Vector Algebra Test

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Vector Algebra Test - 77

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Question 1
1 / -0

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 \(6\).
Question 2
1 / -0

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 \(5\).
Question 3
1 / -0

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,
We know that,
Adj. of Matrix \(|X|=\left[\begin{array}{ll}
A_1 & A_2\\
B_1 & B_2
\end{array}\right]\)
\(|X|=\left(A_1\times B_2-A_2\times B_1\right)\)
\(A ^{-1}=\frac{1}{| A |} \cdot\) 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,
\(B ^{-1} A ^{-1}=\left[\begin{array}{cc}2 & 1 \\ -1 & 0\end{array}\right] \cdot\left[\begin{array}{cc}-1 & 2 \\ 1 & -1\end{array}\right]\)
\(=\left[\begin{array}{cc}-1 & 3 \\ 1 & -2\end{array}\right]\)
Question 4
1 / -0

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 \(A^T\),

In transpose of matrix, rows are converted into columns and vice-versa.

\(A^{T}=\left[\begin{array}{cc} 4 & 1 \\ -3 & 0 \end{array}\right]\)

Now,

\(\begin{aligned} &A+A^{T}=\left[\begin{array}{cc} 4 & -3 \\ 1 & 0 \end{array}\right] + \left[\begin{array}{cc} 4 & 1 \\ -3 & 0 \end{array}\right] \\ &=\left[\begin{array}{cc} 4+4 & -3+1 \\ 1+(-3) & 0+0 \end{array}\right] \\ &=\left[\begin{array}{cc} 8 & -2 \\ -2 & 0 \end{array}\right] \end{aligned}\)
Question 5
1 / -0

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 \(\left[\begin{array}{cc}1 & \frac{4}{3} \\ \frac{5}{3} & 2 \\ \frac{7}{3} & \frac{8}{3}\end{array}\right]\).
Question 6
1 / -0

If \(A=\left(\begin{array}{lll}0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0\end{array}\right)\), then \(A^{-1}=\) ?

A
\(A\)

B
\(-A\)

C
\(1\)

D
\(-1\)

Solution

\(A ^{-1}=\frac{a d j A}{|A|}\)

We are given \(A=\left(\begin{array}{lll}0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0\end{array}\right)\)

On finding its determinant, we get,

\(| A |=0(0-0)-0(0-0)+1(0-1)=-1\)

\(\Rightarrow| A |=-1\).

Now,

Finding the cofactors of A, we get,

\(A _{11}=\left|\begin{array}{ll}1 & 0 \\ 0 & 0\end{array}\right|=0, A _{12}=\left|\begin{array}{ll}0 & 0 \\ 1 & 0\end{array}\right|=0, \quad A _{13}=\left|\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right|=-1\)

\(A_{21}=\left|\begin{array}{ll}0 & 1 \\ 0 & 0\end{array}\right|=0, \quad A_{22}=\left|\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right|=-1, A_{23}=\left|\begin{array}{ll}0 & 0 \\ 1 & 0\end{array}\right|=0\)

\(A _{31}=\left|\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right|=-1, \quad A _{32}=\left|\begin{array}{ll}0 & 1 \\ 0 & 0\end{array}\right|=0, \quad A _{33}=\left|\begin{array}{ll}0 & 0 \\ 0 & 1\end{array}\right|=0\)

i.e. \(\operatorname{adj}( A )=\left(\begin{array}{lll}A_{11} & A_{12} & A_{13} \\ A_{21} & A_{22} & A_{23} \\ A_{31} & A_{32} & A_{33}\end{array}\right)\)

\(\Rightarrow \operatorname{adj}(A)=\left(\begin{array}{ccc}0 & 0 & -1 \\ 0 & -1 & 0 \\ -1 & 0 & 0\end{array}\right)\)

Since, \(A ^{-1}=\frac{a d j A}{|A|}\)

\(\Rightarrow A^{-1}=\frac{1}{-1}\left(\begin{array}{ccc}0 & 0 & -1 \\ 0 & -1 & 0 \\ -1 & 0 & 0\end{array}\right)\)

\(\Rightarrow A ^{-1}=\left(\begin{array}{ccc}0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0\end{array}\right)= A\)
Question 7
1 / -0

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 \(B=p \times q=2 \times 3\)
Question 8
1 / -0

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 \(90^{\circ}\).
Question 9
1 / -0

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 \(\begin{aligned}Q=\left[\begin{array}{cc}
0 & 4 \\
-4 & 0
\end{array}\right]
\end{aligned}\)
Question 10
1 / -0

If \(A=\left[\begin{array}{cc}1 & -1 \\ -1 & 1\end{array}\right]\), then the expression \(A^3-2 A^2\) is:

A
A null matrix

B
An identify matrix

C
Equal to A

D
Equal to -A

Solution

Given,

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

\(A^2=A \cdot A\)

\(=\left[\begin{array}{cc}1 & -1 \\ -1 & 1\end{array}\right] \times\left[\begin{array}{cc}1 & -1 \\ -1 & 1\end{array}\right]\)

\(=\left[\begin{array}{cc}1+1 & -1-1 \\ -1-1 & 1+1\end{array}\right]\)

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

\(A^3=A^2 \cdot A\)

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

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

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

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

Now,

\(A^3=\left[\begin{array}{cc}4 & -4 \\ -4 & 4\end{array}\right]\) and

\(2 A^2=2 \times\left[\begin{array}{cc}2 & -2 \\ -2 & 2\end{array}\right]\)

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

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

So, the expression \(A^3-2 A^2\) is a null matrix.

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Vector Algebra Test - 77

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Vector Algebra Test - 77

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Question 1

6

0

-6

-7

Solution

Question 2

3

4

5

6

Solution

Question 3

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

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

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

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

Solution

Question 4

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

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

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

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

Solution

Question 5

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

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

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

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

Solution

Question 6

\(A\)

\(-A\)

\(1\)

\(-1\)

Solution

Question 7

\(3 \times 3\)

\(2 \times 3\)

\(4 \times 2\)

\(3 \times 2\)

Solution

Question 8

\(0^{\circ}\)

\(45^{\circ}\)

\(90^{\circ}\)

\(60^{\circ}\)

Solution

Question 9

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

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

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

None of these

Solution

Question 10

A null matrix

An identify matrix

Equal to A

Equal to -A

Solution

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