Matrices Test

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Matrices Test - 44

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Weekly Quiz Competition

Question 1
1 / -0

If $$ \begin{bmatrix} 1/25 & 0 \\ x & 1/25 \end{bmatrix}\quad =\quad \begin{bmatrix} 5 & 0 \\ -a & 5 \end{bmatrix}^{ -2 } $$, then the value of x is

A
$$ a / 125 $$

B
$$ 2a / 125 $$

C
$$ 2a / 25 $$

D
None of these

Solution

Let $$ A = \begin{bmatrix} 5 & 0 \\ -a & 5 \end{bmatrix} $$
$$ \Rightarrow (A)= \begin{bmatrix} 5 & 0 \\ -a & 5 \end{bmatrix} $$
$$ \Rightarrow A^{-1} = \frac {1}{ |A| } \begin{bmatrix} 5 & 0 \\ a & 5 \end{bmatrix} = \frac {1}{25} \begin{bmatrix} 5 & 0 \\ a & 5 \end{bmatrix} $$
$$ \Rightarrow A^{-2} =(A^{-1})^2 = \frac {1}{25} \begin{bmatrix} 5 & 0 \\ a & 5 \end{bmatrix} \frac {1}{25} \begin{bmatrix} 5 & 0 \\ a & 5 \end{bmatrix} $$
$$ =\frac {1}{625} \begin{bmatrix}25 & 0 \\10a&25 \end{bmatrix} $$
Now ,
$$ \begin{bmatrix} 1/2 & 0 \\ x & 1/25 \end{bmatrix}=\begin{bmatrix} \frac {1}{25} &0 \\ \frac {2a}{125} & \frac {1}{25} \end{bmatrix} $$
$$ \Rightarrow x = 2a /125 $$
Question 2
1 / -0

The matrix $$\begin{bmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 4 \end{bmatrix}$$ is a

A
Identity matrix

B
Symmetric matrix

C
Skew symmetric matrix

D
None of these

Solution

Let $$A=\begin{bmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 4 \end{bmatrix}$$
$$A' =\begin{bmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 4 \end{bmatrix} =A$$
So, the given matrix is a symmetric matrix.
[Since, in a square $$A$$. If $$A'=A$$, then $$A$$ is called symmetric matrix]
Question 3
1 / -0

If $$A=\dfrac { 1 }{ \pi } \begin{bmatrix} \sin ^{ -1 }{ \left( x\pi \right) } & \tan ^{ -1 }{ \left( \dfrac { x }{ \pi } \right) } \\ \sin ^{ -1 }{ \left( \dfrac { x }{ \pi } \right) } & \cot ^{ -1 }{ \left( \pi x \right) } \end{bmatrix} B=\begin{bmatrix} -\cos ^{ -1 }{ \left( x\pi \right) } & \tan ^{ -1 }{ \left( \dfrac { x }{ \pi } \right) } \\ \sin ^{ -1 }{ \left( \dfrac { x }{ \pi } \right) } & -\tan ^{ -1 }{ \left( \pi x \right) } \end{bmatrix}$$ then $$A-B$$ equal to

A
$$I$$

B
$$0$$

C
$$2I$$

D
$$\dfrac 12 I$$

Solution

We have, $$A= \begin{bmatrix} \dfrac { 1 }{ \pi } \sin ^{ -1 }{ \left( x\pi \right) } & \dfrac { 1 }{ \pi } \tan ^{ -1 }{ \left( \dfrac { x }{ \pi } \right) } \\ \dfrac { 1 }{ \pi }\sin ^{ -1 }{ \left( \dfrac { x }{ \pi } \right) } & \dfrac { 1 }{ \pi } \cot ^{ -1 }{ \left( \pi x \right) } \end{bmatrix} B=\begin{bmatrix} -\dfrac { 1 }{ \pi }\cos ^{ -1 }{ \left( x\pi \right) } & \dfrac { 1 }{ \pi }\tan ^{ -1 }{ \left( \dfrac { x }{ \pi } \right) } \\ \dfrac { 1 }{ \pi }\sin ^{ -1 }{ \left( \dfrac { x }{ \pi } \right) } & -\dfrac { 1 }{ \pi }\tan ^{ -1 }{ \left( \pi x \right) } \end{bmatrix}$$

$$\therefore A-B=\begin{bmatrix} \dfrac { 1 }{ \pi } \left( \sin ^{ -1 }{ x\pi } +\cos ^{ -1 }{ x\pi } \right) & \dfrac { 1 }{ \pi } \left( \tan ^{ -1 }{ \dfrac { x }{ \pi } } -\tan ^{ -1 }{ \dfrac { x }{ \pi } } \right) \\ \dfrac { 1 }{ \pi } \left( \sin ^{ -1 }{ \dfrac { x }{ \pi } } -\sin ^{ -1 }{ \dfrac { x }{ \pi } } \right) & \dfrac { 1 }{ \pi } \left( \cot ^{ -1 }{ \pi x } \right) +\tan ^{ -1 }{ \pi x } \end{bmatrix}$$

$$\begin{bmatrix} \dfrac { 1 }{ \pi } .\dfrac { \pi }{ 2 } & 0 \\ 0 & \dfrac { 1 }{ \pi } .\dfrac { \pi }{ 2 } \end{bmatrix}\begin{bmatrix} \because \sin ^{ -1 }{ x } +\cos ^{ -1 }{ x } =\dfrac { \pi }{ 2 } \\ and\tan ^{ -1 }{ x+ } \cot ^{ -1 }{ x } =\dfrac { \pi }{ 2 } \end{bmatrix}$$

$$=\dfrac { 1 }{ 2 } \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}=\dfrac { 1 }{ 2 } I$$
Question 4
1 / -0

If matrix $$A=[a_{ij}]_{2\times 2}$$, where $$a_{ij} *1$$ if $$1*j$$ and $$0$$ if $$i=j$$ then $$A^2$$ is equal to

A
$$I$$

B
$$A$$

C
$$0$$

D
$$None\ of\ these$$

Solution

We have, $$A=[a_{ij}]_{2\times 2}$$. where $$a_{ij}=1$$ if $$1\neq j$$ and if $$i=j$$
$$\therefore A=\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$$
And $$A^2=\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}=\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}=I$$
Question 5
1 / -0

If $$A$$ and $$B$$ are square matrices of the same order, then $$(A+B)(A-B)$$ is equal to

A
$$A^2-B^2$$

B
$$A^2-BA-AB-B^2$$

C
$$A^2-B^2+BA-AB$$

D
$$A^2-BA+B^2+AB$$

Solution

$$(A+B)(A-B)=A(A-B)+B(A-B)$$.........because matrix product is distributive.
$$=A^2-AB+BA-B^2$$
And as matrix product is not commutative, so $$AB\neq BA$$
Hence option $$C$$ is correct.
Question 6
1 / -0

If $$A=\begin{bmatrix} 2 & 1 & 3 \\ 4 & 5 & 1 \end{bmatrix}$$ and $$B=\begin{bmatrix} 2 & 3 \\ 4 & 2 \\ 1 & 5 \end{bmatrix}$$, then

A
Only $$AB$$ is defined

B
only $$BA$$ is defined

C
$$AB$$ and $$BA$$ both are defined

D
$$AB$$ and $$BA$$ both are not defined

Solution

Let $$A=[a_{ij}]_{2\times 3}, B=[b_{ij}]_{3\times 2}$$.
For, $$AB$$ columns of $$A=3$$ is equal to rows of $$B=3$$. So, $$AB$$ is defined.
For, $$BA$$ columns of $$B=2$$ is equal to rows of $$A=2$$. So, $$BA$$ is defined.
So, Both $$AB$$ and $$BA$$ are defined.
$$(C)$$ is the correct answer.
Question 7
1 / -0

If $$3\begin{bmatrix}2 & 3\\ -4 & 1\end{bmatrix} - 2 \begin{bmatrix} x& y\\ 3 & 4\end{bmatrix} = \begin{bmatrix}10 & 11\\ z & -5\end{bmatrix}$$, then $$x + y - z =$$

A
$$-21$$

B
$$-15$$

C
$$0$$

D
$$15$$

E
$$21$$
Question 8
1 / -0

If the matrix $$A = \begin{bmatrix}2 & 0 & 0 \\ 0 & 2 & 0 \\ 2 & 0 & 2\end{bmatrix}$$, then $$A^n=\begin{bmatrix}a & 0 & 0 \\ 0 & a & 0 \\ b & 0 & a\end{bmatrix}. n \in N$$ where

A
$$a = 2n, b = 2^n$$

B
$$a = 2^n, b = 2n$$

C
$$a = 2^n, b = n2^{n-1}$$

D
$$a = 2^n, b = n2^n$$

Solution

Given $$A=\begin{bmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 2 & 0 & 2 \end{bmatrix}\quad =2\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 1 & 0 & 1 \end{bmatrix}$$
$$\Rightarrow {A}^{2} =2^2\begin{bmatrix}1&0&0\\0&1&0\\1&0&1 \end{bmatrix}\begin{bmatrix}1&0&0\\0&1&0\\1&0&1\end{bmatrix}= 2^{2}\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 2 & 0 & 1 \end{bmatrix}$$
$$\Rightarrow {A}^{3} = 2^{3}\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 3 & 0 & 1 \end{bmatrix}$$ and so on
we can observe that $${A}^{n} = 2^{n}\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ n& 0 & 1 \end{bmatrix}$$
By comparing we get $$ a= 2^{n} , b= n2^{n} $$
Question 9
1 / -0

$$A $$ is of order $$m \times n$$ and $$B$$ is of order $$p \times q,$$ addition of $$A$$ and $$B$$ is possible only if

A
$$m = p$$

B
$$n = q$$

C
$$n = p$$

D
$$m = p, n = q$$

Solution

Since, the order of $$A$$ is $$m\times n $$ and order of $$ B$$ is $$p\times q$$
Then $$A+B$$ is only possible if $$m=p$$ $$\&$$ $$n=q.$$
Hence, the answer is $$m=p$$ $$\&$$ $$n=q.$$
Question 10
1 / -0

What is the inverse of the matrix
$$A=\begin{bmatrix} \cos { \theta } & \sin { \theta } & 0 \\ -\sin { \theta } & \cos { \theta } & 0 \\ 0 & 0 & 1 \end{bmatrix}$$ ?

A
$$\begin{bmatrix} \cos { \theta } & -\sin { \theta } & 0 \\ \sin { \theta } & \cos { \theta } & 0 \\ 0 & 0 & 1 \end{bmatrix}$$

B
$$\begin{bmatrix} \cos { \theta } & 0 & -\sin { \theta } \\ 0 & 1 & 0 \\ \sin { \theta } & 0 & \cos { \theta } \end{bmatrix}$$

C
$$\begin{bmatrix} 1 & 0 & 0 \\ 0 & \cos { \theta } & -\sin { \theta } \\ 0 & \sin { \theta } & \cos { \theta } \end{bmatrix}$$

D
$$\begin{bmatrix} \cos { \theta } & \sin { \theta } & 0 \\ -\sin { \theta } & \cos { \theta } & 0 \\ 0 & 0 & 1 \end{bmatrix}$$

Solution

$$A = \begin{bmatrix} \cos \theta & \sin \theta & 0 \\ -\sin \theta & \cos \theta & 0 \\ 0 & 0 & 1 \end{bmatrix}$$
Calculate first minors.
$$M_{11} = \cos \theta , M_{13} = 0, M_{22} = \cos \theta$$
$$M_{12} = -\sin \theta, M_{21} = \sin \theta, M_{23} = 0$$
$$M_{31} = 0, M_{32} = 0, M_{33} = \cos^{2}\theta + \sin^{2}\theta = 1$$
Cofactor Matrix $$= \begin{bmatrix} \cos \theta & \sin \theta & 0 \\ -\sin \theta & \cos \theta & 0\\ 0 & 0 & 1 \end{bmatrix} = C$$
$$det|A| = \cos^{2}\theta + \sin^{2}\theta = 1$$

$$adj (A) = C^{T} = \begin{bmatrix} \cos \theta & -\sin \theta & 0\\ \sin \theta & \cos \theta & 0\\ 0 & 0 & 1\end{bmatrix}$$

$$A^{-1} = \dfrac {adj(A)}{(A)} = \begin{bmatrix} \cos \theta & -\sin \theta & 0\\ \sin \theta & \cos \theta & 0\\ 0& 0 & 1\end{bmatrix}$$.

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Re-Attempt Weekly Quiz Competition

Matrices Test - 44

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Matrices Test - 44

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SHARING IS CARING

Question 1

$$ a / 125 $$

$$ 2a / 125 $$

$$ 2a / 25 $$

None of these

Solution

Question 2

Identity matrix

Symmetric matrix

Skew symmetric matrix

None of these

Solution

Question 3

$$I$$

$$0$$

$$2I$$

$$\dfrac 12 I$$

Solution

Question 4

$$I$$

$$A$$

$$0$$

$$None\ of\ these$$

Solution

Question 5

$$A^2-B^2$$

$$A^2-BA-AB-B^2$$

$$A^2-B^2+BA-AB$$

$$A^2-BA+B^2+AB$$

Solution

Question 6

Only $$AB$$ is defined

only $$BA$$ is defined

$$AB$$ and $$BA$$ both are defined

$$AB$$ and $$BA$$ both are not defined

Solution

Question 7

$$-21$$

$$-15$$

$$0$$

$$15$$

$$21$$

Question 8

$$a = 2n, b = 2^n$$

$$a = 2^n, b = 2n$$

$$a = 2^n, b = n2^{n-1}$$

$$a = 2^n, b = n2^n$$

Solution

Question 9

$$m = p$$

$$n = q$$

$$n = p$$

$$m = p, n = q$$

Solution

Question 10

$$\begin{bmatrix} \cos { \theta } & -\sin { \theta } & 0 \\ \sin { \theta } & \cos { \theta } & 0 \\ 0 & 0 & 1 \end{bmatrix}$$

$$\begin{bmatrix} \cos { \theta } & 0 & -\sin { \theta } \\ 0 & 1 & 0 \\ \sin { \theta } & 0 & \cos { \theta } \end{bmatrix}$$

$$\begin{bmatrix} 1 & 0 & 0 \\ 0 & \cos { \theta } & -\sin { \theta } \\ 0 & \sin { \theta } & \cos { \theta } \end{bmatrix}$$

$$\begin{bmatrix} \cos { \theta } & \sin { \theta } & 0 \\ -\sin { \theta } & \cos { \theta } & 0 \\ 0 & 0 & 1 \end{bmatrix}$$

Solution

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