Matrices Test

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Matrices Test - 32

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Weekly Quiz Competition

Question 1
1 / -0

If $$A =\begin{bmatrix} ab&b^2 \\-a^2 &-ab \end{bmatrix}$$, then $$A^2$$ is equal

A
$$O$$

B
$$I$$

C
$$-I$$

D
none of these

Solution

$$A =\begin{bmatrix} ab&b^2 \\-a^2 &-ab \end{bmatrix}$$
$$A^2=\begin{bmatrix} ab&b^2 \\-a^2 &-ab \end{bmatrix}\begin{bmatrix} ab&b^2 \\-a^2 &-ab \end{bmatrix}$$
$$=\begin{bmatrix} a^{ 2 }b^{ 2 }-a^{ 2 }b^{ 2 } & ab^{ 3 }-ab^{ 3 } \\ -a^{ 3 }b+a^{ 3 }b & -a^{ 2 }b^{ 2 }+a^{ 2 }b^{ 2 } \end{bmatrix}=\begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$$
$$\therefore A^2=\begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} =O$$
Hence, option A.
Question 2
1 / -0

If $$\displaystyle A=\begin{bmatrix} 1 & \frac { 1 }{ 2 } \\ 0 & 1 \end{bmatrix}$$ then $${ A }^{ 64 }$$ is

A
$$\begin{bmatrix} 1 & 32 \\ 32 & 1 \end{bmatrix}$$

B
$$\begin{bmatrix} 1 & 0 \\ 32 & 1 \end{bmatrix}$$

C
$$\begin{bmatrix} 1 & 32 \\ 0 & 1 \end{bmatrix}$$

D
none of these

Solution

$$\displaystyle { A }^{ 2 }=\begin{bmatrix} 1 & \frac { 1 }{ 2 } \\ 0 & 1 \end{bmatrix}\begin{bmatrix} 1 & \frac { 1 }{ 2 } \\ 0 & 1 \end{bmatrix}=\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}\\ { A }^{ 4 }=\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}=\begin{bmatrix} 1 & 2 \\ 0 & 1 \end{bmatrix}\\ { A }^{ 8 }=\begin{bmatrix} 1 & 2 \\ 0 & 1 \end{bmatrix}\begin{bmatrix} 1 & 2 \\ 0 & 1 \end{bmatrix}=\begin{bmatrix} 1 & 4 \\ 0 & 1 \end{bmatrix}$$
Similarly, $${ A }^{ 64 }=\begin{bmatrix} 1 & 32 \\ 0 & 1 \end{bmatrix}$$
Question 3
1 / -0

Find the value of $$x$$ and $$y$$ that satisfy the equation:
$$\begin{bmatrix} 3&-2 \\3 &0 \\2 &4 \end{bmatrix}
\begin{bmatrix} y&y \\x &x \end{bmatrix}=\begin{bmatrix} 3&3 \\3y &3y \\10 &10 \end{bmatrix}$$

A
$$x=3/2, \:y = 2$$

B
$$x=2, \:y = 3/2$$

C
$$x=3, \:y = 2$$

D
$$x=2, \:y = 3$$

Solution

$$\left[ \begin{matrix} 3\quad & -2 \\ 3 & \quad 0 \\ 2 & \quad 4 \end{matrix} \right] \begin{bmatrix} y\quad & y \\ x\quad & x \end{bmatrix}=\left[ \begin{matrix} 3\quad & 3 \\ 3y\quad & 3y \\ 10\quad & 10 \end{matrix} \right] \\ \Rightarrow \left[ \begin{matrix} 3y-2x\quad & 3y-2x \\ 3y\quad & 3y \\ 2y+4x\quad & 2y+4x \end{matrix} \right] =\left[ \begin{matrix} 3 & \quad 3 \\ 3y\quad & 3y \\ 10\quad & 10 \end{matrix} \right] $$
$$\therefore 3y-2x=3$$ and $$2y+4x=10$$
$$\displaystyle \Rightarrow x=\frac { 3 }{ 2 } ,y=2$$
Question 4
1 / -0

Let $$A$$ be square matrix. Then which of the following is not a symmetric matrix.

A
$$A+A'$$

B
$$AA'$$

C
$$A'A$$

D
$$A-A'$$

Solution

If $$A$$ is a square matrix, then and if $$A'$$ represents its transpose, then
$$A+A'$$ is symmetric and $$A-A'$$ is skew symmetric.
Hence matrix A can be written as
$$A=(\dfrac{A+A'}{2})+(\dfrac{A-A'}{2})$$

Therefore of all the above matrix,
$$A-A'$$ is not symmetric.
Question 5
1 / -0

If A is any square matrix then (1/2) $$\displaystyle \left ( A+A^{T} \right )$$ is a _____ matrix

A
symmetric

B
skew symmetric

C
scalar

D
identity

Solution

Given $$\cfrac { 1 }{ 2 } \left( A+{ A }^{ T } \right) $$
Let $$A+{ A }^{ T }=P$$
Then $${ P }^{ T }={ \left( A+{ A }^{ T } \right) }^{ T }=A+{ A }^{ T }=P$$
$$\because { P }^{ T }=P$$
$$\Rightarrow { \left( A+{ A }^{ T } \right) }^{ T }=A+{ A }^{ T }$$
$$\Rightarrow \cfrac { 1 }{ 2 } \left( A+{ A }^{ T } \right) $$is symmetric
OPTION A
Question 6
1 / -0

If $$A=\begin{bmatrix} 4 & 2 \\ -1 & 1 \end{bmatrix}$$, then $$(A-2I)(A-3I)=$$

A
$$0$$

B
$$A$$

C
$$I$$

D
$$5I$$

Solution

$$A=\begin{bmatrix} 4 & 2 \\ -1 & 1 \end{bmatrix}$$
$$\Rightarrow$$ $$(A-2I)(A-3I)={A}^{2}-5AI+6{I}^{2}$$
$${A}^{2}-5A+6I$$
by Hemilton rule $$\begin{vmatrix} 4-\lambda & 2 \\ -1 & 1-\lambda \end{vmatrix}=0$$
after expanding determinant we got;
$$\Rightarrow 4-5\lambda +{ \lambda }^{ 2 }+2=0$$
$$\Rightarrow { \lambda }^{ 2 }-5\lambda +6=0$$
$$\Rightarrow { A }^{ 2 }-5A+6I=0$$
Ans: 0
Or
It can be done using scalar multiplication and followed by matrix multiplication.
Question 7
1 / -0

Inverse of a diagonal matrix is

A
Symmetric

B
A skew-symmetric

C
A diagonal matrix

D
None of these

Solution

Consider a diagonal
$$\begin{bmatrix} 2 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 3 \end{bmatrix}$$
Minor of the matrix
$$\begin{bmatrix} 4 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 2 \end{bmatrix}$$
Co factor of matrix
$$\begin{bmatrix} 4 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 2 \end{bmatrix}\begin{bmatrix} + & - & + \\ - & + & - \\ + & - & + \end{bmatrix}$$
$$\begin{bmatrix} 4 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 2 \end{bmatrix}$$
Adjoint of matrix$$=$$ Transpose of co factor$$=\begin{bmatrix} 4 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 2 \end{bmatrix}$$
Set of matrix considered$$=6$$
$${ A }^{ -1 }=\cfrac { Adjoint\quad (A) }{ det(A) } $$
$$=\cfrac { 1 }{ 6 } \begin{bmatrix} 4 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 2 \end{bmatrix}$$
Clearly $${ A }^{ -1 }$$ is also diagonal matrix
$$\therefore $$ Inverse of a diagonal matrix is a diagonal matrix
Option C is correct
Question 8
1 / -0

If $$A+I=\begin{bmatrix} 3 & -2 \\ 4 & 1 \end{bmatrix}$$, then $$\left( A+I \right) \cdot \left( A-I \right) $$ is equal to

A
$$\begin{bmatrix} -5 & -4 \\ 8 & -9 \end{bmatrix}$$

B
$$\begin{bmatrix} -5 & 4 \\ -8 & 9 \end{bmatrix}$$

C
$$\begin{bmatrix} 5 & 4 \\ 8 & 9 \end{bmatrix}$$

D
$$\begin{bmatrix} -5 & -4 \\ -8 & -9 \end{bmatrix}$$

Solution

Given, $$A+I=\begin{bmatrix} 3 & -2 \\ 4 & 1 \end{bmatrix}$$
$$\therefore A-I=A+I-2I$$
$$=\begin{bmatrix} 3 & -2 \\ 4 & 1 \end{bmatrix}-\begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix}$$
$$=\begin{bmatrix} 1 & -2 \\ 4 & -1 \end{bmatrix}$$
$$\therefore \left( A+I \right) \cdot \left( A-I \right) =\begin{bmatrix} 3 & -2 \\ 4 & 1 \end{bmatrix}\begin{bmatrix} 1 & -2 \\ 4 & -1 \end{bmatrix}$$
$$=\begin{bmatrix} 3-8 & -6+2 \\ 4+4 & -8-1 \end{bmatrix}$$
$$=\begin{bmatrix} -5 & -4 \\ 8 & -9 \end{bmatrix}$$
Question 9
1 / -0

Two matrices $$A$$ and $$B$$ are multiplied to get $$AB$$ if

A
both are rectangular

B
both have same order

C
number of columns of $$A$$ is equal to rows of $$B$$

D
number of rows of $$A$$ is equal to no of columns of $$B$$

Solution

According to condition of multiplication of two matrices:
The columns in first matrix should be same in number as the rows in the second matrix [For $$AB$$ to be defined, we must have columns in $$A=$$ rows in $$B$$]
Hence option C is true.
Question 10
1 / -0

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

A
Null matrix

B
Itself $$A$$

C
Unit matrix

D
Scalar matrix

Solution

Given, $$A=\begin{bmatrix} 2 & -1 & 1 \\ -2 & 3 & -2 \\ -4 & 4 & -3 \end{bmatrix}$$
$$\therefore { A }^{ 2 }=\begin{bmatrix} 2 & -1 & 1 \\ -2 & 3 & -2 \\ -4 & 4 & -3 \end{bmatrix}\begin{bmatrix} 2 & -1 & 1 \\ -2 & 3 & -2 \\ -4 & 4 & -3 \end{bmatrix}$$
$$=\begin{bmatrix} 4+2-4 & -2-3+4 & 2+2-3 \\ -4-6+8 & 2+9-8 & -2-6+6 \\ -8-8+12 & 4+12-12 & -4-8+9 \end{bmatrix}$$
$$=\begin{bmatrix} 2 & -1 & 1 \\ -2 & 3 & -2 \\ -4 & 4 & -3 \end{bmatrix}=A$$

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Matrices Test - 32

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Matrices Test - 32

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SHARING IS CARING

Question 1

$$O$$

$$I$$

$$-I$$

none of these

Solution

Question 2

$$\begin{bmatrix} 1 & 32 \\ 32 & 1 \end{bmatrix}$$

$$\begin{bmatrix} 1 & 0 \\ 32 & 1 \end{bmatrix}$$

$$\begin{bmatrix} 1 & 32 \\ 0 & 1 \end{bmatrix}$$

none of these

Solution

Question 3

$$x=3/2, \:y = 2$$

$$x=2, \:y = 3/2$$

$$x=3, \:y = 2$$

$$x=2, \:y = 3$$

Solution

Question 4

$$A+A'$$

$$AA'$$

$$A'A$$

$$A-A'$$

Solution

Question 5

symmetric

skew symmetric

scalar

identity

Solution

Question 6

$$0$$

$$A$$

$$I$$

$$5I$$

Solution

Question 7

Symmetric

A skew-symmetric

A diagonal matrix

None of these

Solution

Question 8

$$\begin{bmatrix} -5 & -4 \\ 8 & -9 \end{bmatrix}$$

$$\begin{bmatrix} -5 & 4 \\ -8 & 9 \end{bmatrix}$$

$$\begin{bmatrix} 5 & 4 \\ 8 & 9 \end{bmatrix}$$

$$\begin{bmatrix} -5 & -4 \\ -8 & -9 \end{bmatrix}$$

Solution

Question 9

both are rectangular

both have same order

number of columns of $$A$$ is equal to rows of $$B$$

number of rows of $$A$$ is equal to no of columns of $$B$$

Solution

Question 10

Null matrix

Itself $$A$$

Unit matrix

Scalar matrix

Solution

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