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CBSE Questions for Class 12 Commerce Maths Matrices Quiz 5 - MCQExams.com
CBSE
Class 12 Commerce Maths
Matrices
Quiz 5
lf $$\mathrm{A}=\left\{\begin{array}{lll}
1 & 1 & 3\\
5 & 2 & 6\\
-2 & -1 & -3
\end{array}\right\},$$ then $$\mathrm{A}^{3}$$ is a/an
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diagonal matrix
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square matrix
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null matrix
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unit Matrix
Explanation
$$A^2$$$$=$$$$\left\{\begin{array}{lll} 1 & 1 & 3\\ 5 & 2 & 6\\ -2 & -1 & -3 \end{array}\right\}\times\left\{\begin{array}{lll} 1 & 1 & 3\\ 5 & 2 & 6\\ -2 & -1 & -3 \end{array}\right\}$$$$=$$$$\left\{\begin{array}{lll} 0 & 0 & 0\\ 3 & 3 & 9\\ -1 & -1 & -3 \end{array}\right\}$$
$$A^3$$$$=$$$${A^2}\times{A}$$
$$A^3$$$$=$$$$\left\{\begin{array}{lll} 0 & 0 & 0\\ 3 & 3 & 9\\ -1 & -1 & -3 \end{array}\right\}\times\left\{\begin{array}{lll} 1 & 1 & 3\\ 5 & 2 & 6\\ -2 & -1 & -3 \end{array}\right\}$$$$=$$$$\left\{\begin{array}{lll} 0 & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & 0 \end{array}\right\}$$
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 $$x=$$ .......... $$, y $$ $$=$$ ........
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$$7,4$$
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$$4,7$$
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$$8,9$$
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$$9,8$$
Explanation
$$\begin{bmatrix} x & -3 \\ -9 & y \end{bmatrix}\begin{bmatrix} 4 & -3 \\ 9 & 7 \end{bmatrix}=\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\\ \Rightarrow \begin{bmatrix} 4x-27 & -3x+(-21) \\ -36+9y & +27+7y \end{bmatrix}=\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$$
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 $$x=7$$ & $$y=4$$
State true/false:
If
$$A=\begin{bmatrix} 1 & -1 & 1 \\ 2 & -1 & 0 \\ 1 & 0 & 0 \end{bmatrix},$$ then solve is
$${ A }^{ 3 }=I$$?
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True
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False
Explanation
Given
$$A=\begin{bmatrix} 1 & -1 & 1 \\ 2 & -1 & 0 \\ 1 & 0 & 0 \end{bmatrix}$$
Calculating $${ A }^{ 3 }$$
$${ A }^{ 2 }=\begin{bmatrix} 1 & -1 & 1 \\ 2 & -1 & 0 \\ 1 & 0 & 0 \end{bmatrix}\begin{bmatrix} 1 & -1 & 1 \\ 2 & -1 & 0 \\ 1 & 0 & 0 \end{bmatrix}$$
$${ A }^{ 2 }=\begin{bmatrix} \left( 1\times 1+2\times -1+1\times 1 \right) & \left( 1\times -1+\left( -1\times -1 \right) +\left( 1\times 0 \right) \right) & 1\times 1+-1\times 0+1\times 0 \\ 2\times 1+-1\times 2+0\times 1 & 2\times -1-1\times -1+0\times 0 & 1\times 1+1\times 0+0\times 0 \\ 1\times 1+2\times 0+0\times 1 & -1\times 1+-1\times 0+0\times 0 & 1\times 1+0\times 0+0\times 0 \end{bmatrix}$$
$${ A }^{ 2 }=\begin{bmatrix} 1-2+1 & -1+1+0 & 1+0+0 \\ 2-2+0 & -2+1+0 & 2+0+0 \\ 1+0+0 & -1+0+0 & 1+0+0 \end{bmatrix}=\begin{bmatrix} 0 & 0 & 1 \\ 0 & -1 & 2 \\ 1 & -1 & 1 \end{bmatrix}$$
$${ A }^{ 3 }={ A }^{ 2 }\times A=\begin{bmatrix} 0 & 0 & 1 \\ 0 & -1 & 2 \\ 1 & -1 & 1 \end{bmatrix}\begin{bmatrix} 1 & -1 & 1 \\ 2 & -1 & 0 \\ 1 & 0 & 0 \end{bmatrix}$$
$${ A }^{ 3 }=\begin{bmatrix} 0\times 1+0\times 2+1\times 1 & 0\times -1+0\times -1+1\times 0 & 0\times 1+0\times 0+1\times 0 \\ 0\times 1+-1\times 2+2\times 1 & 0\times -1+-1\times -1+2\times 0 & 0\times 1+(-1\times 0)+2\times 0 \\ 1\times 1+2\times -1+1\times 1 & -1\times 1+(-1\times -1)+1\times 0 & 1\times 1+(-1\times 0)+(1\times 0) \end{bmatrix}$$
$${ A }^{ 3 }=\begin{bmatrix} 0+0+1 & 0+0+0 & 0+0+0 \\ 0-2+2 & 0+1+0 & 0+0+0 \\ 1-2+1 & -1+1+0 & 1+0+0 \end{bmatrix}=\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}=I$$
$$\Rightarrow { A }^{ 3 }=I$$
Hence, given statement is true.
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 $$\left ( AB \right )\times C=$$
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$$\begin{bmatrix}
8 & 11 \\
18 & 23
\end{bmatrix}$$
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$$\begin{bmatrix}
11 & 30 \\
23 & 60
\end{bmatrix}$$
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$$\begin{bmatrix}
17 & 47 \\
13 & 38
\end{bmatrix}$$
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$$\begin{bmatrix}
0 & 1 \\
1 &2
\end{bmatrix}$$
If $$A=\begin{bmatrix} 2 & 1 \\ 1 & 3 \end{bmatrix}$$ , $$B=\begin{bmatrix} 3 & 2 & 0 \\ 1 & 0 & 4 \end{bmatrix}$$, then $$AB=$$ is
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$$\begin{bmatrix} 6 & 2 & 5 \\ 7 & 2 & 3 \end{bmatrix}$$
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$$\begin{bmatrix} 2 & 1 & 5 \\ 0 & 1 & 7 \end{bmatrix}$$
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$$\begin{bmatrix} -5 & 2 & 5 \\ 1 & 2 & 11 \end{bmatrix}$$
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$$\begin{bmatrix} 7 & 4 & 4 \\ 6 & 2 & 12 \end{bmatrix}$$
Explanation
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 $$(D)$$
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 $$y = $$ _____.
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$$- 9, -8$$
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$$8, 9$$
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$$- 8, -9$$
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$$9, 8$$
Explanation
$$ A = \begin{bmatrix} 3 & 4 \\ 3 & 5 \end{bmatrix} $$;
$$ B = \begin{bmatrix} x & y \\ 6 & 5 \end{bmatrix} $$
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.
If $$P=\begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$$ and if
$$PQ=\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$$,then $$Q=$$
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$$\begin{bmatrix} -4 & 2 \\ 3 & -1 \end{bmatrix}$$
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$$\begin{bmatrix} 4 & -2 \\ 3 & 1 \end{bmatrix}$$
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$$\begin{bmatrix} 5 & 2 \\ 3 & 1 \end{bmatrix}$$
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$$\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$$
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 $$AB=$$
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$$\begin{bmatrix} 5 & 3 & 11 \\ 1 & 2 & 2 \\ 1 & 3 & 5 \end{bmatrix}$$
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$$\begin{bmatrix} 5 & 9 & 13 \\ -1 & 2 & 4 \\ -2 & 2 & 4 \end{bmatrix}$$
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$$\begin{bmatrix} 5 & 8 & 11 \\ 1 & 2 & 3 \\ 2 & 2 & -3 \end{bmatrix}$$
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$$\begin{bmatrix} 5 & 3 & 13 \\ -1 & 2 & 4 \\ -2 & 3 & -5 \end{bmatrix}$$
Explanation
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 $$(B)$$
If $$\displaystyle A=\left [ a_{ij} \right ]$$ is a square matrix of even order such that $$\displaystyle \left [ a_{ij} \right ]=i^{2}-j^{2}$$, then
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$$A$$ is a skew-symmetric matrix and $$\displaystyle \left | A \right |=0$$
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$$A$$ is symmetric matrix and $$\displaystyle \left | A \right |$$ is a square
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$$A$$ is symmetric matrix and $$\displaystyle \left | A \right |=0$$
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none of these
Explanation
Given $$\displaystyle \left [ a_{ij} \right ]=i^{2}-j^{2}$$
$$\Rightarrow a_{ii}=0$$ for all i.
Also, for $$i \ne j, a_{ij}=i^{2}-j^{2}=-(j^{2}-i^{2})=-a_{ji}$$
$$\Rightarrow a_{ij}=-a_{ji}$$
Hence, A is skew-symmetric matrix.
For order 2, $$A=\begin{bmatrix} 0 & -3 \\ 3 & 0 \end{bmatrix}$$
$$|A|=9\ne 0$$
Hence, A is a skew-symmetric matrix and |A| is a square
$$A=\begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{bmatrix}, \text{then} \,A^{3}-4A^{2}-6A=$$
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$$0$$
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$$A$$
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$$-A$$
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$$I$$
Explanation
$$A^{2}=\begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{bmatrix}\begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{bmatrix}=\begin{bmatrix} 9 & 8 & 8 \\ 8 & 9 & 8 \\ 8 & 8 & 9 \end{bmatrix}$$
$$A^{3}=A^{2}\times A=\begin{bmatrix} 9 & 8 & 8 \\ 8 & 9 & 8 \\ 8 & 8 & 9 \end{bmatrix}\begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{bmatrix}=\begin{bmatrix} 41 & 42 & 42 \\ 42 & 41 & 42 \\ 42 & 42 & 41 \end{bmatrix}$$
$$A^{3}-4A^{2}-6A=\begin{bmatrix} 41 & 42 & 42 \\ 42 & 41 & 42 \\ 42 & 42 & 41 \end{bmatrix}-\begin{bmatrix} 36 & 32 & 32 \\ 32 & 36 & 32 \\ 32 & 32 & 36 \end{bmatrix}\begin{bmatrix} 6 & 12 & 12 \\ 12 & 6 & 12 \\ 12 & 12 & 6 \end{bmatrix}$$
$$=\begin{bmatrix} -1 & -2 & -2 \\ -2 & -1 & -2 \\ -2 & -2 & -1 \end{bmatrix}=-A$$
If $$A=\begin{bmatrix} 1 & tanx \\ -tanx & 1 \end{bmatrix}$$, then $${ A }^{ T }{ A }^{ -1 }$$ is
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$$\begin{bmatrix} -cos2x & sin2x \\ -sin2x & cos2x \end{bmatrix}$$
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$$\begin{bmatrix} cos2x & -sin2x \\ sin2x & cos2x \end{bmatrix}$$
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$$\begin{bmatrix} cos2x & cos2x \\ cos2x & sin2x \end{bmatrix}$$
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none of these
Explanation
$$A=\begin{bmatrix} 1 & tanx \\ -tanx & 1 \end{bmatrix}$$
$$\Rightarrow |A|=\sec^{2}x$$
$$adj A=C^{T}={\begin{bmatrix} 1 & \tan x \\ -\tan x & 1 \end{bmatrix}}^{T}$$
$$\Rightarrow adj A=\begin{bmatrix} \cos ^{ 2 }{ x } & -\sin { x } \cos { x } \\ \sin { x } \cos { x } & \cos ^{ 2 }{ x } \end{bmatrix}$$
$$A^{-1}=\displaystyle \frac {adj A}{|A|}$$
$$=\begin{bmatrix} \cos ^{ 2 }{ x } & -\sin { x } \cos { x } \\ \sin { x } \cos { x } & \cos ^{ 2 }{ x } \end{bmatrix}$$
Now, $$A^{T}A^{-1}=\begin{bmatrix} 1 & -\tan x \\ \tan x & 1 \end{bmatrix} \begin{bmatrix} \cos ^{ 2 }{ x } & -\sin { x } \cos { x } \\ \sin { x } \cos { x } & \cos ^{ 2 }{ x } \end{bmatrix}$$
$$\Rightarrow A^{T}A^{-1}=\begin{bmatrix} \cos { 2x } & -\sin { 2x } \\ \sin { 2x } & \cos { 2x } \end{bmatrix}$$
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
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$$0$$
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$$1$$
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$$2$$
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$$-3$$
Explanation
$$\begin{bmatrix} 2 & -1 \\ 1 & 0 \\ -3 & 4 \end{bmatrix}A=\begin{bmatrix} -1 & -8 & -10 \\ 1 & -2 & -5 \\ 9 & 22 & 15 \end{bmatrix}$$
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.
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:
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there exists exactly one B such that $$AB=BA$$
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there exist exactly infinitely many B's such that $$AB=BA$$
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there cannot exist any B such that $$AB=BA$$
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there exist more than one but finite number of B's such that $$AB=BA$$
Explanation
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
If $$\begin{bmatrix} x & 1 \\ 1 & 0 \end{bmatrix}$$ and $$ A^{2}=I$$, then $$x=$$
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$$0$$
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$$1$$
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$$-1$$
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$$2$$
Explanation
Given $$A=\begin{bmatrix} x & 1\\ 1 &0\end{bmatrix}$$
Also, $$A^2=I$$
$$\Rightarrow \begin{bmatrix} x &1 \\1 & 0 \end{bmatrix} \begin{bmatrix} x & 1 \\ 1& 0\end{bmatrix}= \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$$
$$\Rightarrow \begin{bmatrix} x^2+1 & x \\ x & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0\\ 0 &1 \end{bmatrix}$$
For the above two matrices to be equal each of the first matrix must be equal to the corresponding element of the second matrix.
Therefore, $$x^2+1=1 ...(i)$$ and $$x=0...(ii)$$.
From equation $$(i)$$ and equation $$(ii)$$ we get,
$$x=0$$
Hence, the correct option is $$(A)$$
If $$[1\ 2\ 3] B = [3\ 4],$$ then order of the matrix $$B$$ is
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$$3 \times 1$$
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$$1 \times 3$$
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$$2 \times 3$$
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$$3 \times 2$$
Explanation
Given, $$[1 \ 2\ 3] $$ $$B = [3 \ 4]$$
or $${[1 \ 2\ 3]}$$ $$_{1\times3}$$ $$B={[3\ 4]}$$ $$_{1\times2}$$
So, order of $$B $$ should be $$3\times2.$$
Hence, option 'D' is correct.
The inverse of the matrix $$\begin{bmatrix}2 & 1\\ 1 & 3\end{bmatrix}$$ is
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$$\displaystyle \frac{1}{5} \begin{bmatrix}2 & 1\\ 1 & 3\end{bmatrix}$$
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$$\displaystyle \frac{1}{5} \begin{bmatrix}3 & -1\\ -1 & 2\end{bmatrix}$$
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$$\displaystyle \frac{1}{5} \begin{bmatrix}-3 & 1\\ 1 & 2\end{bmatrix}$$
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$$\displaystyle \frac{1}{5} \begin{bmatrix}-3 & -1\\ 1 & 2\end{bmatrix}$$
Explanation
Given, $$A=\begin{bmatrix}2 & 1\\ 1 & 3\end{bmatrix}$$
$$|A|=5$$
Now, $$adj A=C^{T}=\begin{bmatrix} 3 & -1 \\ -1 & 2 \end{bmatrix}^{ T }$$
$$\Rightarrow adj A=\begin{bmatrix} 3 & -1 \\ -1 & 2 \end{bmatrix}$$
$$A^{-1}=\displaystyle \frac{1}{5}\begin{bmatrix} 3 & -1 \\ -1 & 2 \end{bmatrix}$$
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
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$$0$$
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$$-1$$
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$$2$$
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none of these
Explanation
$$\displaystyle \:A= \left [ \begin{matrix}1 &2 &x \\0 &1 &0 \\0 &0 &1 \end{matrix} \right ]$$ 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.
Let A be a square matrix.
Which of the following is/are not symmetric matrix/matrices?
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$$A+A^{T}$$
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$$AA^{T}$$
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$$A-A^{T}$$
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$$A^{T}A$$
Explanation
We know a square matrix $$A$$ is called symmetric if $$A^T = A$$
. Let $$P = A-A^T \Rightarrow P^T = (A-A^T)^T=A^T-A=-P\Rightarrow $$ This is not a symmetric matrix, it is skew symmetric matrix.
All other matrices in the other option are symmetric.
If $$A = \begin{bmatrix}1 & 2 & 2\\ 2 & 1 & -2\\ a & 2 & b\end{bmatrix}$$ is a matrix satisfying $$AA^T = 9 I_3$$, then the values of $$a$$ and $$b$$ are
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$$a = -2, b = - 1$$
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$$a = -2, b = 1$$
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$$a = 2, b = - 1$$
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No values of a,b satisfy given conditions
Explanation
We have,
$$A = \begin{bmatrix}1 & 2 & 2\\ 2 & 1 &
-2\\ a & 2 & b\end{bmatrix} \Rightarrow A^T = \begin{bmatrix}1
& 2 & a\\ 2 & 1 & 2\\ 2 & -2 & b\end{bmatrix}$$
$$\therefore AA^T = 9I_3$$
$$\Rightarrow \begin{bmatrix}1
& 2 & 2\\ 2 & 1 & -2\\ a & 2 &
b\end{bmatrix} \begin{bmatrix}1 & 2 & a\\ 2 & 1 & 2\\ 2
& -2 & b\end{bmatrix} = 9 \begin{bmatrix}1 & 0 & 0\\ 0
& 1 & 0\\ 0 & 0 & 1\end{bmatrix}$$
or
$$\begin{bmatrix}9 & 0 & a+2b + 4\\ 0 & 9 & 2a + 2 -
2b\\ a+2b +4 & 2a + 2 - 2b & a^2 + 4 + b^2\end{bmatrix}
= \begin{bmatrix}9 & 0 & 0\\ 0 & 9 & 0\\ 0 & 0 &
9\end{bmatrix}$$
On comparing above matrices, we get
$$a + 2b + 4 = 0, 2a + 2 - 2b = 0$$ and $$a^2 + 4 + b^2 = 9$$
or $$a + 2b + 4 = 0, a - b + 1= 0$$
and $$a^2 + b^2 = 5$$
Solving $$a + 2b + 4 =0$$ and $$a - b + 1 = 0$$, we get
$$a = -2, b = -1$$. Clearly, these values satisfy $$a^2 + b^2 = 5$$.
$$\therefore$$ $$a = - 2$$ and $$b = - 1$$.
Using elementary transformation, find the inverse of the matrix $$A =\begin{bmatrix}a & b\\ c & \left ( \frac{1 + bc}{a} \right )\end{bmatrix}$$.
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$$\Rightarrow A^{-1} = \begin{bmatrix} \frac{1 + bc}{a}& b\\ -c & a\end{bmatrix} $$
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$$\Rightarrow A^{-1} = \begin{bmatrix} \frac{1 + bc}{a}& - b\\ c & a\end{bmatrix} $$
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$$\Rightarrow A^{-1} = \begin{bmatrix} \frac{1 + bc}{a}& b\\ c & a\end{bmatrix} $$
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None of these.
Explanation
$$A = \begin{bmatrix}a & b\\ c & \left ( \frac{1 + bc}{a} \right ) \end{bmatrix}$$
We
write, $$A = I.A$$
$$\Rightarrow \begin{bmatrix}a & b\\ c & \left ( \frac{1 + bc}{a}
\right ) \end{bmatrix} = \begin{bmatrix}1 & 0\\ 0 & 1
\end{bmatrix}A$$
$$\Rightarrow \begin{bmatrix}1 & \frac{b}{a}\\ c
& \left ( \frac{1 + bc}{a} \right ) \end{bmatrix} =
\begin{bmatrix}\frac{1}{a} & 0\\ 0 & 1 \end{bmatrix} A$$
$$\left ( R_1 \rightarrow \frac{R_1}{a} \right )$$
or $$
\begin{bmatrix}1 & \frac{b}{a}\\ 0 & \frac{1}{a} \end{bmatrix} =
\begin{bmatrix} \frac{1}{a} & 0\\ \frac{-c}{a} &
1\end{bmatrix} A$$ $$(R_2 \rightarrow R_2 - c R_1)$$
or
$$ \begin{bmatrix}1 & \frac{b}{a}\\ 0 & 1\end{bmatrix} =
\begin{bmatrix}\frac{1}{a} & 0\\ -c & a\end{bmatrix}A $$
$$(R_2 \rightarrow aR_2)$$
or $$\begin{bmatrix}1 & 0\\ 0 &
1\end{bmatrix} = \begin{bmatrix} \frac{1 + bc}{a}& -b\\ -c &
a\end{bmatrix} A$$ $$\left ( R_1 \rightarrow R_1 -
\frac{b}{a} R_2 \right )$$
$$\Rightarrow A^{-1} = \begin{bmatrix} \frac{1 + bc}{a}& - b\\ -c & a\end{bmatrix} $$
If $$A = \begin{bmatrix}1 & -2 & 3\\ -4 & 2 & 5\end{bmatrix}$$ and $$B = \begin{bmatrix}2 & 3\\ 4 & 5\\ 2 & 1\end{bmatrix}$$, then the product of AB and BA is
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$$\begin{bmatrix}-10 & 2 & 21\\ -16 & 2 & 37\\ -2 & -2 & 11\end{bmatrix}$$
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$$\begin{bmatrix}0 & -4\\ 10 & 3\end{bmatrix}$$
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$$\begin{bmatrix}-64 & -8\\ -148 & 26\end{bmatrix}$$
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$$Cannot\space be\space computed$$
Explanation
$$AB = \begin{bmatrix} 1& -2 & 3\\ -4 & 2 &
5\end{bmatrix} \begin{bmatrix}2 &3 \\ 4 & 5\\ 2 &
1\end{bmatrix}$$
$$\begin{bmatrix}2 - 8 + 6 & 3 - 10 + 3\\ -8 + 8
+ 10 & -12 + 10 + 5\end{bmatrix}= \begin{bmatrix}0 & -4\\ 10
& 3\end{bmatrix}$$
$$BA = \begin{bmatrix}2 & 3\\ 4 & 5\\ 2
& 1\end{bmatrix} \begin{bmatrix}1 & -2 & 3\\ -4 & 2
& 5\end{bmatrix}$$
$$\begin{bmatrix}2 - 12 & -4 + 6 & 6 +
15\\ 4 - 20 & - 8 + 10 & 12 + 25\\ 2 - 4 & -4 + 2 & 6 +
5\end{bmatrix}= \begin{bmatrix}-10 & 2 & 21\\ -16 & 2 &
37\\ -2 & -2 & 11\end{bmatrix}$$
Thus, order of AB is $$2\times2$$ and order of $$BA$$ is $$3\times3.
$$
So, the product AB and BA can not be computed.
The value of x is such that matrix product $$\begin{bmatrix}2 & 0 & 7\\ 0 & 1 & 0\\ 1 & -2 & 1\end{bmatrix} \begin{bmatrix}-x & 14x & 7x\\ 0 & 1 & 0\\ x & -4x & -2x\end{bmatrix}$$ equals an identity matrix. Then the value of 20x is
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0%
4
0%
5
0%
1
0%
7
Explanation
$$\begin{bmatrix}2 & 0 & 7\\ 0 & 1 & 0\\ 1 &
-2 & 1\end{bmatrix} \begin{bmatrix}-x & 14x & 7x\\ 0 & 1
& 0\\ x & -4x & -2x\end{bmatrix}$$
$$=\begin{bmatrix}5x
& 0 & 0\\ 0 & 1 & 0\\ 0 & 10x-2 &
5x\end{bmatrix}= \begin{bmatrix}1 & 0 & 0\\ 0 & 1 & 0\\ 0
& 0 & 1\end{bmatrix}$$ (given)
Comparing elements we get,
$$\Rightarrow \displaystyle 5x = 1, 10 x - 2 = 0, \therefore x = \frac{1}{5}$$
If A is a skew-symmetric matrix and n is odd positive integer, then $$A^n$$ is
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a skew-symmetric matrix
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a symmetric matrix
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a diagonal matrix
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none of these
Explanation
Given, $$A$$ is a skew symmetric matrix. Therefore, $$A^T=-A$$. $$n$$ is an odd positive integer.
Now, to check whether $$A^n$$ is a skew symmetric matrix or not.
$$(A^n)^T=(A.A^{n-1})^T$$
$$= (A^{n-1})^T.A^T$$
$$= (A.A^{n-2})^T.(-A)$$
$$= (A^{n-2})^T.A^T.(-A)$$
$$= (A.A^{n-3})^T(-A)(-A)$$
$$= (A^{n-3})^T.A^T.(-A)^2$$ and so on....
$$= (-A)^n$$
$$= (-1)^nA^n$$
$$ = -A^n$$
Therefore, $$A^n$$ is also a skew symmetric matrix if $$A$$ is a skew symmetric matrix
If $$\displaystyle A=\begin{bmatrix} 1 & 0 \\ \cfrac { 1 }{ 2 } & 1 \end{bmatrix},$$ then $${A}^{50}$$ is
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$$\begin{bmatrix} 1 & 0 \\ 0 & 50 \end{bmatrix}$$
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$$\begin{bmatrix} 1 & 0 \\ 50 & 1 \end{bmatrix}$$
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$$\begin{bmatrix} 1 & 25 \\ 0 & 1 \end{bmatrix}$$
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None of these
Explanation
$$\displaystyle A=\begin{bmatrix} 1 & 0 \\ \cfrac { 1 }{ 2 } & 1 \end{bmatrix}$$
$$\displaystyle{ A }^{ 2 }=\begin{bmatrix} 1 & 0 \\ \cfrac { 1 }{ 2 } & 1 \end{bmatrix}\begin{bmatrix} 1 & 0 \\ \cfrac { 1 }{ 2 } & 1 \end{bmatrix}=\begin{bmatrix} 1 & 0 \\ 2\left( \cfrac { 1 }{ 2 } \right) & 1 \end{bmatrix}=\begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix}$$
$${ A }^{ 4 }=\begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix}\begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix}=\begin{bmatrix} 1 & 0 \\ 2 & 1 \end{bmatrix}\\ { A }^{ 8 }=\begin{bmatrix} 1 & 0 \\ 2 & 1 \end{bmatrix}\begin{bmatrix} 1 & 0 \\ 2 & 1 \end{bmatrix}=\begin{bmatrix} 1 & 0 \\ 4 & 1 \end{bmatrix}$$
Similarly $$\displaystyle{ A }^{ n }=\begin{bmatrix} 1 & 0 \\ \cfrac { n }{ 2 } & 1 \end{bmatrix}$$ for even $$n$$
$$\Rightarrow { A }^{ 50 }=\begin{bmatrix} 1 & 0 \\ 25 & 1 \end{bmatrix}$$
State true or false:
If $$A = \begin{bmatrix}2 & -1 \\ 0 & 1\end{bmatrix}$$ and $$B = \begin{bmatrix}1 & 0 \\ -1 & -1\end{bmatrix},$$ then $$(A+B)^2 = A^2 + AB + BA + B^2 = A^2 + 2AB + B^2$$
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True
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False
Explanation
Given $$A = \begin{bmatrix}2 & -1 \\ 0 & 1\end{bmatrix}$$ and $$B = \begin{bmatrix}1 & 0 \\ -1 & -1\end{bmatrix}$$
$$(A+B)^2 = A^2 + AB + BA + B^2 = A^2 + 2AB + B^2$$ is true only when $$AB=BA$$
$$AB=\begin{bmatrix}2 & -1 \\ 0 & 1\end{bmatrix}\begin{bmatrix}1 & 0 \\ -1 & -1\end{bmatrix}=\begin{bmatrix}3 & 1 \\ -1 & -1\end{bmatrix}$$
$$BA=\begin{bmatrix}1 & 0 \\ -1 & -1\end{bmatrix}\begin{bmatrix}2 & -1 \\ 0 & 1\end{bmatrix}=\begin{bmatrix}2 & -1 \\ -2 & 0\end{bmatrix}$$
$$AB\neq BA$$
$$\therefore$$ Given statement is false.
If $$\displaystyle \begin{bmatrix} 2 & -3 \\ 1 & \lambda \end{bmatrix} \times \begin{bmatrix} 1 & 5 & \mu \\ 0 & 2 & -3 \end{bmatrix} = \begin{bmatrix} 2 & 4 & 1 \\ 1 & -1 & 13 \end{bmatrix}$$ then
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$$\displaystyle \lambda = 3, \mu = 4$$
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$$\displaystyle \lambda = 4, \mu = 3$$
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$$\lambda = 2 , \mu = 5$$
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none of these
Explanation
Since, $$\begin{bmatrix} 2 & -3 \\ 1 & \lambda \end{bmatrix}\times \begin{bmatrix} 1 & 5 & \mu \\ 0 & 2 & -3 \end{bmatrix}=\begin{bmatrix} 2 & 4 & 1 \\ 1 & -1 & 13 \end{bmatrix}$$
$$\Rightarrow \begin{bmatrix} 2-0 & 10-6 & 2\mu +9 \\ 1+0 & 5+2\lambda & \mu -3\lambda \end{bmatrix}=\begin{bmatrix} 2 & 4 & 1 \\ 1 & -1 & 13 \end{bmatrix}$$
$$\Rightarrow \begin{bmatrix} 2 & 4 & 2\mu +9 \\ 1 & 5+2\lambda & \mu -3\lambda \end{bmatrix}=\begin{bmatrix} 2 & 4 & 1 \\ 1 & -1 & 13 \end{bmatrix}$$
$$\Rightarrow 2\mu +9=1$$ and $$5+2\lambda =-1$$
$$\Rightarrow \mu =-4,\lambda =-3$$
But these value does not satisfy $$\mu -3\lambda =13$$
If the matrix $$\displaystyle A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$$ then $$\displaystyle A^2$$ is
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$$\displaystyle \begin{bmatrix} a^2 & b^2 \\ c^2 & d^2 \end{bmatrix}$$
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$$\displaystyle \begin{bmatrix} a^2 + bc & ab + bd \\ ac + dc & bc + d^2 \end{bmatrix}$$
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nonexistent
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none of these
Explanation
Given $$\displaystyle A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$$
$$A^{2}=\begin{bmatrix} a & b \\ c & d \end{bmatrix} \begin{bmatrix} a & b \\ c & d \end{bmatrix}$$
$$=\displaystyle \begin{bmatrix} a^2 + bc & ab + bd \\ ac + dc & bc + d^2 \end{bmatrix}$$
State true or false.
If $$A, B, C$$ are three matrices such that
$$A = \begin{bmatrix}x & y & z\end{bmatrix},$$ $$B = \begin{bmatrix} a & h & g \\ h & b & f \\ g & f & c\end{bmatrix}$$ and $$C = \begin{bmatrix}x \\ y \\ z\end{bmatrix},$$ then $$ABC = \begin{bmatrix}ax^2 + by^2 + cz^2 + 2hxy + 2gzx + 2fyz\end{bmatrix}$$
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0%
True
0%
False
Explanation
Given,
$$A, B, C$$ are three matrices such that
$$A = \begin{bmatrix}x & y & z\end{bmatrix},$$ $$B = \begin{bmatrix} a & h & g \\ h & b & f \\ g & f & c\end{bmatrix}$$ and $$C = \begin{bmatrix}x \\ y \\ z\end{bmatrix}$$
Consider, $$AB=\left[ xyz \right] \begin{bmatrix} a & h & g \\ h & b & f \\ g & f & c \end{bmatrix}$$
$$\Rightarrow$$ $$= \left[ ax+yh+zg\quad xh+yb+zc\quad xg+yf+zc \right] $$
Now, $$ABC=\left[ ax+yh+zg\quad xh+yb+zc\quad xg+yf+zc \right] \begin{bmatrix} x \\ y \\ z \end{bmatrix}$$
$$\Rightarrow$$ $$ =\left[ { ax }^{ 2 }+{ xy }h+zgx+xhy+{ y }^{ 2 }h+zcy+xgz+yfz+{ z }^{ 2 }c \right] $$
Hence it's true
If $$A = \begin{bmatrix}1 & 1 & -1 \\ 2 & -3 & 4 \\ 3 & -2 & 3 \end{bmatrix},$$ $$B = \begin{bmatrix} -1 & -2 & -1 \\ 6 & 12 & 6 \\ 5 & 10 & 5 \end{bmatrix}$$ and $$C = \begin{bmatrix} -1 & -1 & 1 \\ 2 & 2 & -2 \\ -3 & -3 & 3 \end{bmatrix},$$
which if the following are null matrices ?
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$$CA$$
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$$AB$$
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$$BA$$
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$$AC$$
Explanation
Given $$A = \begin{bmatrix}1 & 1 & -1 \\ 2 & -3 & 4 \\ 3 & -2 & 3 \end{bmatrix},$$ $$B = \begin{bmatrix} -1 & -2 & -1 \\ 6 & 12 & 6 \\ 5 & 10 & 5 \end{bmatrix}$$ and $$C = \begin{bmatrix} -1 & -1 & 1 \\ 2 & 2 & -2 \\ -3 & -3 & 3 \end{bmatrix}$$
$$CA=\begin{bmatrix} -1 & -1 & 1 \\ 2 & 2 & -2 \\ -3 & -3 & 3 \end{bmatrix}\begin{bmatrix}1 & 1 & -1 \\ 2 & -3 & 4 \\ 3 & -2 & 3 \end{bmatrix}=\begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}=O$$
$$AB=\begin{bmatrix}1 & 1 & -1 \\ 2 & -3 & 4 \\ 3 & -2 & 3 \end{bmatrix}\begin{bmatrix} -1 & -2 & -1 \\ 6 & 12 & 6 \\ 5 & 10 & 5 \end{bmatrix}=\begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}=O$$
$$BA=\begin{bmatrix} -1 & -2 & -1 \\ 6 & 12 & 6 \\ 5 & 10 & 5 \end{bmatrix}\begin{bmatrix}1 & 1 & -1 \\ 2 & -3 & 4 \\ 3 & -2 & 3 \end{bmatrix}=\begin{bmatrix} -8 & 7 & -10 \\ 48 & -42 & 60 \\ 40 & -35 & 50 \end{bmatrix}\neq O$$
$$AC=\begin{bmatrix}1 & 1 & -1 \\ 2 & -3 & 4 \\ 3 & -2 & 3 \end{bmatrix}\begin{bmatrix} -1 & -1 & 1 \\ 2 & 2 & -2 \\ -3 & -3 & 3 \end{bmatrix}=\begin{bmatrix} 4 & 4 & -4 \\ -20 & -20 & 20 \\ -16 & -16 & 16 \end{bmatrix}\neq O$$
Hence, options A and B.
If $$A = \begin{bmatrix} 0 & 2 & 3 \\ 3 & 5 & 7 \end{bmatrix},$$ $$B = \begin{bmatrix} 1 & 3 & 7 \\ 2 & 4 & 1 \end{bmatrix}$$ and $$A+B = \begin{bmatrix} 1 & 5 & 10 \\ 5 & k & 8 \end{bmatrix},\\ $$ then find the value of $$k .$$
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9
0%
4
0%
5
0%
1
Explanation
Given $$A = \begin{bmatrix} 0 & 2 & 3 \\ 3 & 5 &
7 \end{bmatrix}$$ and $$B = \begin{bmatrix} 1 & 3 & 7 \\ 2 &
4 & 1 \end{bmatrix}$$
$$\therefore A+B = \mbox{sum of respective elements of A and B}=\begin{bmatrix} 1 & 5
& 10 \\ 5 & 9 & 8 \end{bmatrix} $$
Hence the value of $$k$$ is $$9$$
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Practice Class 12 Commerce Maths Quiz Questions and Answers
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