Matrices Test

Result

Matrices Test - 19

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

If $$A = \begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ a & b & -1\end{bmatrix}$$, then $$A^2$$ is equal to

A
$$A$$

B
$$-A$$

C
null matrix

D
$$I$$

Solution

Given, $$A = \begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ a & b & -1\end{bmatrix}$$

$$A^{2}=\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ a & b & -1 \end{bmatrix}\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ a & b & -1 \end{bmatrix}$$

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

$$\Rightarrow A^{2}=I$$
Question 2
1 / -0

If $$A = \begin{bmatrix}3 & 1 \\ -1 & 2\end{bmatrix}$$, Then $$A^2$$ =

A
$$\begin{bmatrix}8 & -5 \\ -5 & 3\end{bmatrix}$$

B
$$\begin{bmatrix}8 & -5 \\ 5 & 3\end{bmatrix}$$

C
$$\begin{bmatrix}8 & -5 \\ -5 & -3\end{bmatrix}$$

D
$$\begin{bmatrix}8 & 5 \\ -5 & 3\end{bmatrix}$$

Solution

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

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

$$=\begin{bmatrix} 8 & 5 \\ -5 & 3 \end{bmatrix}$$
Question 3
1 / -0

If $$\displaystyle A=\begin{bmatrix}x &y \\z  &w \end{bmatrix},B=\begin{bmatrix}x &-y \\-z  &w \end{bmatrix}$$ and $$C=\begin{bmatrix}-2x &0 \\0  &-2w \end{bmatrix},$$ then $$A+B+C$$ is a:

A
identity matrix

B
null matrix

C
row matrix

D
column matrix

Solution

Given:
$$\displaystyle A=\begin{bmatrix}x &y \\z  &w \end{bmatrix},B=\begin{bmatrix}x &-y \\-z  &w \end{bmatrix}$$ and $$C=\begin{bmatrix}-2x &0 \\0  &-2w \end{bmatrix}$$

Consider, $$\displaystyle A+B+C=\begin{bmatrix}X &Y \\Z  &W \end{bmatrix}+\begin{bmatrix}X &-Y \\-Z  &W \end{bmatrix}+\begin{bmatrix}-2X &0 \\0  &-2W \end{bmatrix}$$

$$=\begin{bmatrix}X+X+\left ( -2X \right ) &Y+\left ( -Y \right )+0 \\Z+\left ( -Z \right )+0  &W+W+\left ( -2W \right ) \end{bmatrix}$$

$$=\begin{bmatrix}0 &0 \\0  &0 \end{bmatrix}$$
Hence, $$A+B+C$$ is a null matrix.
Question 4
1 / -0

If $$\begin{bmatrix}3 & -1 \\ 2 & 5\end{bmatrix}\begin{bmatrix}x \\ y \end{bmatrix} = \begin{bmatrix}4 \\ -3\end{bmatrix},$$ find $$x$$ and $$y$$

A
$$x=3,\space y = -1$$

B
$$x=2,\space y = 5$$

C
$$x=1,\space y = -1$$

D
$$x=-1,\space y = 1$$

Solution

$$\begin{bmatrix}3 & -1 \\ 2 & 5\end{bmatrix}\begin{bmatrix}x \\ y \end{bmatrix} = \begin{bmatrix}4 \\ -3\end{bmatrix}$$
$$\Rightarrow \begin{bmatrix} 3x-y \\ 2x+5y \end{bmatrix}=\begin{bmatrix} 4 \\ -3 \end{bmatrix}$$
$$\Rightarrow 3x-y=4, 2x+5y=-3$$
On solving we get $$x=1, y=-1$$
Question 5
1 / -0

If $$\displaystyle \:A= \left [ \begin{matrix}4 &x+2 \\2x-3  &x+1 \end{matrix} \right ]$$ is symmetric, then x=

A
3

B
5

C
2

D
4

Solution

Given $$\displaystyle \:A= \left [ \begin{matrix}4 &x+2 \\2x-3  &x+1 \end{matrix} \right ]$$ is symmteric.

$$A^{T}=A$$
$$\Rightarrow \left[ \begin{matrix} 4 & 2x-3 \\ x+2 & x+1 \end{matrix} \right] =\left [ \begin{matrix}4 &x+2 \\2x-3  &x+1 \end{matrix} \right ]$$

$$\Rightarrow x+2=2x-3$$
$$\Rightarrow x=5$$
Question 6
1 / -0

If $$\displaystyle A=\left [ a_{ij} \right ]_{m\times\:n}, B=\left [ b_{ij} \right ]_{m\times\:n},$$ then the element $$\displaystyle C_{23}$$ of the matrix $$C=A+B$$ is

A
$$\displaystyle C_{32}$$

B
$$\displaystyle a_{23}+b_{32}$$

C
$$\displaystyle a_{23}+b_{23}$$

D
$$\displaystyle a_{32}+b_{23}$$

Solution

$$\displaystyle A=\left [ a_{ij} \right ]_{m\times\:n}$$ and $$B=\left [ b_{ij} \right ]_{m\times\:n}$$

$$\therefore C=A+B$$

$$\Rightarrow \left [ c_{ij} \right ]_{m\times\:n}=\left [ a_{ij}+b_{ij} \right ]_{m\times\:n}$$

Hence, $$c_{23}=a_{23}+b_{23}$$
Question 7
1 / -0

If $$A-A'=0$$, then $$A'$$ is

A
orthogonal matrix

B
symmetric matrix

C
skew-symmetric matrix

D
triangular matrix

Solution

It is fundamental concept that,
If $$A = A'$$ then $$A=A'$$ is called symmetric matrix.
Question 8
1 / -0

For any square matrix $$A, \ A+{A}^{T}$$ is-

A
unit matrix

B
symmetric matrix

C
skew symmetric matrix

D
zero matrix

Solution

Let, $$B = A+A^T$$
$$\Rightarrow B^T = (A+A^T)^T =A^T+(A^T)^T =A^T+A =B$$
Hence $$B = A+A^T$$ is a symmetric matrix.
Question 9
1 / -0

If $$\displaystyle \begin{bmatrix} x & y \\ 1 & 6 \end{bmatrix} $$ = $$\displaystyle \begin{bmatrix} 1 & 8 \\ 1 & 6 \end{bmatrix},$$ then $$x+2y=$$

A
$$13$$

B
$$17$$

C
$$19$$

D
None of these

Solution

By equating given matrices, we have
$$x=1$$ and $$y=8$$
$$\displaystyle \therefore x+2y=17$$
Question 10
1 / -0

Given that $$\displaystyle M=\begin{bmatrix}3 &-2 \\-4 &0 \end{bmatrix}\:and\:N=\begin{bmatrix}-2 &2 \\5 &0 \end{bmatrix}$$, then $$M+N$$ is a

A
null matrix

B
unit matrix

C
$$\displaystyle \begin{bmatrix}

1 & 0 \\1

&0

\end{bmatrix}$$

D
$$\displaystyle \begin{bmatrix}

0 &1 \\1

&1

\end{bmatrix}$$

Solution

Given: $$\displaystyle M=\begin{bmatrix}3 &-2 \\-4 &0 \end{bmatrix}\:and\:N=\begin{bmatrix}-2 &2 \\5 &0 \end{bmatrix}$$
Adding $$M$$ and $$N$$, we get
$$M+N=\begin{bmatrix} 3 & -2 \\ -4 & 0 \end{bmatrix}+\begin{bmatrix} -2 & 2 \\ 5 & 0 \end{bmatrix}=\begin{bmatrix} 1 & 0 \\ 1 & 0 \end{bmatrix}$$

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Matrices Test - 19

Result

Matrices Test - 19

Score

-

Rank

-

SHARING IS CARING

Question 1

$$A$$

$$-A$$

null matrix

$$I$$

Solution

Question 2

$$\begin{bmatrix}8 & -5 \\ -5 & 3\end{bmatrix}$$

$$\begin{bmatrix}8 & -5 \\ 5 & 3\end{bmatrix}$$

$$\begin{bmatrix}8 & -5 \\ -5 & -3\end{bmatrix}$$

$$\begin{bmatrix}8 & 5 \\ -5 & 3\end{bmatrix}$$

Solution

Question 3

identity matrix

null matrix

row matrix

column matrix

Solution

Question 4

$$x=3,\space y = -1$$

$$x=2,\space y = 5$$

$$x=1,\space y = -1$$

$$x=-1,\space y = 1$$

Solution

Question 5

3

5

2

4

Solution

Question 6

$$\displaystyle C_{32}$$

$$\displaystyle a_{23}+b_{32}$$

$$\displaystyle a_{23}+b_{23}$$

$$\displaystyle a_{32}+b_{23}$$

Solution

Question 7

orthogonal matrix

symmetric matrix

skew-symmetric matrix

triangular matrix

Solution

Question 8

unit matrix

symmetric matrix

skew symmetric matrix

zero matrix

Solution

Question 9

$$13$$

$$17$$

$$19$$

None of these

Solution

Question 10

null matrix

unit matrix

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

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

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

-

Results Announcement on: coming Monday

Stay Tuned: We’ll update you on your result via email or SMS.

$$\displaystyle \begin{bmatrix}

1 & 0 \\1

&0

\end{bmatrix}$$

$$\displaystyle \begin{bmatrix}

0 &1 \\1

&1

\end{bmatrix}$$