Matrices Test

Result

Matrices Test - 28

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 $$\begin{bmatrix} x & -3 \\ -9 & y \end{bmatrix}\begin{bmatrix} 4 & -3 \\ 9 & 7 \end{bmatrix}=\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix},$$ then $$x=$$ .......... $$, y $$ $$=$$ ........

A
$$7,4$$

B
$$4,7$$

C
$$8,9$$

D
$$9,8$$

Solution

$$\begin{bmatrix} x & -3 \\ -9 & y \end{bmatrix}\begin{bmatrix} 4 & -3 \\ 9 & 7 \end{bmatrix}=\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\\ \Rightarrow \begin{bmatrix} 4x-27 & -3x+(-21) \\ -36+9y & +27+7y \end{bmatrix}=\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$$
Using conditions for equality of matrices we can write
$$4x-27=1\\ \Rightarrow x=7\quad \quad -(i)\\ -(3x+21)=0\\ \Rightarrow x=-7\quad \quad -(ii)\\ -36+9y=0\\ \Rightarrow y=4\quad \quad -(iii)\\ 27+7y=1\\ \Rightarrow y=-4\quad \quad -(iv)$$
Possible answer is $$x=7$$ & $$y=4$$
Question 2
1 / -0

If $$A= \begin{bmatrix}
2 & 3   \\
3 & 2
\end{bmatrix},$$ $$B= \begin{bmatrix}
2 & 1   \\
3 & 5
\end{bmatrix}$$ and $$C= \begin{bmatrix}
0 & 1   \\
1 & 2
\end{bmatrix},$$ then $$\left ( AB \right )\times C=$$

A
$$\begin{bmatrix}

8 & 11 \\

18 & 23

\end{bmatrix}$$

B
$$\begin{bmatrix}

11 & 30 \\

23 & 60

\end{bmatrix}$$

C
$$\begin{bmatrix}

17 & 47 \\

13 & 38

\end{bmatrix}$$

D
$$\begin{bmatrix}

0 & 1 \\

1 &2

\end{bmatrix}$$

Solution
Question 3
1 / -0

If $$A=\begin{bmatrix} 2 & 1 \\ 1 & 3 \end{bmatrix}$$ , $$B=\begin{bmatrix} 3 & 2 & 0 \\ 1 & 0 & 4 \end{bmatrix}$$, then $$AB=$$ is

A
$$\begin{bmatrix} 6 & 2 & 5 \\ 7 & 2 & 3 \end{bmatrix}$$

B
$$\begin{bmatrix} 2 & 1 & 5 \\ 0 & 1 & 7 \end{bmatrix}$$

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

D
$$\begin{bmatrix} 7 & 4 & 4 \\ 6 & 2 & 12 \end{bmatrix}$$

Solution

Given $$A= \begin{bmatrix} 2 &1 \\1 & 3 \end{bmatrix}$$ and $$B=\begin{bmatrix} 3&2&0\\1&0&4 \end{bmatrix}$$
$$AB=\begin{bmatrix} 2\times 3 +1\times1 & 2\times 2+1\times0 & 2\times0+1\times4 \\ 1\times3+3\times1 & 1\times2+3\times0 & 1\times0+3\times4 \end{bmatrix}$$
$$ =\begin{bmatrix} 6+1 & 4+0 & 0+4 \\ 3+3 & 2+0 & 0+12 \end{bmatrix}$$
$$=\begin{bmatrix} 7 & 4 & 4 \\ 6 & 2 & 12 \end{bmatrix}$$

Therefore the correct option is $$(D)$$
Question 4
1 / -0

Let $$A=\begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}\ and\ B=\begin{pmatrix} a & 0 \\ 0 & b \end{pmatrix}, a,b\in N.$$ Then:

A
there exists exactly one B such that $$AB=BA$$

B
there exist exactly infinitely many B's such that $$AB=BA$$

C
there cannot exist any B such that $$AB=BA$$

D
there exist more than one but finite number of B's such that $$AB=BA$$

Solution

Given
$$A=\begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}B=\begin{bmatrix} a & 0 \\ 0 & b \end{bmatrix}$$
Finding AB
$$AB=\begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}\begin{bmatrix} a & 0 \\ 0 & b \end{bmatrix}=\begin{bmatrix} a+2\times 0 & 1\times 0+2\times b \\ 3\times a+4\times 0 & 3\times 0+4\times b \end{bmatrix}$$
$$AB=\begin{bmatrix} a & 2b \\ 3a & 4b \end{bmatrix}$$
Finding BA
$$BA=\begin{bmatrix} a & 0 \\ 0 & b \end{bmatrix}\begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}=\begin{bmatrix} a+0\times 3 & 2a+0\times 4 \\ 0\times 1+3b & 0\times 2+4b \end{bmatrix}$$
$$BA=\begin{bmatrix} a & 2a \\ 3b & 4b \end{bmatrix}$$
If $$AB=BA$$
$$\begin{bmatrix} a & 2b \\ 3a & 4b \end{bmatrix}=\begin{bmatrix} a & 2a \\ 3b & 4b \end{bmatrix}$$
Comparing each element we get
$$a=a,2b=2a$$
$$\Rightarrow a=b$$
$$\therefore $$There are infinitely many b's
for which $$AB=BA$$
Answer=B
Question 5
1 / -0

If A = $$\begin{bmatrix}
2\ \ \ 4 \\
3\ \ 5
\end{bmatrix},$$ $$B= \begin{bmatrix}
x\ \ \ y \\
6\ \ \ 5
\end{bmatrix}$$ and $$AB= \begin{bmatrix}8\ \ \ 2 \\
6\ \ \ -2
\end{bmatrix}$$
then $$x= $$ ______, and $$y = $$ _____.

A
$$- 9, -8$$

B
$$8, 9$$

C
$$- 8, -9$$

D
$$9, 8$$

Solution

$$ A = \begin{bmatrix} 3 & 4 \\ 3 & 5 \end{bmatrix} $$;
$$ B = \begin{bmatrix} x & y \\ 6 & 5 \end{bmatrix} $$
and $$ AB = \begin{bmatrix} 8 & 2 \\ 6 & -2 \end{bmatrix} $$
Now,
$$ AB = \begin{bmatrix} 2 & 4 \\ 3 & 5 \end{bmatrix} \begin{bmatrix} x & y \\ 6 & 5 \end{bmatrix} \\ = \begin{bmatrix} 2x+24 & 2y + 20 \\ 3x + 30 & 3y + 25 \end{bmatrix} \, \, \textrm{ [By the property of multiplication of matrix] }$$

Now, comparing the given value we get,

$$ \begin{bmatrix} 8 & 2 \\ 6 & -2 \end{bmatrix} = \begin{bmatrix} 2x+24 & 2y + 20 \\ 3x+30 & 3y + 25 \end{bmatrix} $$

Equating each element we get,

$$ 2x + 24 = 8 \Rightarrow x = -8 \\ 2y + 20 = 2 \\ \Rightarrow y = -9 $$
$$ \therefore $$ option C is correct.
Question 6
1 / -0

If $$A=\begin{bmatrix} 1 & 3 & 0 \\ -1 & 2 & 1 \\ 0 & 0 & 2 \end{bmatrix}, B=\begin{bmatrix} 2 & 3 & 4 \\ 1 & 2 & 3 \\ -1 & 1 & 2 \end{bmatrix}$$ then $$AB=$$

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

B
$$\begin{bmatrix} 5 & 9 & 13 \\ -1 & 2 & 4 \\ -2 & 2 & 4 \end{bmatrix}$$

C
$$\begin{bmatrix} 5 & 8 & 11 \\ 1 & 2 & 3 \\ 2 & 2 & -3 \end{bmatrix}$$

D
$$\begin{bmatrix} 5 & 3 & 13 \\ -1 & 2 & 4 \\ -2 & 3 & -5 \end{bmatrix}$$

Solution

Given,
$$A=\begin{bmatrix} 1 & 3 & 0 \\ -1 & 2 & 1 \\ 0 & 0 & 2 \end{bmatrix}$$ and $$B=\begin{bmatrix} 2 & 3 & 4 \\ 1 & 2 & 3 \\ -1 & 1 & 2\end{bmatrix}$$

Then, $$AB=\begin{bmatrix} 1 & 3 & 0 \\-1 & 2 & 1\\0&0&2 \end{bmatrix} \begin{bmatrix} 2&3&4 \\1& 2&3 \\-1&1&2 \end{bmatrix}$$

$$= \begin{bmatrix} 1\times2+3\times1+0\times(-1) & 1\times3+3\times2+0\times1& 1\times4+3\times3+0\times2 \\ (-1)\times2+2\times1+1\times(-1) & (-1)\times3+2\times2+1\times1 & (-1)\times4+2\times3+1\times2 \\ 0\times2+0\times1+2\times(-1) & 0\times3+0\times2+2\times1 & 0\times4+0\times3+2\times2 \end{bmatrix}$$

$$= \begin{bmatrix} 5 & 9 & 13 \\ -1 & 2 & 4 \\ -2 & 2 & 4\end{bmatrix}$$

Hence, the correct option is $$(B)$$
Question 7
1 / -0

$$A=\begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{bmatrix}, \text{then} \,A^{3}-4A^{2}-6A=$$

A
$$0$$

B
$$A$$

C
$$-A$$

D
$$I$$

Solution

$$A^{2}=\begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{bmatrix}\begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{bmatrix}=\begin{bmatrix} 9 & 8 & 8 \\ 8 & 9 & 8 \\ 8 & 8 & 9 \end{bmatrix}$$

$$A^{3}=A^{2}\times A=\begin{bmatrix} 9 & 8 & 8 \\ 8 & 9 & 8 \\ 8 & 8 & 9 \end{bmatrix}\begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{bmatrix}=\begin{bmatrix} 41 & 42 & 42 \\ 42 & 41 & 42 \\ 42 & 42 & 41 \end{bmatrix}$$

$$A^{3}-4A^{2}-6A=\begin{bmatrix} 41 & 42 & 42 \\ 42 & 41 & 42 \\ 42 & 42 & 41 \end{bmatrix}-\begin{bmatrix} 36 & 32 & 32 \\ 32 & 36 & 32 \\ 32 & 32 & 36 \end{bmatrix}\begin{bmatrix} 6 & 12 & 12 \\ 12 & 6 & 12 \\ 12 & 12 & 6 \end{bmatrix}$$

$$=\begin{bmatrix} -1 & -2 & -2 \\ -2 & -1 & -2 \\ -2 & -2 & -1 \end{bmatrix}=-A$$
Question 8
1 / -0

If $$\displaystyle \:A= \left [ \begin{matrix}1 &2 &x \\0 &1 &0 \\0 &0 &1 \end{matrix} \right ]and \ B\left [ \begin{matrix}1 &-2 &y \\0 &1 &0 \\0 &0 &1 \end{matrix} \right ]$$ and $$\displaystyle \:AB= I,$$ then $$x+y$$ equals

A
$$0$$

B
$$-1$$

C
$$2$$

D
none of these

Solution

$$\displaystyle \:A= \left [ \begin{matrix}1 &2 &x \\0 &1 &0 \\0 &0 &1 \end{matrix} \right ]$$ and $$B=\left [ \begin{matrix}1 &-2 &y \\0 &1 &0 \\0 &0 &1 \end{matrix} \right ]$$
Also given, $$\displaystyle \:AB= I$$
$$\left[ \begin{matrix} 1 & 2 & x \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} \right] \left[ \begin{matrix} 1 & -2 & y \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} \right] =\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$$

$$\Rightarrow \begin{bmatrix} 1 & 0 & x+y \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}=\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$$
$$\Rightarrow x+y=0$$
Hence, option 'A' is correct.
Question 9
1 / -0

If $$A=\begin{bmatrix} 1 & tanx \\ -tanx & 1 \end{bmatrix}$$, then $${ A }^{ T }{ A }^{ -1 }$$ is

A
$$\begin{bmatrix} -cos2x & sin2x \\ -sin2x & cos2x \end{bmatrix}$$

B
$$\begin{bmatrix} cos2x & -sin2x \\ sin2x & cos2x \end{bmatrix}$$

C
$$\begin{bmatrix} cos2x & cos2x \\ cos2x & sin2x \end{bmatrix}$$

D
none of these

Solution

$$A=\begin{bmatrix} 1 & tanx \\ -tanx & 1 \end{bmatrix}$$

$$\Rightarrow |A|=\sec^{2}x$$

$$adj A=C^{T}={\begin{bmatrix} 1 & \tan x \\ -\tan x & 1 \end{bmatrix}}^{T}$$

$$\Rightarrow adj A=\begin{bmatrix} \cos ^{ 2 }{ x } & -\sin { x } \cos { x } \\ \sin { x } \cos { x } & \cos ^{ 2 }{ x } \end{bmatrix}$$

$$A^{-1}=\displaystyle \frac {adj A}{|A|}$$
$$=\begin{bmatrix} \cos ^{ 2 }{ x } & -\sin { x } \cos { x } \\ \sin { x } \cos { x } & \cos ^{ 2 }{ x } \end{bmatrix}$$

Now, $$A^{T}A^{-1}=\begin{bmatrix} 1 & -\tan x \\ \tan x & 1 \end{bmatrix} \begin{bmatrix} \cos ^{ 2 }{ x } & -\sin { x } \cos { x } \\ \sin { x } \cos { x } & \cos ^{ 2 }{ x } \end{bmatrix}$$

$$\Rightarrow A^{T}A^{-1}=\begin{bmatrix} \cos { 2x } & -\sin { 2x } \\ \sin { 2x } & \cos { 2x } \end{bmatrix}$$
Question 10
1 / -0

If $$\begin{bmatrix} 2 & -1 \\ 1 & 0 \\ -3 & 4 \end{bmatrix}A=\begin{bmatrix} -1 & -8 & -10 \\ 1 & -2 & -5 \\ 9 & 22 & 15 \end{bmatrix}$$, then sum of all the elements of matrix $$A$$ is

A
$$0$$

B
$$1$$

C
$$2$$

D
$$-3$$

Solution

$$\begin{bmatrix} 2 & -1 \\ 1 & 0 \\ -3 & 4 \end{bmatrix}A=\begin{bmatrix} -1 & -8 & -10 \\ 1 & -2 & -5 \\ 9 & 22 & 15 \end{bmatrix}$$
let $$ A=\begin{bmatrix} a & b & c \\ d & e & f \end{bmatrix}$$
$$\Rightarrow \begin{bmatrix} 2 & -1 \\ 1 & 0 \\ -3 & 4 \end{bmatrix}\begin{bmatrix} a & b & c \\ d & e & f \end{bmatrix}=\begin{bmatrix} -1 & -8 & -10 \\ 1 & -2 & -5 \\ 9 & 22 & 15 \end{bmatrix}$$
$$\Rightarrow \begin{bmatrix} 2a-d & 2b-e & 2c-f \\ a & b & c \\ -3a+4d & -3b+4e & -3c+4f \end{bmatrix}=\begin{bmatrix} -1 & -8 & -10 \\ 1 & -2 & -5 \\ 9 & 22 & 15 \end{bmatrix}$$
$$\Rightarrow 2a-d=-1,2b-e=-8,2c-f=-10$$ -----(1)
and $$a=1,b=-2,c=-5$$ ----(2)
from (1) and (2)
$$a+b+c=-6$$ and $$d+e+f=7$$
$$\therefore$$ Sum of all the elements in the matrix $$a+b+c+d+e+f=1$$
Hence, option B.

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Matrices Test - 28

Result

Matrices Test - 28

Score

-

Rank

-

SHARING IS CARING

Question 1

$$7,4$$

$$4,7$$

$$8,9$$

$$9,8$$

Solution

Question 2

$$\begin{bmatrix} 8 & 11 \\ 18 & 23 \end{bmatrix}$$

$$\begin{bmatrix} 11 & 30 \\ 23 & 60 \end{bmatrix}$$

$$\begin{bmatrix} 17 & 47 \\ 13 & 38 \end{bmatrix}$$

$$\begin{bmatrix} 0 & 1 \\ 1 &2 \end{bmatrix}$$

Solution

Question 3

$$\begin{bmatrix} 6 & 2 & 5 \\ 7 & 2 & 3 \end{bmatrix}$$

$$\begin{bmatrix} 2 & 1 & 5 \\ 0 & 1 & 7 \end{bmatrix}$$

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

$$\begin{bmatrix} 7 & 4 & 4 \\ 6 & 2 & 12 \end{bmatrix}$$

Solution

Question 4

there exists exactly one B such that $$AB=BA$$

there exist exactly infinitely many B's such that $$AB=BA$$

there cannot exist any B such that $$AB=BA$$

there exist more than one but finite number of B's such that $$AB=BA$$

Solution

Question 5

$$- 9, -8$$

$$8, 9$$

$$- 8, -9$$

$$9, 8$$

Solution

Question 6

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

$$\begin{bmatrix} 5 & 9 & 13 \\ -1 & 2 & 4 \\ -2 & 2 & 4 \end{bmatrix}$$

$$\begin{bmatrix} 5 & 8 & 11 \\ 1 & 2 & 3 \\ 2 & 2 & -3 \end{bmatrix}$$

$$\begin{bmatrix} 5 & 3 & 13 \\ -1 & 2 & 4 \\ -2 & 3 & -5 \end{bmatrix}$$

Solution

Question 7

$$0$$

$$A$$

$$-A$$

$$I$$

Solution

Question 8

$$0$$

$$-1$$

$$2$$

none of these

Solution

Question 9

$$\begin{bmatrix} -cos2x & sin2x \\ -sin2x & cos2x \end{bmatrix}$$

$$\begin{bmatrix} cos2x & -sin2x \\ sin2x & cos2x \end{bmatrix}$$

$$\begin{bmatrix} cos2x & cos2x \\ cos2x & sin2x \end{bmatrix}$$

none of these

Solution

Question 10

$$0$$

$$1$$

$$2$$

$$-3$$

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.

$$\begin{bmatrix}

8 & 11 \\

18 & 23

\end{bmatrix}$$

$$\begin{bmatrix}

11 & 30 \\

23 & 60

\end{bmatrix}$$

$$\begin{bmatrix}

17 & 47 \\

13 & 38

\end{bmatrix}$$

$$\begin{bmatrix}

0 & 1 \\

1 &2

\end{bmatrix}$$