Matrices Test

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Matrices Test - 15

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Question 1
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If \(A=\left[\begin{array}{cc}\cos 2 \theta & -\sin 2 \theta \\ \sin 2 \theta & \cos 2 \theta\end{array}\right]\) and \(A + A ^{T}= I\) Where \(I\) is the unit matrix of \(2 \times 2\) and \(A ^{T}\) is the transpose of \(A\), then the value of \(\theta\) is equal to:

A
\(\frac{3 \pi}{2}\)

B
\(\pi\)

C
\(\frac{ \pi}{3}\)

D
\(\frac{ \pi}{6}\)

Solution

Given,
\(A =\left[\begin{array}{cc}\cos 2 \theta & -\sin 2 \theta \\ \sin 2 \theta & \cos 2 \theta\end{array}\right]\)
\(A + A ^{T}=1\)
The transpose of matrix \(A\) is given by, \(
\begin{aligned}
& A ^{ T }=\left[\begin{array}{cc}
\cos 2 \theta & \sin 2 \theta \\
-\sin 2 \theta & \cos 2 \theta
\end{array}\right] \\
\end{aligned}\)
As, \(A + A ^{T}=I\)
\(\begin{aligned}
\left[\begin{array}{cc}
\cos 2 \theta & -\sin 2 \theta \\
\sin 2 \theta & \cos 2 \theta
\end{array}\right]+\left[\begin{array}{cc}
\cos 2 \theta & \sin 2 \theta \\
-\sin 2 \theta & \cos 2 \theta
\end{array}\right]= I \end{aligned}\)
\(\begin{aligned}
&\Rightarrow
{\left[\begin{array}{cc}
2 \cos 2 \theta & 0 \\
0 & 2 \cos 2 \theta
\end{array}\right]
=\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]} \\\end{aligned}\)
As we know that,
If two matrices \(A\) and \(B\) are equal then their corresponding elements are also equal.
\(\therefore 2 \cos 2 \theta=1\)
\(\Rightarrow \cos 2 \theta=\frac{1}{2}\)
\(\Rightarrow \cos 2 \theta=\cos 60^\circ\)
\(\Rightarrow 2 \theta=\frac{\pi}{3}\quad[\because \cos 60^\circ=\frac{\pi}{3}]\)
\(\therefore \theta=\frac{\pi}{6}\)
So, the value of \(\theta\) is \(\frac{\pi}{6}\).
Hence, the correct option is (D).
Question 2
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Consider the following statements in respect of the matrix \(A=\left[\begin{array}{ccc}0 & 1 & 2 \\ -1 & 0 & -3 \\ -2 & 3 & 0\end{array}\right]\)
1) The matrix \(A\) is skew-symmetric.
2) The matrix \(A\) is symmetric.
3) The matrix \(A\) is Iinvertible.
Which of the above statements is/are correct?

A
1 only

B
3 only

C
1 and 3

D
2 and 3

Solution

Given,
\(\begin{aligned}
& A =\left[\begin{array}{ccc}
0 & 1 & 2 \\
-1 & 0 & -3 \\
-2 & 3 & 0
\end{array}\right] \end{aligned}\)
The transpose of matrix \(A\) is given by,
\(\begin{aligned} A ^{ T }=\left[\begin{array}{ccc}
0 & -1 & -2 \\
1 & 0 & 3 \\
2 & -3 & 0
\end{array}\right]
\end{aligned}\)
As we know,
If the transpose of matrix is equal to it, the matrix is known as symmetric matrix.
If the transpose of matrix is equal to the negative of itself, the matrix is said to be skew symmetric matrix.
If the determinant of a square matrix \(n \times n\) is zero, then A is not invertible.
Here, transpose of \(A\) is negative of \(A\), so matrix \(A\) is a skew symmetric matrix.
\(\begin{aligned}
&A=\left[\begin{array}{ccc}
0 & 1 & 2 \\
-1 & 0 & -3 \\
-2 & 3 & 0
\end{array}\right] \\
&\text { Det } A=0-1(0-6)+2(-3-0) \\
&=6-6 \\
&=0
\end{aligned}\)
So, A can not be an invertible matrix.
Hence, the correct option is (A).
Question 3
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If \(\left[\begin{array}{ccc}1 & -4 & 3 \\ 0 & 6 & -7 \\ 2 & 4 & \lambda\end{array}\right]\) is not an invertible matrix, then what is the value of \(\lambda\)?

A
0

B
6

C
-8

D
None of these

Solution

Given,
\(A=\left[\begin{array}{ccc}0 & 6 & -7 \\ 2 & 4 & \lambda\end{array}\right]\) is not an invertible matrix.
\(\therefore A\) is a singular matrix.
As we know,
If \(A\) is a singular matrix. then \(|A|=0\).
\(\begin{aligned}
&| A |=0 \\
&\Rightarrow\left|\begin{array}{ccc}
1 & -4 & 3 \\
0 & 6 & -7 \\
2 & 4 & \lambda
\end{array}\right|=0 \\
&\Rightarrow 1(6 \lambda+28)+4(0+14)+3(0-12)=0 \\
&\Rightarrow 6 \lambda+28+56-36=0 \\
&\Rightarrow 6 \lambda+48=0 \\
&\Rightarrow \lambda=-8
\end{aligned}\)
So, if \(\left[\begin{array}{ccc}1 & -4 & 3 \\ 0 & 6 & -7 \\ 2 & 4 & \lambda\end{array}\right]\) is not an invertible matrix, then the value of \(\lambda=-8\)
Hence, the correct option is (C).
Question 4
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If \(x+2 y=\left[\begin{array}{cc}2 & -3 \\ 1 & 5\end{array}\right]\) and \(2 x+5 y=\left[\begin{array}{ll}7 & 5 \\ 2 & 3\end{array}\right]\), then \(y\) is equal to:

A
\(\left[\begin{array}{cc}3 & 11 \\ 0 & 7\end{array}\right]\)

B
\(\left[\begin{array}{cc}3 & 5 \\ 0 & -7\end{array}\right]\)

C
\(\left[\begin{array}{cc}3 & 11 \\ 0 & -7\end{array}\right]\)

D
\(\left[\begin{array}{ll}3 & 5 \\ 0 & 7\end{array}\right]\)

Solution

Given,
\(\begin{aligned}
&x+2 y=\left[\begin{array}{cc}
2 & -3 \\
1 & 5
\end{array}\right] \end{aligned}\)...(1)
\(\begin{aligned} &2 x+5 y=\left[\begin{array}{ll}
7 & 5 \\
2 & 3
\end{array}\right]
\end{aligned}\)...(2)
Multiplying by 2 in the equation (1), we get
\( 2 x+4 y=\left[\begin{array}{cc}
4 & -6 \\
2 & 10
\end{array}\right]\)...(3)
Subtracting equation (3) from equation (2), we get
\(\begin{aligned}
&(2 x+5 y)-(2 x+4 y)=\left[\begin{array}{ll}
7 & 5 \\
2 & 3
\end{array}\right]-\left[\begin{array}{cc}
4 & -6 \\
2 & 10
\end{array}\right] \\
&\therefore y=\left[\begin{array}{cc}
3 & 11 \\
0 & -7
\end{array}\right]
\end{aligned}\)
Hence, the correct option is (C).
Question 5
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If \(A=\left[\begin{array}{ccc}1 & 3+x & 2 \\ 1-x & 2 & y+1 \\ 2 & 5-y & 3\end{array}\right]\) is a symmetric matrix, then \(3 x+y\) is equal to:

A
-1

B
0

C
1

D
None of these

Solution

Given,
\(\begin{aligned}
&A=\left[\begin{array}{ccc}
1 & 3+x & 2 \\
1-x & 2 & y+1 \\
2 & 5-y & 3
\end{array}\right] \\\end{aligned}\)
As we know,
Any real square matrix \(A= (a_{ij})\) is said to be a symmetric matrix if \(a_{ij}=a_{ji}\) or in other words if \(A\) is a real square matrix such that \(A=A^t\) then \(A\) is said to be a symmetric matrix.
\(A=A^{t}\)
\(\therefore a_{ij}=a_{ji}\)
\(\begin{aligned}\therefore A^{t}=\left[\begin{array}{ccc}
1 & 1-x & 2 \\
3+x & 2 & 5-y \\
2 & y+1 & 3
\end{array}\right]=\left[\begin{array}{ccc}
1 & 3+x & 2 \\
1-x & 2 & y+1 \\
2 & 5-y & 3
\end{array}\right]=A
\end{aligned}\)
On comparing, we get
\(3+x=1-x \)
\(\Rightarrow x+x=1-3\)
\(\Rightarrow 2x=-2\)
\(\Rightarrow x=-1\)
And, \(y+1=5-y\)
\(\Rightarrow y+y=5-1\)
\(\Rightarrow 2y=4\)
\(\Rightarrow y=2\)
Now,
\(3 x+y=3\times (-1)+2 \)
\(\Rightarrow 3 x+y=-3+2\)
\(\Rightarrow 3x+y=-1\)
Hence, the correct option is (A).
Question 6
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The inverse of the matrix \(\left[\begin{array}{ccc}2 & 5 & 0 \\ 0 & 1 & 1 \\ -1 & 0 & 3\end{array}\right]\) is:

A
\(\left[\begin{array}{ccc}3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2\end{array}\right]\)

B
\(\left[\begin{array}{ccc}3 & -5 & 5 \\ -1 & -6 & -2 \\ 1 & -5 & 2\end{array}\right]\)

C
\(\left[\begin{array}{ccc}3 & -15 & 5 \\ -1 & 6 & -2 \\ 1 & -5 & 2\end{array}\right]\)

D
\(\left[\begin{array}{ccc}3 & -15 & 5 \\ -1 & 6 & -2 \\ 1 & -5 & -2\end{array}\right]\)

Solution

Let,
\(\begin{aligned}A =\left[\begin{array}{ccc}
2 & 5 & 0 \\
0 & 1 & 1 \\
-1 & 0 & 3
\end{array}\right] \end{aligned}\)
\(|A |=2(3)-5(0+1) \)
\(=6-5 \)
\(=1\)
Cofactor of \(A=\left[\begin{array}{ccc}3 & 1 & 1 \\ -15 & 6 & 5 \\ 5 & -2 & 2\end{array}\right]\)
And adjoint of \(A =\left[\begin{array}{ccc}3 & 1 & 1 \\ -15 & 6 & 5 \\ 5 & -2 & 2\end{array}\right]^{ T }=\left[\begin{array}{ccc}3 & -15 & 5 \\ -1 & 6 & -2 \\ 1 & -5 & 2\end{array}\right]\)
\(\therefore\) Inverse of \(A =\begin{aligned}
&\frac{1}{| A |}\left[\begin{array}{ccc}
3 & -15 & 5 \\
-1 & 6 & -2 \\
1 & -5 & 2
\end{array}\right] \end{aligned}\)
\(\begin{aligned}
&=\frac{1}{1}\left[\begin{array}{ccc}
3 & -15 & 5 \\
-1 & 6 & -2 \\
1 & -5 & 2
\end{array}\right] \\
&=\left[\begin{array}{ccc}
3 & -15 & 5 \\
-1 & 6 & -2 \\
1 & -5 & 2
\end{array}\right]
\end{aligned}\)
Hence, the correct option is (C).
Question 7
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Find the value of \(x\) and \(y\) such that \(\left[\begin{array}{l}x-y \\ x+y\end{array}\right]=\left[\begin{array}{c}2 \\ 16\end{array}\right]\).

A
\(x=7\) and \(y=9\)

B
\(x=9\) and \(y=7\)

C
\(x=8\) and \(y=5\)

D
\(x=13\) and \(y=3\)

Solution

Given,
\(\left[\begin{array}{l}
x-y \\
x+y
\end{array}\right]=\left[\begin{array}{c}
2 \\
16
\end{array}\right]\)
As we know,
If two matrices and are said to be equal, then
Order of matrix \(A=\) Order of matrix \(B\)
Corresponding element of matrix \(A=\) Corresponding element of \(B\)
So,
\( x-y=2\)...(1)
\( x+y=16\)...(2)
On adding equation (1) and (2) we get,
\(2x=18\)
\(\Rightarrow x=9\)
Substituting \(x =9\) in equation (2) we get
\(9+y=16\)
\(\Rightarrow y=16-9\)
\(\Rightarrow y=7\)
So, the value of \(x=9\) and \(y=7\)
Hence, the correct option is (B).
Question 8
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If \(A =\left[\begin{array}{cc}\cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha\end{array}\right]\) then \(AA ^{T}\) is equal to: (where \(A ^{ T }\) is the transpose of \(\left.A \right)\)

A
Null matrix

B
Identify matrix

C
A

D
-A

Solution

Given,
\(\begin{aligned}A=\left[\begin{array}{cc}
\cos \alpha & \sin \alpha \\
-\sin \alpha & \cos \alpha
\end{array}\right] \\ \end{aligned}\)
The transpose of matrix \(A\) is given by,
\(\begin{aligned} A ^{ T }=\left[\begin{array}{cc}
\cos \alpha & -\sin \alpha \\
\sin \alpha & \cos \alpha
\end{array}\right]
\end{aligned}\)
Now,
\(\begin{aligned}
& AA ^{ T }=\left[\begin{array}{cc}
\cos \alpha & \sin \alpha \\
-\sin \alpha & \cos \alpha
\end{array}\right] \times\left[\begin{array}{cc}
\cos \alpha & -\sin \alpha \\
\sin \alpha & \cos \alpha
\end{array}\right] \\
&\Rightarrow AA ^{ T }=\left[\begin{array}{cc}
\cos ^{2} \alpha+\sin ^{2} \alpha & -\cos \alpha \sin \alpha+\cos \alpha \sin \alpha \\
-\cos \alpha \sin \alpha+\cos \alpha \sin \alpha & \sin ^{2} \alpha+\cos ^{2} \alpha
\end{array}\right] \\
&\Rightarrow AA ^{ T }=\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]
\end{aligned}\)
\(\Rightarrow AA ^{ T }=I\)
So, \(A A^{T}\) is a identify matrix.
Hence, the correct option is (B).
Question 9
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Find \(2 X-Y\) matrix such as \(X+Y=\left[\begin{array}{ll}7 & 5 \\ 3 & 4\end{array}\right]\) and \(X-Y=\left[\begin{array}{cc}1 & -3 \\ 3 & 0\end{array}\right]\).

A
\(\left[\begin{array}{cc}3 & 4 \\ 0 & -2\end{array}\right]\)

B
\(\left[\begin{array}{cc}5 & -2 \\ 6 & 2\end{array}\right]\)

C
\(\left[\begin{array}{cc}5 & 4 \\ 3 & -2\end{array}\right]\)

D
\(\left[\begin{array}{cc}-3 & 4 \\ 0 & -2\end{array}\right]\)

Solution

Given,
\(X+Y=\left[\begin{array}{ll}7 & 5 \\ 3 & 4\end{array}\right] \)...(i)
\(X-Y=\left[\begin{array}{cc}
1 & -3 \\
3 & 0
\end{array}\right]\)...(ii)
Adding the equations (i) and (ii), we get
\(\begin{aligned}
&2 X=\left[\begin{array}{ll}
8 & 2 \\
6 & 4
\end{array}\right] \\
&\Rightarrow X=\left[\begin{array}{ll}
4 & 1 \\
3 & 2
\end{array}\right]
\end{aligned}\)
Substracting (ii) from (i), we get
\(\begin{aligned}
&\text { 2Y }=\left[\begin{array}{ll}
6 & 8 \\
0 & 4
\end{array}\right] \\
&\Rightarrow Y=\left[\begin{array}{ll}
3 & 4 \\
0 & 2
\end{array}\right]\end{aligned} \)
Let \(A =2 X - Y\)
\(A=2 \times\left[\begin{array}{ll}4 & 1 \\ 3 & 2\end{array}\right]-\left[\begin{array}{ll}3 & 4 \\ 0 & 2\end{array}\right]\)
\(\Rightarrow A=\left[\begin{array}{ll}8 & 2 \\ 6 & 4\end{array}\right]-\left[\begin{array}{ll}3 & 4 \\ 0 & 2\end{array}\right]\)
\(\Rightarrow A=\left[\begin{array}{cc}5 & -2 \\ 6 & 2\end{array}\right]\)
\(\therefore 2X-Y =\left[\begin{array}{cc}5 & -2 \\ 6 & 2\end{array}\right]\)
Hence, the correct option is (B).
Question 10
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Find the value of \(y-x\) from the following equation:
\(2\left[\begin{array}{cc}
x & 5 \\
7 & y-3
\end{array}\right]+\left[\begin{array}{cc}
3 & -4 \\
1 & 2
\end{array}\right]=\left[\begin{array}{cc}
7 & 6 \\
15 & 14
\end{array}\right]\)

A
\(7\)

B
\(-7\)

C
\(6\)

D
\(-6\)

Solution

Given,
\(\begin{aligned} 2\left[\begin{array}{cc}
x & 5 \\
7 & y-3
\end{array}\right]+\left[\begin{array}{cc}
3 & -4 \\
1 & 2
\end{array}\right]=\left[\begin{array}{cc}
7 & 6 \\
15 & 14
\end{array}\right] \end{aligned}\)
\(\begin{aligned}\Rightarrow\left[\begin{array}{cc}
2 x & 10 \\
14 & 2 y -6
\end{array}\right]+\left[\begin{array}{cc}
3 & -4 \\
1 & 2
\end{array}\right]=\left[\begin{array}{cc}
7 & 6 \\
15 & 14
\end{array}\right] \end{aligned}\)
\(\begin{aligned} \Rightarrow\left[\begin{array}{cc}
2 x +3 & 6 \\
15 & 2 y -4
\end{array}\right]=\left[\begin{array}{cc}
7 & 6 \\
15 & 14
\end{array}\right]
\end{aligned}
\)
As we know that,
If two matrices \(A\) and \(B\) are equal then their corresponding elements are also equal.
\(\therefore 2 x+3=7 \)
\(\Rightarrow 2x=7-3\)
\(\Rightarrow 2x=4\)
\(\Rightarrow x=2\)
And \(2 y-4=14\)
\(\Rightarrow 2y=14+4\)
\(\Rightarrow 2y=18\)
\(\Rightarrow y=9\)
Now,
\(y-x=9-2=7\)
So, the value of \(y-x\) is \(7\).
Hence, the correct option is (C).
Question 11
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If \(2\left[\begin{array}{ll}1 & 3 \\ 0 & x \end{array}\right]+\left[\begin{array}{ll} y & 0 \\ 1 & 2\end{array}\right]=\left[\begin{array}{ll}5 & 6 \\ 1 & 8\end{array}\right]\) then find the value of \(x+y\).

A
6

B
0

C
-6

D
-7

Solution

Given,
\(\begin{aligned} 2\left[\begin{array}{cc}
1 & 3 \\
0 & x
\end{array}\right]+\left[\begin{array}{ll}
y & 0 \\
1 & 2
\end{array}\right]=\left[\begin{array}{ll}
5 & 6 \\
1 & 8 \end{array}\right]\end{aligned}\)
\(\begin{aligned}\Rightarrow\left[\begin{array}{cc}
2 & 6 \\
0 & 2 x
\end{array}\right]+\left[\begin{array}{ll}
y & 0 \\
1 & 2
\end{array}\right]=\left[\begin{array}{ll}
5 & 6 \\
1 & 8
\end{array}\right]\end{aligned}\)
\(\begin{aligned}\Rightarrow\left[\begin{array}{cc}
2+y & 6 \\
1 & 2 x+2
\end{array}\right]=\left[\begin{array}{ll}
5 & 6 \\
1 & 8
\end{array}\right]
\end{aligned}
\)
As we know,
If two matrices \(A\) and \(B\) are said to be equal, then
Order of matrix \(A=\) Order of matrix \(B\)
Corresponding element of matrix \(A=\) Corresponding element of \(B\)
So, equating the corresponding elements:
\(\therefore 2+y=5 \)
\(\Rightarrow y=5-2\)
\(\Rightarrow y=3\)
And \(2 x+2=8\)
\(\Rightarrow 2x=8-2\)
\(\Rightarrow 2x=6\)
\(\Rightarrow x=3\)
Now, \(x+y=3+3=6\)
So, the value of \(x+y\) is \(6\).
Hence, the correct option is (A).
Question 12
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If \(\left[\begin{array}{cc}2 x+y & 4 x \\ 5 x-7 & 4 x\end{array}\right]=\left[\begin{array}{cc}7 & 7 y-13 \\ y & x+6\end{array}\right]\), then \(x+y=\)

A
3

B
4

C
5

D
6

Solution

As we know,
When two matrices are equal, all the corresponding elements must be equal.
Given,
\(\left[\begin{array}{cc}2 x+y & 4 x \\ 5 x-7 & 4 x\end{array}\right]=\left[\begin{array}{cc}7 & 7 y-13 \\ y & x+6\end{array}\right]\)
Comparing two elements of first column,
\(2 x+y=7 \)
\(\Rightarrow y= 7-2x\)...(1)
\(5 x-7=y \)...(2)
Compairing value of \(y\) from equation (1) and (2) we get,
\(7-2 x=5 x-7\)
\(\Rightarrow 7+7=5x+2x \)
\(\Rightarrow 14=7x \)
\(\therefore x=2\)
Putting value of \(x\) in equation (1), we get
\(y=7-2\times 2\)
\(\Rightarrow y=7-4\)
\(\Rightarrow y=3\)
Now,
\(x+y=2+3=5\)
So, the value of \(x+y\) is \(5\).
Hence, the correct option is (C).
Question 13
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The order of the given matrix is:
\(A=\left[\begin{array}{cc}
2 & 4 \\
-1 & 0 \\
6 & 5
\end{array}\right]\)

A
\(3 \times 3\)

B
\(3 \times 2\)

C
\(2 \times 3\)

D
\(2 \times 2\)

Solution

As we know,
A matrix having \(m\) rows and \(n\) columns is called a matrix of order \(m \times n\) or simply \(m \times n\) matrix.
Given,
\(A=\left[\begin{array}{cc}2 & 4 \\ -1 & 0 \\ 6 & 5\end{array}\right]\)
As we can see that the given matrix \(A\) has \(3\) rows and \(2\) columns.
\(\therefore\) The order of the given matrix \(A\) is \(3 \times 2\).
Hence, the correct option is (B).
Question 14
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If \(A =\left[\begin{array}{ll}1 & 2 \\ 1 & 1\end{array}\right]\) and \(B =\left[\begin{array}{cc}0 & -1 \\ 1 & 2\end{array}\right]\), then \(B ^{-1} A ^{-1}\) is equal to:

A
\(\left[\begin{array}{cc}1 & -3 \\ -1 & 2\end{array}\right]\)

B
\(\left[\begin{array}{cc}-1 & 3 \\ 1 & -2\end{array}\right]\)

C
\(\left[\begin{array}{cc}-1 & 3 \\ -1 & -2\end{array}\right]\)

D
\(\left[\begin{array}{cc}-1 & -3 \\ 1 & -2\end{array}\right]\)

Solution

As we know,
The inverse of matrix \(A\) is given by, \(A ^{-1}=\frac{1}{| A |} .\) adj. A
\(\begin{aligned}
A=\left(\begin{array}{ll}
a & b \\
c & d
\end{array}\right) \end{aligned}\)
\(\begin{aligned} \operatorname{adj} A=\left(\begin{array}{cc}
d & -b \\
-c & a
\end{array}\right)
\end{aligned}
\)
Interchange the diagonal elements and change the sign of the remaining elements.
Given,
Matrix \(A =\left[\begin{array}{ll}1 & 2 \\ 1 & 1\end{array}\right]\)
The inverse of matrix \(A\) is given by,
\(A ^{-1}=\frac{1}{| A |} \cdot\) adj. \(A\)
\(\begin{aligned}
&\Rightarrow| A |=-1 \cdot \text { adj. } A =\left[\begin{array}{cc}
1 & -2 \\
-1 & 1
\end{array}\right] \\
&\Rightarrow A ^{-1}=\left[\begin{array}{cc}
-1 & 2 \\
1 & -1
\end{array}\right] \end{aligned}\)
Now,
\(\begin{aligned} &B =\left[\begin{array}{cc}
0 & -1 \\
1 & 2
\end{array}\right] \\
&| B |=1 \cdot \text { Adj. } B =\left[\begin{array}{cc}
2 & 1 \\
-1 & 0
\end{array}\right] \\
&\Rightarrow B ^{-1}=\left[\begin{array}{cc}
2 & 1 \\
-1 & 0
\end{array}\right]
\end{aligned}\)
Now,
\(B ^{-1} A ^{-1}=\left[\begin{array}{cc}2 & 1 \\ -1 & 0\end{array}\right] \cdot\left[\begin{array}{cc}-1 & 2 \\ 1 & -1\end{array}\right]\)
\(=\left[\begin{array}{cc}-1 & 3 \\ 1 & -2\end{array}\right]\)
Hence, the correct option is (B).
Question 15
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If \(A=\left[\begin{array}{cc}4 & -3 \\ 1 & 0\end{array}\right]\) then \(A+A^{T}\) is equal to:

A
\(\left[\begin{array}{cc}4 & -2 \\ -3 & 0\end{array}\right]\)

B
\(\left[\begin{array}{cc}8 & -2 \\ -3 & 0\end{array}\right]\)

C
\(\left[\begin{array}{cc}8 & -2 \\ -2 & 0\end{array}\right]\)

D
\(\left[\begin{array}{cc}8 & -2 \\ -2 & 3\end{array}\right]\)

Solution

Given,
\(A=\left[\begin{array}{cc}
4 & -3 \\
1 & 0
\end{array}\right]\)
The transpose of a matrix \(A\) is given by,
\(A^{T}=\left[\begin{array}{cc}
4 & 1 \\
-3 & 0
\end{array}\right]\)
Now,
\(\begin{aligned}
&A+A^{T}=\left[\begin{array}{cc}
4 & -3 \\
1 & 0
\end{array}\right]+\left[\begin{array}{cc}
4 & 1 \\
-3 & 0
\end{array}\right] \\
&=\left[\begin{array}{cc}
4+4 & -3+1 \\
1+(-3) & 0+0
\end{array}\right] \\
&=\left[\begin{array}{cc}
8 & -2 \\
-2 & 0
\end{array}\right]
\end{aligned}\)
Hence, the correct option is (C).
Question 16
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Construct a \(3 \times 2\) matrix whose elements are given by \(a _{ ij }=\frac{1}{3}|2 i + j |\).

A
\(\left[\begin{array}{ll}1 & \frac{5}{3} \\ \frac{4}{3} & 2 \\ \frac{7}{3} & \frac{8}{3}\end{array}\right]\)

B
\(\left[\begin{array}{cc}1 & \frac{4}{3} \\ \frac{7}{3} & 2 \\ \frac{5}{3} & \frac{8}{3}\end{array}\right]\)

C
\(\left[\begin{array}{cc}1 & \frac{4}{3} \\ \frac{5}{3} & 2 \\ \frac{8}{3} & \frac{7}{3}\end{array}\right]\)

D
\(\left[\begin{array}{cc}1 & \frac{4}{3} \\ \frac{5}{3} & 2 \\ \frac{7}{3} & \frac{8}{3}\end{array}\right]\)

Solution

As we know,
In general, a \(3 \times 2\) matrix is given by,
\(A=\left[\begin{array}{ll}a_{11} & a_{12} \\ a_{21} & a_{22} \\ a_{31} & a_{32}\end{array}\right]\)
Given,
\(a _{ ij }=\frac{1}{3}|2 i + j |\)
Here \( i =1,2,3\) and \( j =1,2 \)
\(\begin{aligned}& a _{11}=\frac{1}{3}|2+1|=1, a _{12}=\frac{1}{3}|2+2|=\frac{4}{3} \\
& a _{21}=\frac{1}{3}|4+1|=\frac{5}{3}, a _{22}=\frac{1}{3}|4+2|=\frac{6}{3}=2 \\
& a _{31}=\frac{1}{3}|6+1|=\frac{7}{3}, a _{32}=\frac{1}{3}|6+2|=\frac{8}{3}
\end{aligned}
\)
So, the required matrix is \(\left[\begin{array}{cc}1 & \frac{4}{3} \\ \frac{5}{3} & 2 \\ \frac{7}{3} & \frac{8}{3}\end{array}\right]\).
Hence, the correct option is (D).
Question 17
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If \(A B^{T}\) is defined as a square matrix then what is the order of the matrix \(B\), where matrix \(A\) has order \(2 \times 3\)?

A
\(3 \times 3\)

B
\(2 \times 3\)

C
\(4 \times 2\)

D
\(3 \times 2\)

Solution

Let the matrix \(B\) has an order \(p \times q\), i.e, \(p\) rows and \(q\) columns.
Transpose of \(B\) is \(B'\) will have order \(q \times p\).
For matrix, \(A B^{T}=\left[A_{(2 \times 3)} \times B_{(q \times p)}^{T }\right]\) is defined.
\(\therefore q=3\)
Given,
\(AB ^{T}\) is a square matrix.
\(AB ^{T}=\left[ A _{(2 \times 3)} \times B _{(3 \times p )}^{T}\right]\)
\(\therefore p=2\)
So, the order of \(B=p \times q=2 \times 3\)
Hence, the correct option is (B).
Question 18
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If \(A=\left[\begin{array}{cc}\sin \alpha & -\cos \alpha \\ \cos \alpha & \sin \alpha\end{array}\right]\), then for what value of \(\alpha, A\) is an identity matrix?

A
\(0^{\circ}\)

B
\(45^{\circ}\)

C
\(90^{\circ}\)

D
\(60^{\circ}\)

Solution

Given,
\(A=\left[\begin{array}{cc}\sin \alpha & -\cos \alpha \\ \cos \alpha & \sin \alpha\end{array}\right]\)
As we know,
Any square matrix in which all the elements are zero except those in the principal diagonal is called a diagonal matrix. A diagonal matrix in which all the principal diagonal elements are equal to \(1\) is called an identity matrix. It is also known as a unit matrix whereas an identity matrix of order \(n\) is denoted by \(I\).
Value of \(\alpha\) such that \(A\) is an identity matrix.
i.e., \(A=I\)
\(\begin{aligned}
&\left[\begin{array}{cc}
\sin \alpha & -\cos \alpha \\
\cos \alpha & \sin \alpha
\end{array}\right]=\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\end{aligned}\)
\(\therefore \sin \alpha=1\)
\(\Rightarrow \sin \alpha=\sin 90^\circ\)
\(\Rightarrow \alpha= 90^\circ\)
And \(\cos \alpha=0 \)
\(\Rightarrow \cos \alpha=\cos 90^{\circ}\)
\(\Rightarrow \alpha= 90^{\circ}\)
\(\therefore\) The required value of \(\alpha\) is \(90^{\circ}\).
Hence, the correct option is (C).
Question 19
1 / -0

If \(A =\left[\begin{array}{cc}1 & -5 \\ -3 & 7\end{array}\right]\) and \(B =\left[\begin{array}{ll}8 & 4 \\ 1 & 3\end{array}\right]\) then the value of \(( AB )^{T}\) is:

A
\(\left[\begin{array}{cc}3 & -11 \\ -17 & 9\end{array}\right]\)

B
\(\left[\begin{array}{cc}3 & -17 \\ -11 & 9\end{array}\right]\)

C
\(\left[\begin{array}{cc}-3 & 17 \\ 11 & -9\end{array}\right]\)

D
\(\left[\begin{array}{cc}-6 & 17 \\ 22 & -9\end{array}\right]\)

Solution

Given,
\(A =\left[\begin{array}{cc}1 & -5 \\ -3 & 7\end{array}\right]\) and \(B =\left[\begin{array}{ll}8 & 4 \\ 1 & 3\end{array}\right]\)
\(\begin{aligned}
&A B=\left[\begin{array}{cc}
1 & -5 \\
-3 & 7
\end{array}\right]\left[\begin{array}{ll}
8 & 4 \\
1 & 3
\end{array}\right] \\
&=\left[\begin{array}{ll}
(1 \times 8)+(-5 \times 1) & (1 \times 4)+(-5 \times 3) \\
(-3 \times 8)+(7 \times 1) & (-3 \times 4)+(7 \times 3)
\end{array}\right] \\
&=\left[\begin{array}{cc}
3 & -11 \\
-17 & 9
\end{array}\right]
\end{aligned}\)
Now, we can get transpose of \(A B\) by switching its rows with its columns.
\(\therefore( AB )^{T}=\left[\begin{array}{cc}
3 & -17 \\
-11 & 9
\end{array}\right]\)
Hence, the correct option is (B).
Question 20
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If matrix \(A =\left[\begin{array}{cc}1 & -2 \\ -6 & 4\end{array}\right]\) and \(B =\left[\begin{array}{cc}2 & -1 \\ 1 & 3\end{array}\right]\), that \(A \left( B ^{ T }\right)\) is:

A
\(\left[\begin{array}{cc}4 & -5 \\ -16 & 6\end{array}\right]\)

B
\(\left[\begin{array}{cc}0 & -7 \\ 8 & 12\end{array}\right]\)

C
\(\left[\begin{array}{cc}0 & 7 \\ -8 & 18\end{array}\right]\)

D
\(\left[\begin{array}{cc}0 & -7 \\ -8 & 12\end{array}\right]\)

Solution

Given,
\(A =\left[\begin{array}{cc}1 & -2 \\ -6 & 4\end{array}\right]\) and \(B =\left[\begin{array}{cc}2 & -1 \\ 1 & 3\end{array}\right]\)
Transpose of matrix \(B=\left(B^{T}\right)=\left[\begin{array}{cc}2 & 1 \\ -1 & 3\end{array}\right]\)
Now,
\(\begin{aligned}
& A \left( B ^{T}\right)=\left[\begin{array}{cc}
1 & -2 \\
-6 & 4
\end{array}\right] \times\left[\begin{array}{cc}
2 & 1 \\
-1 & 3
\end{array}\right] \\
& \Rightarrow A \left( B ^{ T }\right)=\left[\begin{array}{cc}
2+2 & 1-6 \\
-12-4 & -6+12
\end{array}\right] \\
& \Rightarrow A \left( B ^{ T }\right)=\left[\begin{array}{cc}
4 & -5 \\
-16 & 6
\end{array}\right]
\end{aligned}\)
Hence, the correct option is (A).
Question 21
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If \(A=\left[\begin{array}{cc}2 & 3 \\ -1 & 2\end{array}\right]=\frac{1}{2}(P+Q)\) where \(P\) is symmetric and \(Q\) is skew symmetric matrix then \(P\) and \(Q\) are:

A
\(P=\left[\begin{array}{ll}4 & 2 \\ 2 & 4\end{array}\right]\) and \(Q=\left[\begin{array}{ll}0 & 4 \\ 4 & 0\end{array}\right]\)

B
\(P=\left[\begin{array}{ll}4 & 2 \\ 2 & 4\end{array}\right]\) and \(Q=\left[\begin{array}{cc}0 & 4 \\ -4 & 0\end{array}\right]\)

C
\(P=\left[\begin{array}{ll}4 & 2 \\ 2 & 4\end{array}\right]\) and \(Q=\left[\begin{array}{cc}0 & -4 \\ -4 & 0\end{array}\right]\)

D
None of these

Solution

Given,
\(A=\left[\begin{array}{cc}
2 & 3 \\
-1 & 2
\end{array}\right]=\frac{1}{2}(P+Q)\)
Where \(P\) is symmetric and \(Q\) is a skew-symmetric matrix.
As we know,
Any square matrix can be be expressed as the sum of the symmetric and skew-symmetric matrix. i.e If \(A\) is a square matrix then \(A\) can be expressed as where \(A + A ^{\prime}\) is symmetric and \(A - A ^{\prime}\) is skew-symmetric matrix.
On comparing \(A=\left[\begin{array}{cc}
2 & 3 \\
-1 & 2
\end{array}\right]=\frac{1}{2}(P+Q)\) with \(A=\frac{1}{2}\left(A+A^{\prime}\right)+\frac{1}{2}\left(A-A^{\prime}\right)\) we get,
\( P=A+A^{\prime}\) and \(Q=A-A^{\prime}\)
As, \(A=\left[\begin{array}{cc}2 & 3 \\ -1 & 2\end{array}\right]\)
\(\therefore A^{\prime}=\left[\begin{array}{cc}2 & -1 \\ 3 & 2\end{array}\right]\)
So, \(\begin{aligned} P=\left[\begin{array}{cc}
2 & 3 \\
-1 & 2
\end{array}\right]+\left[\begin{array}{cc}
2 & -1 \\
3 & 2
\end{array}\right]\end{aligned}\)
\(\begin{aligned}\Rightarrow P=\left[\begin{array}{ll}
4 & 2 \\
2 & 4
\end{array}\right]\end{aligned}\)
Similarly,
\(\begin{aligned}
&Q=\left[\begin{array}{cc}
2 & 3 \\
-1 & 2
\end{array}\right]-\left[\begin{array}{cc}
2 & -1 \\
3 & 2
\end{array}\right] \end{aligned}\)
\(\begin{aligned} \Rightarrow Q=\left[\begin{array}{cc}
0 & 4 \\
-4 & 0
\end{array}\right]
\end{aligned}\)
So,
\(\begin{aligned}
&P=\left[\begin{array}{ll}
4 & 2 \\
2 & 4
\end{array}\right]\end{aligned}\) and \(\begin{aligned}Q=\left[\begin{array}{cc}
0 & 4 \\
-4 & 0
\end{array}\right]
\end{aligned}\)
Hence, the correct option is (B).
Question 22
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If the inverse of the matrix \(A=\left[\begin{array}{lll}3 & 1 & 2 \\ 4 & 2 & 1 \\ 2 & a & 1\end{array}\right]\) does not exist then the value of a is:

A
\(\frac{8}{7}\)

B
\(\frac{4}{5}\)

C
\(\frac{7}{9}\)

D
\(\frac{5}{7}\)

Solution

Given,
\(A=\left[\begin{array}{lll}3 & 1 & 2 \\ 4 & 2 & 1 \\ 2 & a & 1\end{array}\right]\)
For \(A^{-1}\) does not exist if \(|A|=0\).
\(\begin{aligned}
&|A|=\left|\begin{array}{lll}
3 & 1 & 2 \\
4 & 2 & 1 \\
2 & a & 1
\end{array}\right|=0 \\
&\Rightarrow |A|=3(2-a)-1(4-2)+2(4 a-4) \\
& \Rightarrow|A|=6-3 a-2+8 a-8 \\
&\Rightarrow |A|=5 a-4 \\
& \Rightarrow|A|=0 \\
&\Rightarrow5 a-4=0 \\
&\Rightarrow5 a =4\\
&\therefore a=\frac{4}{5}
\end{aligned}\)
Hence, the correct option is (B).
Question 23
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Find the value of \(X\) and \(Y\) if \(X+Y=\left[\begin{array}{cc}10 & 2 \\ 0 & 9\end{array}\right], X-Y=\left[\begin{array}{cc}6 & 12 \\ 0 & -5\end{array}\right]\).

A
\(X =\left[\begin{array}{cc}8 & 7 \\ 0 & -5\end{array}\right], Y =\left[\begin{array}{cc}2 & 5 \\ 7 & -5\end{array}\right]\)

B
\(X =\left[\begin{array}{ll}8 & 7 \\ 0 & 2\end{array}\right], Y =\left[\begin{array}{cc}2 & -5 \\ 0 & 7\end{array}\right]\)

C
\(X =\left[\begin{array}{ll}8 & 7 \\ 0 & 2\end{array}\right], Y =\left[\begin{array}{ll}2 & 5 \\ 0 & 7\end{array}\right]\)

D
\(X =\left[\begin{array}{cc}8 & -7 \\ 0 & 7\end{array}\right], Y =\left[\begin{array}{cc}2 & -5 \\ 0 & 7\end{array}\right]\)

Solution

Given,
\(\begin{aligned}X+Y=\left[\begin{array}{cc}
10 & 2 \\
0 & 9
\end{array}\right] \end{aligned}\)...(1)
\(\begin{aligned}X-Y=\left[\begin{array}{cc}
6 & 12 \\
0 & -5
\end{array}\right]
\end{aligned}\)...(2)
Adding equation (1) and (2), we get
\(\begin{aligned}
&2 X=\left[\begin{array}{cc}
10 & 2 \\
0 & 9
\end{array}\right]+\left[\begin{array}{cc}
6 & 12 \\
0 & -5
\end{array}\right]=\left[\begin{array}{cc}
16 & 14 \\
0 & 4
\end{array}\right] \\
&\therefore X=\left[\begin{array}{ll}
8 & 7 \\
0 & 2
\end{array}\right]
\end{aligned}
\)
Now, subtracting equation (2) from (1), we get
\(\begin{aligned}
&2 Y=\left[\begin{array}{cc}
10 & 2 \\
0 & 9
\end{array}\right]-\left[\begin{array}{cc}
6 & 12 \\
0 & -5
\end{array}\right]=\left[\begin{array}{cc}
4 & -10 \\
0 & 14
\end{array}\right] \\
&\therefore Y=\left[\begin{array}{cc}
2 & -5 \\
0 & 7
\end{array}\right]
\end{aligned}\)
So, the value of \(\begin{aligned} X=\left[\begin{array}{ll}
8 & 7 \\
0 & 2
\end{array}\right]
\end{aligned}\) and \(\begin{aligned} Y=\left[\begin{array}{cc}
2 & -5 \\
0 & 7
\end{array}\right]
\end{aligned}\)
Hence, the correct option is (B).
Question 24
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If \(A=\left[\begin{array}{ccc}2 & x-3 & x-2 \\ 3 & -2 & -1 \\ 4 & -1 & -5\end{array}\right]\) is a symmetric matrix then \(x\):

A
3

B
6

C
8

D
0

Solution

As we know,
Square matrix \(A\) is said to be symmetric if the transpose of matrix \(A\) is equal to matrix \(A\) itself.
\(\therefore A^T=A\) or \(A'=A\) [where \(A^T\) or \(A'\) denotes the transpose of matrix]
Square matrix \(A\) is said to be symmetric if \(a_{ij}=a_{ji}\) where \(a_{ij}\) and \(a_{ji}\) is an element present in matrix.
Given,
\(A\) is a symmetric matrix.
\(\Rightarrow A^{T}=A\) or \(a_{i j}=a_{j i} \)
\(\begin{aligned}A=\left[\begin{array}{ccc}
2 & x-3 & x-2 \\
3 & -2 & -1 \\
4 & -1 & -5
\end{array}\right]
\end{aligned}\)
So, by property of symmetric matrices, we get
\( a _{12}= a _{21} \)
\(\Rightarrow x -3=3 \)
\(\Rightarrow x=3+3\)
\(\therefore x =6\)
Hence, the correct option is (B).
Question 25
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What is the order of \(\left[\begin{array}{lll}4 & 4 & 1\end{array}\right]\left[\begin{array}{lll}3 & 2 & 5 \\ 9 & 7 & 4 \\ 6 & 4 & 1\end{array}\right]\) ?

A
\(1 \times 1\)

B
\(1 \times 3\)

C
\(3 \times 3\)

D
\(3 \times 1\)

Solution

Let \(\left[\begin{array}{lll}4 & 4 & 1\end{array}\right]\left[\begin{array}{lll}3 & 2 & 5 \\ 9 & 7 & 4 \\ 6 & 4 & 1\end{array}\right]= AB\)
Order of matrix \(A\) is \((1 \times 3)\).
Order of matrix \(B\) is \((3 \times 3)\).
As we know,
To multiply an \(m \times n\) matrix by \(n \times p\) matrix, the \(n\) must be the same and the result is an \(m \times p\) matrix.
So, Order of \(A_{(1 \times 3)} B_{(3 \times 3)}\) is \((1 \times 3)\).
\(\therefore\) Order of \(\left[\begin{array}{lll}4 & 4 & 1\end{array}\right]\left[\begin{array}{lll}3 & 2 & 5 \\ 9 & 7 & 4 \\ 6 & 4 & 1\end{array}\right]\) is \((1 \times 3)\).
Hence, the correct option is (B).
Question 26
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Order of \(\left[\begin{array}{lll}2 & 7 & 4 \\ 3 & 1 & 0\end{array}\right]\left[\begin{array}{l}5 \\ 4 \\ 3\end{array}\right]\) is:

A
\(3 \times 1\)

B
\(2 \times 1\)

C
\(2 \times 3\)

D
\(3 \times 3\)

Solution

Given,
First matrix \(=\left[\begin{array}{ccc}2 & 7 & 4 \\ 3 & 1 & 0\end{array}\right]\)
Number of rows \(\left(m_{1}\right)=2\)
Number of columns \(\left(n_{1}\right)=3\)
So, order of first matrix \(=2 \times 3\)
And second matrix \(=\left[\begin{array}{l}5 \\ 4 \\ 3\end{array}\right]\)
Number of rows \(\left(m_{2}\right)=3\)
Number of columns \(\left(n_{2}\right)=1\)
So, order of first matrix \(=3 \times 1\)
As we know,
Multiplication of matrices is possible if
Number of column \((n_1)\) in the first matrix \(=\) Number of rows \((m_2)\) in the second matrix
i.e., \(n_{1}=m_{2}\)
\(\therefore n_{1}=m_{2}=3\)
So, \(\left[\begin{array}{lll}2 & 7 & 4 \\ 3 & 1 & 0\end{array}\right]\left[\begin{array}{l}5 \\ 4 \\ 3\end{array}\right]\) is possible.
As we know,
Multiplication of matrices is possible if the order of the new matrix is a row of the first matrix and column of the second matrix.
i.e., \(m_{1}\times n_{2}\)
Now,
Order of new matrix formed by multiplication \(= m _{1} \times n _{2}=2 \times 1\)
Hence, the correct option is (B).
Question 27
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If \(A =\left[\begin{array}{cc}4 & x +2 \\ 2 x -3 & x +1\end{array}\right]\) is symmetric, then \(x\) is equal to:

A
2

B
3

C
-1

D
5

Solution

Given,
\(A =\left[\begin{array}{cc}
4 & x +2 \\
2 x -3 & x +1
\end{array}\right]\)
A real square matrix \(A=\left(a_{i j}\right)\) is said to be symmetric, if \(A=A^{T}\)
Where \(A^{T}=\) transpose of matrix \(A\)
\(\begin{aligned}
& A ^{ T }=\left[\begin{array}{cc}
4 & 2 x -3 \\
x +2 & x +1
\end{array}\right] \\
&\therefore A = A ^{T} \\
&\Rightarrow\left[\begin{array}{cc}
4 & x +2 \\
2 x -3 & x +1
\end{array}\right]=\left[\begin{array}{cc}
4 & 2 x -3 \\
x +2 & x +1
\end{array}\right]
\end{aligned}\)
On comparing elements of both matrix, we get
\( x+2=2 x-3 \)
\(\Rightarrow 2+3=2x-x\)
\(\therefore x=5\)
Hence, the correct option is (D).
Question 28
1 / -0

If \(A =\left[\begin{array}{rr}0 & - i \\ i & 0\end{array}\right]\) and \(B =\left[\begin{array}{rr}1 & 0 \\ 0 & -1\end{array}\right]\) are matrices, then \(AB + BA\) is:

A
A diagonal matrix

B
An invertible matrix

C
A unit matrix

D
A null matrix

Solution

Given,
\(A =\left[\begin{array}{rr}0 & - i \\ i & 0\end{array}\right]\)
\(B =\left[\begin{array}{rr}1 & 0 \\ 0 & -1\end{array}\right]\)
\(\therefore AB =\left[\begin{array}{rr}
0 & - i \\
i & 0
\end{array}\right] \times\left[\begin{array}{rr}
1 & 0 \\
0 & -1
\end{array}\right]\)
\(=\left[\begin{array}{ll}
0+0 & 0+ i \\
i +0 & 0+0
\end{array}\right]\)
\(=\left[\begin{array}{ll}
0 & i \\
i & 0
\end{array}\right]\)
And \(BA =\left[\begin{array}{rr}1 & 0 \\ 0 & -1\end{array}\right] \times\left[\begin{array}{rr}0 & - i \\ i & 0\end{array}\right]\)
\(=\left[\begin{array}{rr}0+0 & - i +0 \\ 0- i & 0+0\end{array}\right]\)
\(=\left[\begin{array}{rr}0 & - i \\ - i & 0\end{array}\right]\)
\(\therefore A B+B A=\left[\begin{array}{ll}
0 & i \\
i & 0
\end{array}\right]+\left[\begin{array}{rr}
0 & -i \\
-i & 0
\end{array}\right]\)
\(=\left[\begin{array}{cc}
0+0 & i-i \\
i-i & 0+0
\end{array}\right]\)
\(=\left[\begin{array}{ll}
0 & 0 \\
0 & 0
\end{array}\right]\)
So, \(AB+BA\) is a null matrix.
Hence, the correct option is (D).
Question 29
1 / -0

If \(A =\left[\begin{array}{ccc}1 & -1 & 0 \\ 3 & 2 & -1\end{array}\right]\) and \(B =\left[\begin{array}{l}1 \\ 3 \\ 5\end{array}\right]\), find \(( AB )^{T}\).

A
\(\left[\begin{array}{c}-2 \\ 4\end{array}\right]\)

B
\(\left[\begin{array}{cc}-2 & 4\end{array}\right]\)

C
\(\left[\begin{array}{c}2 \\ -4\end{array}\right]\)

D
\(\left[\begin{array}{ll}-2 & -4\end{array}\right]\)

Solution

Given,
\(A =\left[\begin{array}{ccc}1 & -1 & 0 \\ 3 & 2 & -1\end{array}\right]\) and \(B =\left[\begin{array}{l}1 \\ 3 \\ 5\end{array}\right]\)
\(\Rightarrow AB =\left[\begin{array}{ccc}1 & -1 & 0 \\ 3 & 2 & -1\end{array}\right] \times\left[\begin{array}{l}1 \\ 3 \\ 5\end{array}\right]\)
\(\Rightarrow AB =\left[\begin{array}{c}1-3+0 \\ 3+6-5\end{array}\right]\)
\(\Rightarrow AB =\left[\begin{array}{c}-2 \\ 4\end{array}\right]\)
As we know,
The new matrix obtained by interchanging the rows and columns of the original matrix is called as the transpose of the matrix. It is denoted by \(A'\) or \(A^T\).
\(\therefore( AB )^{T}=\left[\begin{array}{ll}-2 & 4\end{array}\right]\)
Hence, the correct option is (B).
Question 30
1 / -0

If \(\left[\begin{array}{ccc}1 & -3 & 2 \\ 2 & -8 & 5 \\ 4 & 2 & \lambda\end{array}\right]\) is not an invertible matrix, then what is the value of \(\lambda\)?

A
-1

B
0

C
1

D
2

Solution

Given,
\(A =\left[\begin{array}{ccc}1 & -3 & 2 \\ 2 & -8 & 5 \\ 4 & 2 & \lambda\end{array}\right]\) is not an invertible matrix.
As we know,
If the matrix \(A\) is non singular matrix then \(| A |=0\).
\(\begin{aligned}
&\therefore \left|\begin{array}{ccc}
1 & -3 & 2 \\
2 & -8 & 5 \\
4 & 2 & \lambda
\end{array}\right|=0 \\
&\Rightarrow 1(-8 \lambda-10)+3(2 \lambda-20)+2(4+32)=0 \\
&\Rightarrow-8 \lambda-10+6 \lambda-60+72=0 \\
&\Rightarrow-2 \lambda+2=0 \\
&\Rightarrow \lambda=1
\end{aligned}\)
So, If \(\left[\begin{array}{ccc}1 & -3 & 2 \\ 2 & -8 & 5 \\ 4 & 2 & \lambda\end{array}\right]\) is not an invertible matrix, then the value of \(\lambda\) is \(1\).
Hence, the correct option is (C).

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Matrices Test - 15

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Matrices Test - 15

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SHARING IS CARING

Question 1

\(\frac{3 \pi}{2}\)

\(\pi\)

\(\frac{ \pi}{3}\)

\(\frac{ \pi}{6}\)

Solution

Question 2

1 only

3 only

1 and 3

2 and 3

Solution

Question 3

0

6

-8

None of these

Solution

Question 4

\(\left[\begin{array}{cc}3 & 11 \\ 0 & 7\end{array}\right]\)

\(\left[\begin{array}{cc}3 & 5 \\ 0 & -7\end{array}\right]\)

\(\left[\begin{array}{cc}3 & 11 \\ 0 & -7\end{array}\right]\)

\(\left[\begin{array}{ll}3 & 5 \\ 0 & 7\end{array}\right]\)

Solution

Question 5

-1

0

1

None of these

Solution

Question 6

\(\left[\begin{array}{ccc}3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2\end{array}\right]\)

\(\left[\begin{array}{ccc}3 & -5 & 5 \\ -1 & -6 & -2 \\ 1 & -5 & 2\end{array}\right]\)

\(\left[\begin{array}{ccc}3 & -15 & 5 \\ -1 & 6 & -2 \\ 1 & -5 & 2\end{array}\right]\)

\(\left[\begin{array}{ccc}3 & -15 & 5 \\ -1 & 6 & -2 \\ 1 & -5 & -2\end{array}\right]\)

Solution

Question 7

\(x=7\) and \(y=9\)

\(x=9\) and \(y=7\)

\(x=8\) and \(y=5\)

\(x=13\) and \(y=3\)

Solution

Question 8

Null matrix

Identify matrix

A

-A

Solution

Question 9

\(\left[\begin{array}{cc}3 & 4 \\ 0 & -2\end{array}\right]\)

\(\left[\begin{array}{cc}5 & -2 \\ 6 & 2\end{array}\right]\)

\(\left[\begin{array}{cc}5 & 4 \\ 3 & -2\end{array}\right]\)

\(\left[\begin{array}{cc}-3 & 4 \\ 0 & -2\end{array}\right]\)

Solution

Question 10

\(7\)

\(-7\)

\(6\)

\(-6\)

Solution

Question 11

6

0

-6

-7

Solution

Question 12

3

4

5

6

Solution