Vector Algebra Test

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Vector Algebra Test - 76

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Question 1
1 / -0

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,
\(A^{T}=\left[\begin{array}{cc}\cos 2 \theta & \sin 2 \theta \\ -\sin 2 \theta & \cos 2 \theta\end{array}\right]\)
As, \(A + A ^{T}=I\)
\(\left[\begin{array}{cc}\cos 2 \theta & -\sin 2 \theta \\ \sin 2 \theta & \cos 2 \theta\end{array}\right]+\left[\begin{array}{cc}\cos 2 \theta & \sin 2 \theta \\ -\sin 2 \theta & \cos 2 \theta\end{array}\right]=I\)
\(\Rightarrow\left[\begin{array}{cc}2 \cos 2 \theta & 0 \\ 0 & 2 \cos 2 \theta\end{array}\right]=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]\)
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 \(\frac{\pi}{6}\).
Question 2
1 / -0

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.
\(\begin{aligned}
&A=\left[\begin{array}{ccc}
0 & 1 & 2 \\
-1 & 0 & -3 \\
-2 & 3 & 0
\end{array}\right] \\
& A=0-1(0-6)+2(-3-0) \\
&=6-6 \\
&=0
\end{aligned}\)
So, A can not be an invertible matrix.
Question 3
1 / -0

If the matrix \(\left[\begin{array}{ccc}\cos \theta & \sin \theta & 0 \\ \sin \theta & \cos \theta & 0 \\ 0 & 0 & 1\end{array}\right]\) is singular, then \(\theta=?\)

A
\(\frac{\pi}{4}\)

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

C
\(\pi\)

D
\(0\)

Solution

A matrix is said to be singular if its determinant is zero i.e. for matrix \(A\) to be singular, \(| A |=0\)
For a singular matrix, the inverse doesn't exist
Given that, matrix \(\left[\begin{array}{ccc}\cos \theta & \sin \theta & 0 \\ \sin \theta & \cos \theta & 0 \\ 0 & 0 & 1\end{array}\right]\) is singular
Then, \(\left|\begin{array}{ccc}\cos \theta & \sin \theta & 0 \\ \sin \theta & \cos \theta & 0 \\ 0 & 0 & 1\end{array}\right|=0\)
\(\Rightarrow\left|\begin{array}{ll}\cos \theta & \sin \theta \\ \sin \theta & \cos \theta\end{array}\right|=0\)
\(\Rightarrow \cos ^2 \theta-\sin ^2 \theta=0\)
\(\Rightarrow \cos 2 \theta=\cos \frac{\pi}{2}\)
\(\therefore \theta=\frac{\pi}{4}\)
Question 4
1 / -0

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 \(|A|=0\).
\(| A |=0 \)
\(\Rightarrow\left|\begin{array}{ccc}
1 & -4 & 3 \\
0 & 6 & -7 \\
2 & 4 & \lambda
\end{array}\right|=0 \)
We know that,
\(X=\left[\begin{array}{ll}
A_1 & A_2 & A_3 \\
B_1 & B_2 & B_3\\
C_1 & C_2 & C_3
\end{array}\right]\)
\(X=A_1(B_2C_3-B_3C_2)-A_2(B_1C_3-B_3C_1)+A_3(B_1C_2-B_2C_1)\)
Then,
\( 1(6 \lambda+28)+4(0+14)+3(0-12)=0 \)
\(\Rightarrow 6 \lambda+28+56-36=0 \)
\(\Rightarrow 6 \lambda+48=0 \)
\(\Rightarrow \lambda=-8\)
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 \(\lambda=-8\)
Question 5
1 / -0

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
\((2 x+5 y)-(2 x+4 y)=\left[\begin{array}{ll}
7 & 5 \\
2 & 3
\end{array}\right]-\left[\begin{array}{cc}
4 & -6 \\
2 & 10
\end{array}\right]\)
We know that,
\(\left[\begin{array}{cc}A_1 & A_2 \\ B_1 & B_2\end{array}\right]-\left[\begin{array}{cc}C_1 & C_2 \\ D_1 & D_2\end{array}\right]=\left[\begin{array}{cc}A_1-C_1 & A_2-C_2 \\ B_1-D_1 & B_2-D_2\end{array}\right]\)
\(\therefore y=\left[\begin{array}{cc}
3 & 11 \\
0 & -7
\end{array}\right]\)
Question 6
1 / -0

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,
\(3 x+y=3\times (-1)+2 \)
\(\Rightarrow 3 x+y=-3+2\)
\(\Rightarrow 3x+y=-1\)
Question 7
1 / -0

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,
\(\begin{aligned}A=\left[\begin{array}{cc}
\cos \alpha & \sin \alpha \\
-\sin \alpha & \cos \alpha
\end{array}\right] \\ \end{aligned}\)
In transpose of matrix, rows are converted into columns and vice-versa.
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,
\(AA ^{ T }=\left[\begin{array}{cc}
\cos \alpha & \sin \alpha \\
-\sin \alpha & \cos \alpha
\end{array}\right] \times\left[\begin{array}{cc}
\cos \alpha & -\sin \alpha \\
\sin \alpha & \cos \alpha
\end{array}\right]\)
We know that,
\(X=\left[\begin{array}{cc}A_1 & A_2 \\ B_1 & B_2\end{array}\right], Y=\left[\begin{array}{cc}C_1 & C_2 \\ D_1 & D_2\end{array}\right]\).
\(XY=\left[\begin{array}{cc}A_1C_1+A_2D_1 & A_1C_2+A_2D_2 \\ B_1C_1+B_2D_1 & B_1C_2+B_2D_2 \end{array}\right]\)
\(\Rightarrow AA ^{ T }=\left[\begin{array}{cc}
\cos ^{2} \alpha+\sin ^{2} \alpha & -\cos \alpha \sin \alpha+\cos \alpha \sin \alpha \\
-\cos \alpha \sin \alpha+\cos \alpha \sin \alpha & \sin ^{2} \alpha+\cos ^{2} \alpha
\end{array}\right] \\
\Rightarrow AA ^{ T }=\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\)
\(\Rightarrow AA ^{ T }=I\)
So, \(A A^{T}\) is a identify matrix.
Question 8
1 / -0

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
\(2 X=\left[\begin{array}{ll}
8 & 2 \\
6 & 4
\end{array}\right]\)
We know that,
When,
\(a\div\left[\begin{array}{cc}A_1 & A_2 \\ B_1 & B_2\end{array}\right]=\left[\begin{array}{cc}A_1\div a & A_2\div a \\ B_1\div a & B_2\div a\end{array}\right]\)
\(\Rightarrow X=\left[\begin{array}{ll}
4 & 1 \\
3 & 2
\end{array}\right]\)
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 \(A =2 X - Y\)
\(A=2 \times\left[\begin{array}{ll}4 & 1 \\ 3 & 2\end{array}\right]-\left[\begin{array}{ll}3 & 4 \\ 0 & 2\end{array}\right]\)
We know that,
\(a\times\left[\begin{array}{cc}A_1 & A_2 \\ B_1 & B_2\end{array}\right]=\left[\begin{array}{cc}A_1\times a & A_2\times a \\ B_1\times a & B_2\times a\end{array}\right]\)
\(\Rightarrow A=\left[\begin{array}{ll}8 & 2 \\ 6 & 4\end{array}\right]-\left[\begin{array}{ll}3 & 4 \\ 0 & 2\end{array}\right]\)
\(\Rightarrow A=\left[\begin{array}{cc}5 & -2 \\ 6 & 2\end{array}\right]\)
\(\therefore 2X-Y =\left[\begin{array}{cc}5 & -2 \\ 6 & 2\end{array}\right]\)
Question 9
1 / -0

If \(3 A +4 B ^{\prime}=\left[\begin{array}{ccc}7 & -10 & 17 \\ 0 & 6 & 31\end{array}\right]\) and \(2 B -3 A ^{\prime}=\left[\begin{array}{rr}-1 & 18 \\ 4 & 0 \\ -5 & -7\end{array}\right]\) then \(B =\) ?

A
\(\left[\begin{array}{rr}1 & 3 \\ -1 & 1 \\ 2 & 4\end{array}\right]\)

B
\(\left[\begin{array}{rr}1 & -3 \\ -1 & 1 \\ 2 & 4\end{array}\right]\)

C
\(\left[\begin{array}{rr}-1 & -18 \\ 4 & -16 \\ -5 & -7\end{array}\right]\)

D
\(\left[\begin{array}{rr}1 & 3 \\ -1 & 1 \\ 2 & -4\end{array}\right]\)

Solution

Properties of Transpose of a Matrix:
The transpose of the transpose of a matrix is the matrix itself: \(\left( A ^{\prime}\right)^{\prime}= A\).
The transposes of equal matrices are also equal:
\(A = B \Rightarrow A^{\prime}= B ^{\prime}\)
The transpose of the sum/difference of two matrices is equivalent to the sumvdifference of their transposes: \(( A \pm B )^{\prime}=A^{\prime} \pm B^{\prime}\)
The transpose of the product of two matrices is equivalent to the product of their transposes in reversed order \(( AB )^{\prime}= B ^{\prime} A ^{\prime}\)
Using the Properties of Transpose of a Matrix:
\(3 A +4 B ^{\prime}=\left[\begin{array}{ccc}7 & -10 & 17 \\ 0 & 6 & 31\end{array}\right]\)
\(\Rightarrow\left(3 A +4 B ^{\prime}\right)^{\prime}=\left[\begin{array}{ccc}7 & -10 & 17 \\ 0 & 6 & 31\end{array}\right]^{\prime}\)
\(\Rightarrow 3 A ^{\prime}+4\left( B ^{\prime}\right)^{\prime}=\left[\begin{array}{ccc}7 & -10 & 17 \\ 0 & 6 & 31\end{array}\right]^{\prime}\)
\(\Rightarrow 3 A ^{\prime}+4 B =\left[\begin{array}{cc}7 & 0 \\ -10 & 6 \\ 17 & 31\end{array}\right]\)
Also,
\(2 B -3 A ^{\prime}=\left[\begin{array}{rr}-1 & 18 \\ 4 & 0 \\ -5 & -7\end{array}\right]\)
Adding equations (1) and (2), we get,
\(\left(3 A ^{\prime}+4 B \right)+\left(2 B -3 A ^{\prime}\right)=\left[\begin{array}{cc}7 & 0 \\ -10 & 6 \\ 17 & 31\end{array}\right]+\left[\begin{array}{rr}-1 & 18 \\ 4 & 0 \\ -5 & -7\end{array}\right]\)
\(\Rightarrow 6 B +0=\left[\begin{array}{cc}7-1 & 0+18 \\ -10+4 & 6+0 \\ 17-5 & 31-7\end{array}\right]\)
\(\Rightarrow B =\left[\begin{array}{cc}1 & 3 \\ -1 & 1 \\ 2 & 4\end{array}\right]\)
Question 10
1 / -0

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 \(7\).

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Vector Algebra Test - 76

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Vector Algebra Test - 76

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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

\(\frac{\pi}{4}\)

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

\(\pi\)

\(0\)

Solution

Question 4

0

6

-8

None of these

Solution

Question 5

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

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

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

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

Solution

Question 6

-1

0

1

None of these

Solution

Question 7

Null matrix

Identify matrix

A

-A

Solution

Question 8

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

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

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

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

Solution

Question 9

\(\left[\begin{array}{rr}1 & 3 \\ -1 & 1 \\ 2 & 4\end{array}\right]\)

\(\left[\begin{array}{rr}1 & -3 \\ -1 & 1 \\ 2 & 4\end{array}\right]\)

\(\left[\begin{array}{rr}-1 & -18 \\ 4 & -16 \\ -5 & -7\end{array}\right]\)

\(\left[\begin{array}{rr}1 & 3 \\ -1 & 1 \\ 2 & -4\end{array}\right]\)

Solution

Question 10

\(7\)

\(-7\)

\(6\)

\(-6\)

Solution

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