Matrices Test

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Matrices Test - 33

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Weekly Quiz Competition

Question 1
1 / -0

If $$\displaystyle A=\left[ \begin{matrix} 2 \\ 0 \\ 0 \end{matrix}\,\,\,\begin{matrix} 0 \\ 2 \\ 0 \end{matrix}\,\,\,\begin{matrix} 0 \\ 0 \\ 2 \end{matrix} \right] $$ then $$\displaystyle { A }^{ 5 }=$$

A
$$\displaystyle 5A$$

B
$$\displaystyle 10A$$

C
$$\displaystyle 16A$$

D
$$\displaystyle 32A$$

Solution

Given, $$A= \displaystyle \left[\begin{matrix} 2 & 0 & 0\\ 0 & 2 & 0\\ 0 & 0 & 2 \end{matrix} \right]$$
     $$\Rightarrow A= \displaystyle 2\left[\begin{matrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{matrix} \right]$$
     $$\Rightarrow A= \displaystyle 2I$$
     $$\Rightarrow A^5= \displaystyle 2^5I^5$$
     $$\Rightarrow A^5= \displaystyle 2^5I$$
     $$\Rightarrow A^5= \displaystyle 2^4.2I$$
     $$\Rightarrow A^5= \displaystyle 2^4. \displaystyle 2\left[\begin{matrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{matrix} \right]$$
   $$\Rightarrow A^5= \displaystyle 16\displaystyle \left[\begin{matrix} 2 & 0 & 0\\ 0 & 2 & 0\\ 0 & 0 & 2 \end{matrix} \right]$$.
$$\Rightarrow A^5= \displaystyle 16A $$
Option $$C$$ is correct.
Question 2
1 / -0

If $$A$$ and $$B$$ are symmetric matrices, then $$ABA$$ is

A
Symmetric

B
Skew-symmetric

C
Diagonal

D
Triangular

Solution

Given $$A$$ and $$B$$ are symmetric matrices.
$$\Rightarrow A^T = A$$ and $$B^T = B$$

Now, take $$(ABA)^T$$
$$\Rightarrow (ABA)^T = A^T B^T A^T$$
$$\Rightarrow (ABA)^T = ABA$$
Hence, $$ABA$$ is also a symmetric matrix.
Question 3
1 / -0

If $$A=\begin{bmatrix}1 & 1\\ 1& 1\end{bmatrix}$$, then $$A^{100}$$ is equal to.

A
$$2^{100}A$$

B
$$2^{99}A$$

C
$$100A$$

D
$$299A$$

Solution

Given, $$A=\begin{bmatrix}1&1\\1&1\end{bmatrix}$$
$$\therefore A^2=A\cdot A=\begin{bmatrix}1&1\\1&1\end{bmatrix}\begin{bmatrix}1&1\\1&1\end{bmatrix}$$
$$=2\begin{bmatrix}1&1\\1&1\end{bmatrix}=2A$$
Now, $$A^4=A^2\cdot A^2=2A\cdot 2A$$
$$=4A^2=4\times 2A$$
$$=8A=2^3A$$
Similarly, $$A^8=2^7A$$
$$\therefore A^{100}=2^{99}A$$
Question 4
1 / -0

If $$\displaystyle A=\begin{bmatrix} 0 & 0 & 1\\ 0 & 1&0 \\ 1& 0 & 0\end{bmatrix}$$, then $$A^{-1}$$ is.

A
$$-A$$

B
$$A$$

C
$$1$$

D
None of these

Solution

We have, $$A=\begin{bmatrix} 0 & 0&1\\ 0 &1 &0\\ 1&0 &0\end{bmatrix}$$
$$\Rightarrow |A|=0(0-0)-0(0-0)+1(0-1)$$
$$\Rightarrow |A|=-1$$
and cofactors of A are
$$A_{11}=0, A_{12}=0, A_{13}=-1,$$
$$A_{21}=0, A_{22}=-1, A_{23}=0,$$
$$A_{31}=-1, A_{32}=0, A_{33}=0$$
$$\therefore A^{-1}=\displaystyle\frac{adj(A)}{|A|}$$
$$=-\displaystyle\frac{1}{1}\begin{bmatrix} 0 & 0 & -1\\0 & -1 &0\\ -1 &0 &0\end{bmatrix}$$
Question 5
1 / -0

If $$A=\begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$$, then $$A^2 - 5A$$ is equal to

A
$$2I$$

B
$$-2I$$

C
$$3I$$

D
Null matrix

Solution

$$ A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} ,$$
Also, $$A^{2} = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} = \begin{bmatrix} 1\times1+2\times 3\,\,\,\,\,1\times 2+2\times 4 \\ 3\times 1+4\times 3\,\,\,\,\,3\times2+4\times 4 \end{bmatrix} $$
$$ \Rightarrow A^{2} = \begin{bmatrix} 7 & 10 \\ 15 & 22 \end{bmatrix} $$
$$ 5A = 5\times \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} = \begin{bmatrix} 5 & 10 \\ 15 & 20 \end{bmatrix} $$
$$ A^{2} -5A = \begin{bmatrix} 7 & 10 \\ 15 & 22 \end{bmatrix}-\begin{bmatrix} 5 & 10 \\ 15 & 20 \end{bmatrix} = \begin{bmatrix} 2 & 0 \\ 0 & +2 \end{bmatrix} = 2 \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} $$
$$ \therefore A^{2}-5A = 2I $$
Question 6
1 / -0

If $$\begin{bmatrix} 1 & 2 \\ 3 & -5 \end{bmatrix}$$, then $${A}^{-1}$$ is equal to

A
$$\begin{bmatrix} \cfrac { 5 }{ 11 } & \cfrac { 2 }{ 11 } \\ \cfrac { 3 }{ 11 } & -\cfrac { 1 }{ 11 } \end{bmatrix}$$

B
$$\begin{bmatrix} -\cfrac { 5 }{ 11 } & -\cfrac { 2 }{ 11 } \\ -\cfrac { 3 }{ 11 } & -\cfrac { 1 }{ 11 } \end{bmatrix}$$

C
$$\begin{bmatrix} \cfrac { 5 }{ 11 } & \cfrac { 2 }{ 11 } \\ \cfrac { 3 }{ 11 } & \cfrac { 1 }{ 11 } \end{bmatrix}$$

D
$$\begin{bmatrix} 5 & 2 \\ 3 & -1 \end{bmatrix}$$

Solution

Since $$A=\begin{bmatrix} 1 & 2 \\ 3 & -5 \end{bmatrix}$$

$$\therefore \left| A \right| =\begin{bmatrix} 1 & 2 \\ 3 & -5 \end{bmatrix}=-5-6=-11$$

and $$adj(A)=\begin{bmatrix} -5 & -2 \\ -3 & 1 \end{bmatrix}$$

$$\therefore { A }^{ -1 }=\cfrac { 1 }{ \left| A \right| } adj(A)$$

$$=-\cfrac { 1 }{ 11 } \begin{bmatrix} -5 & -2 \\ -3 & 1 \end{bmatrix}=\cfrac { 1 }{ 11 } \begin{bmatrix} 5 & 2 \\ 3 & -1 \end{bmatrix}$$

$$\quad =\begin{bmatrix} \cfrac { 5 }{ 11 } & \cfrac { 2 }{ 11 } \\ \cfrac { 3 }{ 11 } & -\cfrac { 1 }{ 11 } \end{bmatrix}$$
Question 7
1 / -0

Inverse of the matrix $$\begin{bmatrix} \cos 2\theta & -\sin 2\theta\\ \sin 2\theta & \cos 2\theta\end{bmatrix}$$ is.

A
$$\begin{bmatrix} \cos 2\theta & -\sin 2\theta\\ \sin 2\theta & \cos 2\theta\end{bmatrix}$$

B
$$\begin{bmatrix} \cos 2\theta & \sin 2\theta\\ \sin 2\theta & -\cos 2\theta\end{bmatrix}$$

C
$$\begin{bmatrix} \cos 2\theta & \sin 2\theta\\ \sin 2\theta & \cos 2\theta\end{bmatrix}$$

D
$$\begin{bmatrix} \cos 2\theta & \sin 2\theta\\ -\sin 2\theta & \cos 2\theta\end{bmatrix}$$

Solution

Let $$A=\begin{bmatrix} \cos 2\theta & -\sin 2\theta\\ \sin 2\theta & \cos 2\theta\end{bmatrix}$$
$$\therefore |A|=\cos^2 2\theta+\sin^2 2\theta=1$$
and $$adj(A)=\begin{bmatrix}\cos 2\theta &\sin 2\theta\\ -\sin 2\theta & \cos 2\theta\end{bmatrix}$$
$$\therefore A^{-1}=\displaystyle\frac{1}{1}\begin{bmatrix}\cos 2\theta & \sin 2\theta\\ -\sin 2\theta &\cos 2\theta\end{bmatrix}$$
$$=\begin{bmatrix}\cos 2\theta & \sin 2\theta\\ -\sin 2\theta & \cos 2\theta\end{bmatrix}$$
Question 8
1 / -0

Let $$A=\begin{bmatrix} 1 & -1 & -1 \\ 2 & 1 & -3 \\ 1 & 1 & 1 \end{bmatrix}$$ and $$10B=\begin{bmatrix} 4 & 2 & 2 \\ -5 & 0 & \alpha \\ 1 & -2 & 3 \end{bmatrix}$$, if $$B$$ is the inverse of matrix $$A$$, then $$\alpha $$ is

A
$$-2$$

B
$$1$$

C
$$2$$

D
$$5$$

Solution

Since, $$B$$ is the inverse of $$A$$.
ie, $$B=10{ A }^{ -1 }$$
$$\therefore \left( 10 \right) { A }^{ -1 }=\begin{bmatrix} 4 & 2 & 2 \\ -5 & 0 & \alpha \\ 1 & -2 & 3 \end{bmatrix}$$
$$\therefore \left( 10 \right) { A }^{ -1 }\cdot A=\begin{bmatrix} 4 & 2 & 2 \\ -5 & 0 & \alpha \\ 1 & -2 & 3 \end{bmatrix}A$$
$$\Rightarrow 10I=\begin{bmatrix} 4 & 2 & 2 \\ -5 & 0 & \alpha \\ 1 & -2 & 3 \end{bmatrix}\begin{bmatrix} 1 & -1 & 1 \\ 2 & 1 & -3 \\ 1 & 1 & 1 \end{bmatrix}$$
$$\Rightarrow \begin{bmatrix} 10 & 0 & 0 \\ 0 & 10 & 0 \\ 0 & 0 & 10 \end{bmatrix}=\begin{bmatrix} 10 & 0 & 0 \\ -5+\alpha & 5+\alpha & -5+\alpha \\ 0 & 0 & 10 \end{bmatrix}$$
$$\Rightarrow 5+\alpha =10$$
$$\Rightarrow \alpha =5$$
Question 9
1 / -0

If $$\begin{bmatrix}1 & 1 &1\\ 1&-2 &-2\\1 & 3 &1\end{bmatrix}$$ $$\begin{bmatrix} x\\ y\\z\end{bmatrix}=\begin{bmatrix} 0 \\ 3\\4\end{bmatrix}$$, then $$\begin{bmatrix} x\\ y\\z\end{bmatrix}$$ is equal to.

A
$$\begin{bmatrix} 0\\1\\1\end{bmatrix}$$

B
$$\begin{bmatrix} 1\\2\\-3\end{bmatrix}$$

C
$$\begin{bmatrix} 5\\-2\\1\end{bmatrix}$$

D
$$\begin{bmatrix} 1\\-2\\3\end{bmatrix}$$

Solution

Given, $$\begin{bmatrix} 1&1&1\\1&-2&-2\\1&2&1\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}0\\3\\4\end{bmatrix}$$
$$\therefore x+y+z=0$$,
$$x-2y-2z=3$$,
$$x+3y=z=4$$
On solving these equations, we get
$$x=1, y=2$$ and $$z=-3$$
$$\therefore \begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}1\\2\\-3\end{bmatrix}$$
Question 10
1 / -0

If $$A=\begin{bmatrix} 1 & -2 & 1 \\ 2 & 1 & 3 \end{bmatrix}$$ and $$B=\begin{bmatrix} 2 & 1 \\ 3 & 2 \\ 1 & 1 \end{bmatrix}$$, then $${ \left( AB \right) }^{ T }$$ is equal to

A
$$\begin{bmatrix} -3 & -2 \\ 10 & 7 \end{bmatrix}$$

B
$$\begin{bmatrix} -3 & 10 \\ -2 & 7 \end{bmatrix}$$

C
$$\begin{bmatrix} -3 & 7 \\ 10 & 2 \end{bmatrix}$$

D
None of these

Solution

Given, $$A=\begin{bmatrix} 1 & -2 & 1 \\ 2 & 1 & 3 \end{bmatrix},B=\begin{bmatrix} 2 & 1 \\ 3 & 2 \\ 1 & 1 \end{bmatrix}$$
$$\therefore AB=\begin{bmatrix} 2-6+1 & 1-4+1 \\ 4+3+3 & 2+2+3 \end{bmatrix}=\begin{bmatrix} -3 & -2 \\ 10 & 7 \end{bmatrix}$$
Now, $${ \left( AB \right) }^{ T }=\begin{bmatrix} -3 & 10 \\ -2 & 7 \end{bmatrix}$$

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Re-Attempt Weekly Quiz Competition

Matrices Test - 33

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Matrices Test - 33

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SHARING IS CARING

Question 1

$$\displaystyle 5A$$

$$\displaystyle 10A$$

$$\displaystyle 16A$$

$$\displaystyle 32A$$

Solution

Question 2

Symmetric

Skew-symmetric

Diagonal

Triangular

Solution

Question 3

$$2^{100}A$$

$$2^{99}A$$

$$100A$$

$$299A$$

Solution

Question 4

$$-A$$

$$A$$

$$1$$

None of these

Solution

Question 5

$$2I$$

$$-2I$$

$$3I$$

Null matrix

Solution

Question 6

$$\begin{bmatrix} \cfrac { 5 }{ 11 } & \cfrac { 2 }{ 11 } \\ \cfrac { 3 }{ 11 } & -\cfrac { 1 }{ 11 } \end{bmatrix}$$

$$\begin{bmatrix} -\cfrac { 5 }{ 11 } & -\cfrac { 2 }{ 11 } \\ -\cfrac { 3 }{ 11 } & -\cfrac { 1 }{ 11 } \end{bmatrix}$$

$$\begin{bmatrix} \cfrac { 5 }{ 11 } & \cfrac { 2 }{ 11 } \\ \cfrac { 3 }{ 11 } & \cfrac { 1 }{ 11 } \end{bmatrix}$$

$$\begin{bmatrix} 5 & 2 \\ 3 & -1 \end{bmatrix}$$

Solution

Question 7

$$\begin{bmatrix} \cos 2\theta & -\sin 2\theta\\ \sin 2\theta & \cos 2\theta\end{bmatrix}$$

$$\begin{bmatrix} \cos 2\theta & \sin 2\theta\\ \sin 2\theta & -\cos 2\theta\end{bmatrix}$$

$$\begin{bmatrix} \cos 2\theta & \sin 2\theta\\ \sin 2\theta & \cos 2\theta\end{bmatrix}$$

$$\begin{bmatrix} \cos 2\theta & \sin 2\theta\\ -\sin 2\theta & \cos 2\theta\end{bmatrix}$$

Solution

Question 8

$$-2$$

$$1$$

$$2$$

$$5$$

Solution

Question 9

$$\begin{bmatrix} 0\\1\\1\end{bmatrix}$$

$$\begin{bmatrix} 1\\2\\-3\end{bmatrix}$$

$$\begin{bmatrix} 5\\-2\\1\end{bmatrix}$$

$$\begin{bmatrix} 1\\-2\\3\end{bmatrix}$$

Solution

Question 10

$$\begin{bmatrix} -3 & -2 \\ 10 & 7 \end{bmatrix}$$

$$\begin{bmatrix} -3 & 10 \\ -2 & 7 \end{bmatrix}$$

$$\begin{bmatrix} -3 & 7 \\ 10 & 2 \end{bmatrix}$$

None of these

Solution

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