array}{rrrr}1 & 9 & 7 & 3 \\ 5 & 4 & -2 & 11 \\ 2 & -1 & -8 & 6\end{array}\right]$ is:
- ✓
$3 \times 4$
- B
$4 \times 3$
- C
$3 \times 3$
- D
$4 \times 4$
AnswerCorrect option: A. $3 \times 4$
(a) $3 \times 4$
Explanation:
In the given matrix,
Number of rows $=3$
and Number of columns $=4$
$\therefore$ Order $=3 \times 4$
View full question & answer→MCQ 231 Mark
If $A=\left[\begin{array}{ll}1 & 1 \\ 8 & 3\end{array}\right]$, then $A^2-4 A=$
- A
$\left[\begin{array}{rr}-3 & -3 \\ 60 & 5\end{array}\right]$
- B
$\left[\begin{array}{rr}-3 & -3 \\ 8 & 3\end{array}\right]$
- ✓
$\left[\begin{array}{ll}5 & 0 \\ 0 & 5\end{array}\right]$
- D
$\left[\begin{array}{cc}8 & 3 \\ 24 & 14\end{array}\right]$
AnswerCorrect option: C. $\left[\begin{array}{ll}5 & 0 \\ 0 & 5\end{array}\right]$
(c) $\left[\begin{array}{ll}5 & 0 \\ 0 & 5\end{array}\right]$
Explanation:
$\begin{array}{l} A ^2-4 A= A \cdot A -4 A \\ =\left[\begin{array}{ll}1 & 1 \\ 8 & 3\end{array}\right]\left[\begin{array}{ll}1 & 1 \\ 8 & 3\end{array}\right]-4\left[\begin{array}{ll}1 & 1 \\ 8 & 3\end{array}\right] \\ =\left[\begin{array}{cc}9 & 4 \\ 32 & 17\end{array}\right]-\left[\begin{array}{cc}4 & 4 \\ 32 & 12\end{array}\right] \\ =\left[\begin{array}{ll}5 & 0 \\ 0 & 5\end{array}\right]\end{array}$
View full question & answer→MCQ 241 Mark
If $M =\left[\begin{array}{ll}1 & 2\end{array}\right]$ and $N =\left[\begin{array}{l}2 \\ 1\end{array}\right]$, then the order of matrix MN is:
- A
$1 \times 2$
- B
$2 \times 1$
- C
$2 \times 2$
- ✓
$1 \times 1$
AnswerCorrect option: D. $1 \times 1$
(d) $1 \times 1$
Explanation:
$MN =\left[\begin{array}{ll}1 & 2]_{1 \times 2}\end{array}\left[\begin{array}{l}2 \\ 1\end{array}\right]_{2 \times 1}=[4]_{1 \times 1}\right.$
View full question & answer→MCQ 251 Mark
If $2\left[\begin{array}{ll}3 & 4 \\ 5 & x\end{array}\right]+\left[\begin{array}{ll}1 & y \\ 0 & 1\end{array}\right]=\left[\begin{array}{cc}7 & 0 \\ 10 & 5\end{array}\right]$, then the values of $x$ and $y$ are:
- A
$x=4, y=-4$
- ✓
$x=2, y=-8$
- C
$x=2, y=-4$
- D
$x=4, y=-8$
AnswerCorrect option: B. $x=2, y=-8$
(b) $x=2, y=-8$
Explanation:
$\begin{array}{l}2\left[\begin{array}{ll}3 & 4 \\ 5 & x\end{array}\right]+\left[\begin{array}{ll}1 & y \\ 0 & 1\end{array}\right]=\left[\begin{array}{cc}7 & 0 \\ 10 & 5\end{array}\right] \\ \Rightarrow\left[\begin{array}{cc}6 & 8 \\ 10 & 2 x\end{array}\right]+\left[\begin{array}{ll}1 & y \\ 0 & 1\end{array}\right]=\left[\begin{array}{cc}7 & 0 \\ 10 & 5\end{array}\right] \\ \Rightarrow\left[\begin{array}{cc}7 & 8+y \\ 10 & 2 x+1\end{array}\right]=\left[\begin{array}{cc}7 & 0 \\ 10 & 5\end{array}\right] \\ \Rightarrow 8+y=0\end{array}$
and $2 x+1=5$
$\begin{array}{l}\Rightarrow y=-8 \\ \text { and } x=2 .\end{array}$
View full question & answer→MCQ 261 Mark
If $A=\left[\begin{array}{rr}-2 & 3 \\ 4 & 5\end{array}\right]$ and $B=\left[\begin{array}{rr}5 & 2 \\ -7 & 3\end{array}\right]$, then the matrix $C$, such that $A+B-C=0$ is:
- ✓
$\left[\begin{array}{rr}3 & 5 \\ -3 & 8\end{array}\right]$
- B
$\left[\begin{array}{rr}-7 & 1 \\ 11 & 2\end{array}\right]$
- C
$\left[\begin{array}{rr}7 & -1 \\ -11 & -2\end{array}\right]$
- D
$\left[\begin{array}{rr}-3 & -5 \\ 3 & -8\end{array}\right]$
AnswerCorrect option: A. $\left[\begin{array}{rr}3 & 5 \\ -3 & 8\end{array}\right]$
(a) $\left[\begin{array}{rr}3 & 5 \\ -3 & 8\end{array}\right]$
Explanation:
Since, $A+B-C=0$
$\begin{array}{l}\Rightarrow C = A + B \\ =\left[\begin{array}{rr}-2 & 3 \\ 4 & 5\end{array}\right]+\left[\begin{array}{rr}5 & 2 \\ -7 & 3\end{array}\right] \\ =\left[\begin{array}{rr}3 & 5 \\ -3 & 8\end{array}\right]\end{array}$
View full question & answer→MCQ 271 Mark
If $A=\left[\begin{array}{ll}3 & 9 \\ 7 & 2\end{array}\right]$ and $I$ is an identity matrix of order 2 , then the value of $A+5 I$ is:
- A
$\left[\begin{array}{cc}3 & 14 \\ 12 & 2\end{array}\right]$
- B
$\left[\begin{array}{ll}3 & 9 \\ 7 & 2\end{array}\right]$
- ✓
$\left[\begin{array}{ll}8 & 9 \\ 7 & 7\end{array}\right]$
- D
$\left[\begin{array}{cc}8 & 14 \\ 12 & 7\end{array}\right]$
AnswerCorrect option: C. $\left[\begin{array}{ll}8 & 9 \\ 7 & 7\end{array}\right]$
(c)$\left[\begin{array}{ll}8 & 9 \\ 7 & 7\end{array}\right]$
Explanation:
$\begin{array}{l}A+5 I=\left[\begin{array}{ll}3 & 9 \\ 7 & 2\end{array}\right]+5\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right] \\ =\left[\begin{array}{ll}3 & 9 \\ 7 & 2\end{array}\right]+\left[\begin{array}{ll}5 & 0 \\ 0 & 5\end{array}\right]=\left[\begin{array}{ll}8 & 9 \\ 7 & 7\end{array}\right]\end{array}$
View full question & answer→MCQ 281 Mark
The simplified form of $\left[\begin{array}{ccc}\cos 45^{\circ} & \sin 30^{\circ} \\ \sqrt{2} \cos 0^{\circ} & \sin 0^{\circ}\end{array}-\left[\begin{array}{ll}\sin 45^{\circ} & \cos 90^{\circ} \\ \sin 90^{\circ} & \cos 45^{\circ}\end{array}\right]\right.$ is:
- A
$\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$
- ✓
$\left[\begin{array}{ll}1 & \frac{1}{2} \\ 1 & 0\end{array}\right]$
- C
$\left[\begin{array}{ll}1 & 1 \\ 1 & 0\end{array}\right]$
- D
$\left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right]$
AnswerCorrect option: B. $\left[\begin{array}{ll}1 & \frac{1}{2} \\ 1 & 0\end{array}\right]$
(b) $\left[\begin{array}{ll}1 & \frac{1}{2} \\ 1 & 0\end{array}\right]$
Explanation:
$\begin{array}{l}{\left[\begin{array}{cc}\cos 45^{\circ} & \sin 30^{\circ} \\ \sqrt{2} \cos 0^{\circ} & \sin 0^{\circ}\end{array}\left[\begin{array}{ll}\sin 45^{\circ} & \cos 90^{\circ} \\ \sin 90^{\circ} & \cos 45^{\circ}\end{array}\right]\right.} \\ \left.=\left[\begin{array}{cc}\frac{1}{\sqrt{2}} & \frac{1}{2} \\ \sqrt{2} \times 1 & 0\end{array}\right] \begin{array}{cc}\frac{1}{\sqrt{2}} & 0 \\ 1 & 1\end{array}\right] \\ =\left[\begin{array}{cc}\frac{1}{2}+\frac{1}{2} & 0+\frac{1}{2} \\ 1+0 & 0+0\end{array}\right]=\left[\begin{array}{ll}1 & \frac{1}{2} \\ 1 & 0\end{array}\right]\end{array}$
View full question & answer→MCQ 291 Mark
If $A =\left[\begin{array}{rr}-1 & 1 \\ a & b\end{array}\right]$ and $A ^2= I _2$, then the values of $a$ and $b$ are:
- A
$a=b=1$
- B
$a=1, b=0$
- C
$a=b=0$
- ✓
$a=0, b=1$
AnswerCorrect option: D. $a=0, b=1$
(d) $a=0, b=1$
Explanation:
Since, $A^2=I$
$\begin{array}{l}\Rightarrow A \cdot A = I \\ \Rightarrow\left[\begin{array}{rr}-1 & 1 \\ a & b\end{array}\right]\left[\begin{array}{rr}-1 & 1 \\ a & b\end{array}\right]=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right] \\ \Rightarrow\left[\begin{array}{rr}1+a & -1+b \\ -a+a b & a+b^2\end{array}\right]=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]\end{array}$
By equality of matrices, we have
$-1+b=0$ and $1+a=1$
$\Rightarrow b=1$ and $a=0$.
View full question & answer→MCQ 301 Mark
If $\left[\begin{array}{ll}1 & 2 \\ 2 & 9\end{array}\right]\left[\begin{array}{l}x \\ y\end{array}\right]-\left[\begin{array}{l}20 \\ 90\end{array}\right]$, then the values of $x$ and $y$ are:
- A
$x=10, y=0$
- B
$x=5, y=4$
- ✓
$x=0, y=10$
- D
$x=4, y=5$
AnswerCorrect option: C. $x=0, y=10$
(c) $x=0, y=10$
Explanation:
$\begin{array}{l}{\left[\begin{array}{ll}1 & 2 \\ 2 & 9\end{array}\right]\left[\begin{array}{l}x \\ y\end{array}\right]=\left[\begin{array}{l}20 \\ 90\end{array}\right]} \\ \Rightarrow\left[\begin{array}{l}1 \times x+2 \times y \\ 2 \times x+9 \times y\end{array}\right]=\left[\begin{array}{l}20 \\ 90\end{array}\right] \\ \Rightarrow\left[\begin{array}{r}x+2 y \\ 2 x+9 y\end{array}\right]=\left[\begin{array}{l}20 \\ 90\end{array}\right] \\ \Rightarrow x+2 y=20 \text { and } 2 x+9 y=90\end{array}$
Solving the two linear equations for $x$ and $y$, we get
x=0, y=10
View full question & answer→MCQ 311 Mark
If $A=\left[\begin{array}{ll}2 & 3 \\ 4 & 2\end{array}\right]$ and $B=\left[\begin{array}{ll}1 & 2 \\ 3 & 4\end{array}\right]$, then the matrix $B A$ is:
- A
$\left[\begin{array}{ll}11 & 16 \\ 10 & 16\end{array}\right]$
- B
$\left[\begin{array}{rr}10 & 22 \\ 7 & 17\end{array}\right]$
- C
$\left[\begin{array}{ll}11 & 10 \\ 16 & 16\end{array}\right]$
- ✓
$\left[\begin{array}{rr}10 & 7 \\ 22 & 17\end{array}\right]$
AnswerCorrect option: D. $\left[\begin{array}{rr}10 & 7 \\ 22 & 17\end{array}\right]$
(d) $\left[\begin{array}{cc}10 & 7 \\ 22 & 17\end{array}\right]$
Explanation:
$\begin{array}{l} BA =\left[\begin{array}{ll}1 & 2 \\ 3 & 4\end{array}\right]\left[\begin{array}{ll}2 & 3 \\ 4 & 2\end{array}\right] \\ =\left[\begin{array}{ll}1 \times 2+2 \times 4 & 1 \times 3+2 \times 2 \\ 3 \times 2+4 \times 4 & 3 \times 3+4 \times 2\end{array}\right] \\ =\left[\begin{array}{cc}10 & 7 \\ 22 & 17\end{array}\right]\end{array}$
View full question & answer→MCQ 321 Mark
If $x\left[\begin{array}{r}-1 \\ 2\end{array}\right]+4\left[\begin{array}{r}2 \\ -y\end{array}\right]-\left[\begin{array}{r}7 \\ -8\end{array}\right]$, then the respective values of $x$ and $y$ are:
- A
$-1, \frac{3}{2}$
- B
$15, \frac{19}{2}$
- ✓
$1, \frac{5}{2}$
- D
$-1, \frac{5}{2}$
AnswerCorrect option: C. $1, \frac{5}{2}$
(c) $1, \frac{5}{2}$
Explanation:
$\begin{array}{l}x\left[\begin{array}{r}-1 \\ 2\end{array}\right]+4\left[\begin{array}{r}2 \\ -y\end{array}\right]=\left[\begin{array}{r}7 \\ -8\end{array}\right] \\ \Rightarrow\left[\begin{array}{r}-x \\ 2
x\end{array}\right]+\left[\begin{array}{c}8 \\ -4 y\end{array}\right]=\left[\begin{array}{r}7 \\ -8\end{array}\right] \\ \Rightarrow\left[\begin{array}{r}-x+8 \\ 2 x-4 y\end{array}\right]=\left[\begin{array}{r}7 \\ -8\end{array}\right] \\ \Rightarrow-x+8=7\end{array}$
and $2 x-4 y=-8$
$\begin{array}{l}\Rightarrow-x=-1 \\ \Rightarrow x=1\end{array}$
So, $2(1)-4 y=-8$
$\begin{array}{l}\Rightarrow-4 y=-10 \\ \Rightarrow y=\frac{5}{2}\end{array}$
View full question & answer→MCQ 331 Mark
If $A=\left[\begin{array}{l}5 \\ 4\end{array}\right], B=\left[\begin{array}{l}2 \\ 0\end{array}\right]$ and matrix $C=2 A+B$, then the matrix $C$ is:
- ✓
$\left[\begin{array}{c}12 \\ 8\end{array}\right]$
- B
$\left[\begin{array}{c}12 \\ 4\end{array}\right]$
- C
$\left[\begin{array}{l}7 \\ 4\end{array}\right]$
- D
$\left[\begin{array}{l}8 \\ 8\end{array}\right]$
AnswerCorrect option: A. $\left[\begin{array}{c}12 \\ 8\end{array}\right]$
(a) $\left[\begin{array}{c}12 \\ 8\end{array}\right]$
Explanation:
C=2 A+B
$\begin{array}{l}=2\left[\begin{array}{l}5 \\ 4\end{array}\right]+\left[\begin{array}{l}2 \\ 0\end{array}\right]=\left[\begin{array}{c}10 \\ 8\end{array}\right]+\left[\begin{array}{l}2 \\ 0\end{array}\right] \\ =\left[\begin{array}{r}10+2 \\ 8+0\end{array}\right]=\left[\begin{array}{c}12 \\ 8\end{array}\right]\end{array}$
View full question & answer→MCQ 341 Mark
A matrix which has only one column is called a:
MCQ 351 Mark
Which of the following is a row matrix?
- ✓
$\left[\begin{array}{lll}1 & 2 & 3\end{array}\right]$
- B
$\left[\begin{array}{l}1 \\ 2 \\ 3\end{array}\right]$
- C
$\left[\begin{array}{ll}1 & 2 \\ 3 & 4\end{array}\right]$
- D
$\left[\begin{array}{ll}1 & 4 \\ 2 & 5 \\ 3 & 6\end{array}\right]$
AnswerCorrect option: A. $\left[\begin{array}{lll}1 & 2 & 3\end{array}\right]$
(a) [1 2 23]
Explanation:
A row matrix has only one row.
View full question & answer→MCQ 361 Mark
If a matrix has 4 elements, then the order of the matrix not possible is:
- A
$2 \times 2$
- B
$1 \times 4$
- ✓
$2 \times 3$
- D
$4 \times 1$
AnswerCorrect option: C. $2 \times 3$
(c) $2 \times 3$
Explanation:
If a matrix has $p$ elements, then the order $m \times n$ of the matrix must be such that $m \times n=p$.
So, out of the given options, $2 \times 3 \neq 4$.
View full question & answer→