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CBSE Questions for Class 12 Commerce Maths Matrices Quiz 8 - MCQExams.com
CBSE
Class 12 Commerce Maths
Matrices
Quiz 8
If $$A$$ is a scalar matrix $$kI$$ with scalar $$k\ne 0$$ of order $$3$$, the $${A}^{-1}$$ is:
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$$\cfrac { 1 }{ { k }^{ 2 } } I$$
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$$\cfrac { 1 }{ { k }^{ 3 } } I$$
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$$\cfrac { 1 }{ { k}^{ } } I$$
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$$kI$$
Explanation
If $$A$$ is a scalar matrix with scalar $$k$$, then $$A=kI$$
Thus $$A^{-1}=(kI)^{-1}=\dfrac{1}{k}I$$
Note: Inverse of identity matrix is Identity matrix itself
And $$(kA)^{-1}=\dfrac{1}{k}I^{-1}=\dfrac1k I$$
If $$A=\begin{bmatrix} 7 & 2 \\ 1 & 3 \end{bmatrix}$$ and $$A+B=\begin{bmatrix} -1 & 0 \\ 2 & -4 \end{bmatrix},$$ then matrix $$B=$$?
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$$\begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix}$$
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$$\begin{bmatrix} 6 & 2 \\ 3 & -1 \end{bmatrix}$$
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$$\begin{bmatrix} -8 & -2 \\ 1 & -7 \end{bmatrix}$$
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$$\begin{bmatrix} 8 & 2 \\ -1 & 7 \end{bmatrix}$$
Explanation
$$A+B = \begin{bmatrix} -1 & 0 \\ 2& -4 \end{bmatrix}$$
$$A= \begin{bmatrix} 7& 2 \\ 1& 3 \end{bmatrix}$$
As we have learnt that we can subtract two matrices of the same order.
(Here, order of both the matrix is $$2 \times 2$$)
So, we will subtract matrix $$A$$ from both sides of the equation.
We get,
$$B= \begin{bmatrix} -1 & 0 \\ 2& -4 \end{bmatrix} - \begin{bmatrix} 7 & 2 \\ 1& 3 \end{bmatrix}$$
$$B = \begin{bmatrix} -1-7 & 0-2 \\ 2-1 & -4-3 \end{bmatrix} $$
(To subtract 2 matrices, we have to subtract their corresponding elements.)
$$B= \begin{bmatrix} -8 & -2 \\ 1 & -7 \end{bmatrix}$$
Hence, Option C is the correct answer
If $$A = \begin{bmatrix} 4& -2\\ 6 & -3\end{bmatrix}$$, then $$A^2$$ is
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$$\begin{bmatrix} 16 & 4\\ 36 & 9\end{bmatrix}$$
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$$\begin{bmatrix} 8 & -4\\ 12 & -6\end{bmatrix}$$
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$$\begin{bmatrix} -4 & 2\\ -6 & 3\end{bmatrix}$$
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$$\begin{bmatrix} 4 & -2\\ 6 & -3\end{bmatrix}$$
Explanation
$$A=\begin{bmatrix} 4 & -2 \\ 6 & -3 \end{bmatrix}$$
$$ { A }^{ 2 }=\begin{bmatrix} 4 & -2 \\ 6 & -3 \end{bmatrix}\times\begin{bmatrix} 4 & -2 \\ 6 & -3 \end{bmatrix}$$
using the multiplication properties of matrices, we get
$${ A }^{ 2 }=\begin{bmatrix} 4\times4+(-2)\times6\quad \quad\quad\quad & 4\times(-2)+(-2)\times(-3) \\ 6\times4+(-3)\times6\quad \quad \quad \quad & 6\times(-2)+(-3)\times(-3) \end{bmatrix}$$
$${ A }^{ 2 }=\begin{bmatrix} 4 & -2 \\ 6 & -3 \end{bmatrix}$$
If $$A+B=\begin{bmatrix} 2 & 3 \\ 4 & 5 \end{bmatrix}$$ and $$A=\begin{bmatrix} 1 & 2 \\ 0 & 3 \end{bmatrix}$$, then matrix $$B$$ is
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$$\begin{bmatrix} 1 & 1 \\ 4 & 2 \end{bmatrix}$$
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$$\begin{bmatrix} 1 & 4 \\ 1 & 2 \end{bmatrix}$$
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$$\begin{bmatrix} 2 & 4 \\ 1 & 1 \end{bmatrix}$$
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$$\begin{bmatrix} 4 & 2 \\ 1 & 1 \end{bmatrix}$$
Explanation
As given $$A+B=\begin{bmatrix} 2 & 3 \\ 4 & 5 \end{bmatrix}$$ and
$$A=\begin{bmatrix} 1 & 2 \\ 0 & 3 \end{bmatrix}$$
Substitute $$A$$ in $$A+B$$ we get,
$$\begin{bmatrix} 1 & 2 \\ 0 & 3 \end{bmatrix} +B=\begin{bmatrix} 2 & 3 \\ 4 & 5 \end{bmatrix}$$
$$\implies$$ $$B=\begin{bmatrix} 2 & 3 \\ 4 & 5 \end{bmatrix}-\begin{bmatrix} 1 & 2 \\ 0 & 3 \end{bmatrix}$$
$$\implies$$ $$B=\begin{bmatrix} (2-1) & (3-2) \\ (4-0) & (5-3) \end{bmatrix}$$
$$\therefore$$ $$B=\begin{bmatrix} 1 & 1 \\ 4 & 2 \end{bmatrix}$$
Hence, option A is correct.
If $$\bigl(\begin{smallmatrix} 1& 2\\ 2 & 1\end{smallmatrix}\bigr) \bigl(\begin{smallmatrix} x \\ y \end{smallmatrix}\bigr) = \bigl(\begin{smallmatrix} 2 \\ 4 \end{smallmatrix}\bigr)$$, then the values of $$x$$ and $$y$$ respectively, are
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$$2, 0$$
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$$0, 2$$
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$$0, -2$$
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$$1, 1$$
Explanation
$$\left[ \begin{matrix} 1 & 2 \\ 2 & 1 \end{matrix} \right] \left[ \begin{matrix} x \\ y \end{matrix} \right] =\left[ \begin{matrix} 2 \\ 4 \end{matrix} \right] $$
$$\Rightarrow \left[ \begin{matrix} x & 2y \\ 2x & y \end{matrix} \right] =\left[ \begin{matrix} 2 \\ 4 \end{matrix} \right] $$
$$\Rightarrow x+2y=2\longrightarrow (1)\;\; \& \;\;2x+y=4\longrightarrow (2)$$
Multiply $$(1)$$ by 2,$$
$$\Rightarrow 2x+4y=4\longrightarrow(3)$$
Solving $$(2)\;\&\;(3),[(2)-(3)],$$ we get
$$\Rightarrow -3y=0$$
$$\Rightarrow y=0$$
Put $$y=0$$ in $$(1),$$ we get
$$\Rightarrow x+0=2$$
$$\Rightarrow x=2$$
$$\Rightarrow y=0$$
Hence, the answer is $$2,0.$$
If $$\begin{bmatrix} 5 & x & 1 \end{bmatrix} \begin{bmatrix} 2 \\ -1 \\ 3 \end{bmatrix} = (20)$$, then the value of $$x$$ is
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$$7$$
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$$-7$$
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$$\dfrac{1}{7}$$
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$$0$$
Explanation
Given expression: $${ \begin{bmatrix} 5 &x&1\end{bmatrix} }_{ 1\times 3 }\times { \begin{bmatrix} 2 \\ -1 \\ 3 \end{bmatrix} }_{ 3\times 1 }={ \begin{pmatrix} 20 \end{pmatrix} }_{ 1\times 1 }\\\Rightarrow { (10+(-x)+3) }_{ 1\times 1 }={ \begin{pmatrix} 20 \end{pmatrix} }_{ 1\times 1 }$$
$$\Rightarrow 13-x=20$$ [by equality of matrix]
$$\Rightarrow x=-7$$
If $$A = \begin{bmatrix}7 &2 \\ 1 & 3\end{bmatrix}$$ and $$A + B = \begin{bmatrix} -1& 0\\ 2 & -4\end{bmatrix}$$, then the matrix $$B =$$
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$$\left[\begin{matrix}1 &0 \\ 0 & 1\end{matrix}\right]$$
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$$\left[\begin{matrix} 6&2 \\ 3 & -1\end{matrix}\right]$$
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$$\left[\begin{matrix} -8& -2\\ 1 & -7\end{matrix}\right]$$
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$$\left[\begin{matrix} 8& 2\\ -1 & 7\end{matrix}\right]$$
Explanation
Given, $$A=\begin{bmatrix} 7 & 2 \\ 1 & 3 \end{bmatrix}$$
Also,
$$A+B=\begin{bmatrix} -1 & 0 \\ 2 & -4 \end{bmatrix}\\ \Rightarrow B=\begin{bmatrix} -1 & 0 \\ 2 & -4 \end{bmatrix}-\begin{bmatrix} 7 & 2 \\ 1 & 3 \end{bmatrix}\\ \Rightarrow B=\begin{bmatrix} -1-7 & 0-2 \\ 2-1 & -4-3 \end{bmatrix}\\ \Rightarrow B=\begin{bmatrix} -8 & -2 \\ 1 & -7 \end{bmatrix}$$
$$\Rightarrow B=\begin{bmatrix} -8 & -2 \\ 1 & -7 \end{bmatrix}$$
Choose the correct statement related to the matrices $$A=\begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}$$ and $$B=\begin{bmatrix}0 & 1 \\ 1 & 0\end{bmatrix}$$
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$$A^3=A, B^3 \ne B$$
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$$A^3\ne A, B^3=B$$
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$$A^3=A, B^3 = B$$
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$$A^3\ne A, B^3 \ne B$$
Explanation
Clearly, $$A$$ is an identity matrix
Now, $${ A }^{ 3 }=$$
$$\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\quad $$
$$= \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}=\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = A$$
$$\implies A^3=A$$
Also, $${ B}^{ 3 }= \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}=\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}=\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} = B$$
$$\implies B^{3}=B$$
Hence, option C is correct.
If $$A = \bigl(\begin{smallmatrix} 4& -2\\ 6 & -3\end{smallmatrix}\bigr)$$, then $$A^2$$ is
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$$\bigl(\begin{smallmatrix} 16 & 4\\ 36 & 9\end{smallmatrix}\bigr)$$
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$$\bigl(\begin{smallmatrix}8 & -4\\ 12 & -6\end{smallmatrix}\bigr)$$
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$$\bigl(\begin{smallmatrix} -4& 2\\ -6 & 3\end{smallmatrix}\bigr)$$
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$$\bigl(\begin{smallmatrix} 4& -2\\ 6 & -3\end{smallmatrix}\bigr)$$
Explanation
$$A=\left[ \begin{matrix} 4 & -2 \\ 6 & -3 \end{matrix} \right] $$
$$\Rightarrow { A }^{ 2 }=A.A=\left[ \begin{matrix} 4 & -2 \\ 6 & -3 \end{matrix} \right] \left[ \begin{matrix} 4 & -2 \\ 6 & -3 \end{matrix} \right] $$
$$=\left[ \begin{matrix} 16-12 & -8+6 \\ 24-18 & -12+9 \end{matrix} \right] $$
$$=\left[ \begin{matrix} 4 & -2 \\ 6 & -3 \end{matrix} \right] $$
Hence, the answer is $$\left[ \begin{matrix} 4 & -2 \\ 6 & -3 \end{matrix} \right] .$$
$$\bigl(\begin{smallmatrix} -1& 0\\ 0 & 1\end{smallmatrix}\bigr) \bigl(\begin{smallmatrix}a & b\\ c & d\end{smallmatrix}\bigr) = \bigl(\begin{smallmatrix} 1& 0\\ 0 & -1\end{smallmatrix}\bigr)$$, then the values of $$a, b, c $$ and $$d $$ respectively are
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$$-1, 0, 0, -1$$
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$$1, 0, 0, 1$$
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$$-1, 0, 1, 0$$
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$$1, 0, 0, 0$$
Explanation
$$\left[ \begin{matrix} -1 & 0 \\ 0 & 1 \end{matrix} \right] \left[ \begin{matrix} a & b \\ c & d \end{matrix} \right] =\left[ \begin{matrix} 1 & 0 \\ 0 & -1 \end{matrix} \right] $$
$$\Rightarrow \left[ \begin{matrix} -a & -b \\ c & d \end{matrix} \right] =\left[ \begin{matrix} 1 & 0 \\ 0 & -1 \end{matrix} \right] $$
On comparing both sides, corresponding columns,
$$\Rightarrow -a=1,$$ $$-b=0$$ and $$c=0$$
$$a=-1,$$ $$b=0$$ and $$d=-1$$
Hence, the answer is $$-1,0,0,-1.$$
If $$\bigl(\begin{smallmatrix}a & 3\\ 1 & 2\end{smallmatrix}\bigr) \bigl(\begin{smallmatrix} 2 \\ -1 \end{smallmatrix}\bigr) = \bigl(\begin{smallmatrix} 5\\ 0 \end{smallmatrix}\bigr)$$, then the value of $$a$$ is
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$$8$$
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$$4$$
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$$2$$
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$$11$$
Explanation
$$\left[ \begin{matrix} a & 3 \\ 1 & 2 \end{matrix} \right] \left[ \begin{matrix} 2 \\ -1 \end{matrix} \right] =\left[ \begin{matrix} 5 \\ 0 \end{matrix} \right] $$
$$\Rightarrow \left[ \begin{matrix} 2a-3 \\ 0 \end{matrix} \right] =\left[ \begin{matrix} 5 \\ 0 \end{matrix} \right] $$
$$\Rightarrow 2a-3=5$$
$$\Rightarrow 2a=8$$
$$\Rightarrow a=4$$
If
$$A=\left[ \begin{matrix} 1 & -2 & 3 \end{matrix} \right] $$ and $$B=\left[ \begin{matrix} -1 \\ 2 \\ -3 \end{matrix} \right] $$
, then $$A + B$$ is
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$$\begin{bmatrix} 0& 0 & 0 \end{bmatrix}$$
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$$\begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$$
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$$\begin{bmatrix} -1& 4 \end{bmatrix}$$
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not defined
Explanation
Given, $$A=\left[ \begin{matrix} 1 & -2 & 3 \end{matrix} \right] $$ and $$B=\left[ \begin{matrix} 1 \\ -2 \\ 3 \end{matrix} \right] $$
Here $$A+B$$ is not possible because the order of the matrices is not same.
Hence, the answer is not defined.
If matrix $$A=\begin{bmatrix} 1 & 2 \\ 4 & 3 \end{bmatrix}$$ such that $$Ax=I$$, then $$x=$$................
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$$\cfrac { 1 }{ 5 } \begin{bmatrix} 1 & 3 \\ 2 & -1 \end{bmatrix}$$
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$$\cfrac { 1 }{ 5 } \begin{bmatrix} 4 & 2 \\ 4 & -1 \end{bmatrix}$$
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$$\cfrac { 1 }{ 5 } \begin{bmatrix} -3 & 2 \\ 4 & -1 \end{bmatrix}$$
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$$\cfrac { 1 }{ 5 } \begin{bmatrix} -1 & 2 \\ -1 & 4 \end{bmatrix}$$
Explanation
Given : $$A=\begin{bmatrix} 1 & 2 \\ 4 & 3 \end{bmatrix}$$ such that $$Ax=I$$ ... $$(i)$$
Since, $$A$$ and $$I$$ are of order $$2\times 2$$. So, $$x$$ will be a matrix of order $$2\times 2$$
Let
$$x=\begin{bmatrix} a & b \\ c & d \end{bmatrix}$$
From $$(i)$$, we get
$$Ax=I$$
$$\implies \begin{bmatrix} 1 & 2 \\ 4 & 3 \end{bmatrix} \begin{bmatrix} a & b \\ c & d \end{bmatrix}=\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$$
$$\implies \begin{bmatrix} a+2c & b+2d \\ 4a+3c & 4b+3d \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$$
$$\implies a+2c=1$$ ...... $$(ii),$$ $$4a+3c=0$$ ........ $$(iii)$$
$$b+2d=0$$ ....... $$(iv)$$ and $$4b+3d=1$$ ....... $$(v)$$
Solving $$(ii)$$ and $$(iii)$$ simultaneously we get
$$a=-\dfrac{3}{5}$$ and $$c=\dfrac{4}{5}$$
Then solving $$(iv)$$ and $$(v)$$ simultaneously we get
$$b=\dfrac{2}{5}$$ and $$d=-\dfrac{1}{5}$$
Substituting all these values in $$x$$, we get
$$x=\begin{bmatrix} -\dfrac{3}{5} & \dfrac{2}{5} \\ \dfrac{4}{5} & -\dfrac{1}{5} \end{bmatrix}$$
$$\therefore x=\dfrac{1}{5} \begin{bmatrix} -3 & 2 \\ 4 & -1 \end{bmatrix}$$
The matrix $$A = \begin{bmatrix} 0& 1 & -1\\ -1 & 0 & 1\\ 1 & -1 & 0\end{bmatrix}$$ is a :
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Diagonal matrix
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Symmetric matrix
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Skew-symmetric matrix
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Identity matrix
Explanation
The matrix $$A=\quad \begin{bmatrix} 0 & \quad \quad 1 & \quad \quad -1 \\ -1 & \quad \quad 0 & \quad \quad 1 \\ 1 & \quad \quad -1 & \quad \quad 0 \end{bmatrix}$$ is a $${ \quad A }^{ T\quad }=\quad \begin{bmatrix} 0 & \quad -1 & \quad \quad 1 \\ 1 & \quad \quad 0 & \quad \quad -1 \\ -1 & \quad \quad 1 & \quad \quad 0 \end{bmatrix}$$ $$Hence \ \ \ { -A }^{ T }\quad =\quad \begin{bmatrix} 0 & \quad \quad 1 & \quad \quad -1 \\ -1 & \quad \quad 0 & \quad \quad 1 \\ 1 & \quad \quad -1 & \quad \quad 0 \end{bmatrix}\quad \\ \\ \quad A\quad =\quad { -A }^{ T }$$
The matrix A is skew symmetric matrix
The inverse of a diagonal matrix is a :
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Symmetric matrix
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Skew-symmetric matrix
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Diagonal matrix
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None of the above
Explanation
A diagonal matrix has elements only in it's diagonal.
So the inverse will also have all non zero elements in the diagonal.
So, it will be symmetric and will also be a diagonal matrix.
Hence, option A and C are correct
If $$\begin{bmatrix} 3 & -1 \\ 0 & 6 \end{bmatrix}\begin{bmatrix} 3x \\ 1 \end{bmatrix}+\begin{bmatrix} -2x \\ 3 \end{bmatrix}=\begin{bmatrix} 8 \\ 9 \end{bmatrix}$$, then the value of $$x$$ is
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$$-\dfrac { 3 }{ 8 } $$
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$$7$$
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$$-\dfrac { 2 }{ 9 } $$
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None of these
Explanation
Given, $$\begin{bmatrix} 3 & -1 \\ 0 & 6 \end{bmatrix}\begin{bmatrix} 3x \\ 1 \end{bmatrix}+\begin{bmatrix} -2x \\ 3 \end{bmatrix}=\begin{bmatrix} 8 \\ 9 \end{bmatrix}$$
LHS $$=\begin{bmatrix} 3 & -1 \\ 0 & 6 \end{bmatrix}\begin{bmatrix} 3x \\ 1 \end{bmatrix}+\begin{bmatrix} -2x \\ 3 \end{bmatrix}$$
$$=\begin{bmatrix} 9x+\left( -1 \right) \\ 0+6 \end{bmatrix}+\begin{bmatrix} -2x \\ 3 \end{bmatrix}=\begin{bmatrix} 9x-1 \\ 6 \end{bmatrix}+\begin{bmatrix} -2x \\ 3 \end{bmatrix}$$
$$=\begin{bmatrix} 9x-1-2x \\ 6+3 \end{bmatrix}=\begin{bmatrix} 7x-1 \\ 9 \end{bmatrix}$$
Now, $$\begin{bmatrix} 7x-1 \\ 9 \end{bmatrix}=\begin{bmatrix} 8 \\ 9 \end{bmatrix}$$
$$\therefore 7x-1=8$$
$$\Rightarrow 7x=9$$
$$\Rightarrow x=\dfrac { 9 }{ 7 } $$.
Let $$A = \begin{bmatrix}x + y & y\\ 2x & x - y\end{bmatrix}, B = \begin{bmatrix} 2& -1\end{bmatrix}$$ and $$C = \begin{bmatrix} 3& 2\end{bmatrix}.$$ If $$AB = C$$, then $$A^{2}$$ is equal to
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$$\begin{bmatrix}
6 & -10\\
4 & 26
\end{bmatrix}$$
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$$\begin{bmatrix}
-10 & 5\\
4 & 24
\end{bmatrix}$$
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$$\begin{bmatrix}
-5 & -6\\
-4 & -20
\end{bmatrix}$$
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None of these.
Explanation
$$A=\begin{bmatrix} x+y & y \\ 2x & x-y \end{bmatrix}, B=\begin{bmatrix} 2 & -1 \end{bmatrix}$$ and $$C=\begin{bmatrix} 3 & 2 \end{bmatrix}$$
Since,
$$AB=C\\ \Longrightarrow \begin{bmatrix} x+y & y \\ 2x & x-y \end{bmatrix}\begin{bmatrix} 2 & -1 \end{bmatrix}=\begin{bmatrix} 3 & 2 \end{bmatrix}$$
The multiplication of $$A$$ and $$B$$ can't be done because it doesn't satisfy the conditions of matrix multipication
If the matrix A is such that $$\begin{bmatrix} 1 & 3\\ 0 & 1\end{bmatrix} A=\begin{bmatrix} 1 & 1 \\ 0 & -1\end{bmatrix}$$, then what is equal to A?
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$$\begin{bmatrix} 1 & 4 \\ 0 & -1\end{bmatrix}$$
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$$\begin{bmatrix} 1 & 4 \\ 0 & 1\end{bmatrix}$$
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$$\begin{bmatrix} -1 & 4\\ 0 & -1\end{bmatrix}$$
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$$\begin{bmatrix} 1 & -4 \\ 0 & -1\end{bmatrix}$$
Explanation
Given
$$\begin{bmatrix} 1 & 3 \\ 0 & 1 \end{bmatrix}A=\begin{bmatrix} 1 & 1 \\ 0 & -1 \end{bmatrix}$$
Let $$A=\begin{bmatrix} \alpha & \beta \\ \gamma & \delta \end{bmatrix}$$
$$\begin{bmatrix} 1 & 3 \\ 0 & 1 \end{bmatrix}\begin{bmatrix} \alpha & \beta \\ \gamma & \delta \end{bmatrix}=\begin{bmatrix} 1 & 1 \\ 0 & -1 \end{bmatrix}$$
$$\begin{bmatrix} \alpha +3\gamma & \beta +4\delta \\ \gamma & \delta \end{bmatrix}=\begin{bmatrix} 1 & 1 \\ 0 & -1 \end{bmatrix}$$
Two matrices are equal if corresponding elements are equal
$$\Rightarrow \alpha +3\gamma =1, \gamma =0,$$
$$\beta +3\delta =1$$ and $$ \delta =-1$$
$$\Rightarrow \alpha =1,$$
$$\beta =4$$
$$\Rightarrow A=\begin{bmatrix} 1 & 4 \\ 0 & -1 \end{bmatrix}$$
If $$A=\begin{bmatrix}1&1&-1\\2&-3&4\\3&-2&3\end{bmatrix}$$ and $$B=\begin{bmatrix}-1&-2&-1\\6&12&6\\5&10&5\end{bmatrix}$$, then which of the following is/are correct?
A and B commute.
AB is null matrix.
Select the correct answer using the code given below :
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1 only
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2 only
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Both 1 and 2
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Neither 1 nor 2
Explanation
We have, $$A=\begin{bmatrix}1&1&-1\\2&-3&4\\3&-2&3\end{bmatrix}$$ and $$B=\begin{bmatrix}-1&-2&-1\\6&12&6\\5&10&5\end{bmatrix}$$
$$AB=\begin{bmatrix}-1+6-5&-2+12-10&-1+6-5\\-2-18+20&-4-36+40&-2-18+20\\-3-12+15&-6-24+30&-3-12+15\end{bmatrix}$$
$$=\begin{bmatrix}0&0&0\\0&0&0\\0&0&0\end{bmatrix}$$
Hence, $$AB$$ is a null matrix.
and $$AB\neq BA$$
Hence, B is the correct option.
If $$\begin{pmatrix} 2 & 3 \\ 4 & 1 \end{pmatrix}\times \begin{pmatrix} 5 & -2 \\ -3 & 1 \end{pmatrix}=\begin{pmatrix} 1 & -1 \\ 17 & \lambda \end{pmatrix}$$ then what is $$\lambda $$ equal to?
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$$7$$
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$$-7$$
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$$9$$
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$$-9$$
Explanation
GIven,
$$\begin{pmatrix} 2 & 3 \\ 4 & 1 \end{pmatrix}\times \begin{pmatrix} 5 & -2 \\ -3 & 1 \end{pmatrix}=\begin{pmatrix} 1 & -1 \\ 17 & \lambda \end{pmatrix}$$
Now,
$$\begin{pmatrix} 2 & 3 \\ 4 & 1 \end{pmatrix}\times \begin{pmatrix} 5 & -2 \\ -3 & 1 \end{pmatrix}=\begin{pmatrix} 10-9 & -4+3 \\ 20-3 & -8+1 \end{pmatrix}=\begin{pmatrix} 1 & -1 \\ 17 & -7 \end{pmatrix}$$
So comparing both, we get $$\lambda =-7$$
If $$\quad A=\begin{pmatrix} 1 & 3 \\ 4 & 5 \end{pmatrix}$$ then $${ A }^{ -1 }$$ equals
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$$\cfrac { 1 }{ 7 } \left( A+6I \right) $$
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$$\cfrac { 1 }{ 7 } \left( A-6I \right) $$
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$$\cfrac { 1 }{ 7 } \left( 6I-A \right) $$
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None of these
Explanation
$$A=\begin{pmatrix} 1 & 3 \\ 4 & 5 \end{pmatrix}$$
for every square matrix $$A$$, the characterisitc of $$A$$ is given $$\left| A-\lambda I \right| =0$$
$$\Rightarrow \begin{vmatrix} 1-\lambda & 3 \\ 4 & 5-\lambda \end{vmatrix}=0$$
$$\Rightarrow { \lambda }^{ 2 }-6\lambda -7=0\quad \Rightarrow { A }^{ 2 }-6A-7I=0\Rightarrow 7I={ A }^{ 2 }-6A\quad \quad $$
$$\Rightarrow { A }^{ -1 }=1/7(A-6I)\quad $$
If $$[1\,x\,1] \begin{bmatrix} 1&3&2 \\ 0&5&1\\0&2&0 \end{bmatrix}$$ $$\begin{bmatrix} 1 \\ 1 \\ x \end{bmatrix}=0$$, then the values of $$x$$ are:
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$$1,5$$
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$$-1,-5$$
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$$1,6$$
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$$-1,-6$$
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$$3,3$$
Explanation
Since, $$[1\,x\,1] \begin{bmatrix} 1&3&2 \\ 0&5&1\\0&2&0 \end{bmatrix}$$ $$\begin{bmatrix} 1 \\ 1 \\ x \end{bmatrix}=0$$
$$\Rightarrow \begin{bmatrix} 1 & 3+5x+2 & 2+x+0 \end{bmatrix}\begin{bmatrix} 1 \\ 1 \\ x \end{bmatrix}=0$$
$$\Rightarrow \begin{bmatrix} 1 &5+5x & 2+x \end{bmatrix}\begin{bmatrix} 1 \\ 1 \\ x \end{bmatrix}=0$$
$$\Rightarrow \left[ 1+5+5x+2x+{ x }^{ 2 } \right] =0$$
$$\Rightarrow { x }^{ 2 }+7x+6=0$$
$$\Rightarrow x^2+6x+x+6=0$$
$$\Rightarrow (x+1)(x+6)=0$$
$$\therefore x=-1,-6$$
If $$A = \begin{bmatrix} 2& 3\\ -1 & 2\end{bmatrix}$$, then $$A^{3} + 3A^{2} - 4A + 1$$ is equal to
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$$\begin{bmatrix} 1& 1\\ 1 & 0\end{bmatrix}$$
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$$\begin{bmatrix} -14& 51\\ -17 & -14\end{bmatrix}$$
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$$\begin{bmatrix} -14& -51\\ -17 & -14\end{bmatrix}$$
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$$\begin{bmatrix} -1& -1\\ -1 & 0\end{bmatrix}$$
Explanation
$$A = \begin{pmatrix} 2& 3\\ -1 & 2\end{pmatrix}$$
$$\therefore A^{2} = \begin{pmatrix} 1& 12\\ -4 & 1\end{pmatrix}, A^{3} = \begin{pmatrix} -10& 27\\ -9 & -10\end{pmatrix}$$
$$\therefore A^{3} + 3A^{2} - 4A + t = \begin{pmatrix} -10& 27\\ -9 & -10\end{pmatrix} + \begin{pmatrix} 3& 36\\ -12 & 3\end{pmatrix} + \begin{pmatrix} -8& -12\\ 4 & -8\end{pmatrix} + \begin{pmatrix} 1& 0\\ 0 & 1\end{pmatrix}$$
$$= \begin{pmatrix} -14& 51\\ -17 & -14\end{pmatrix}$$
Hence (b) is correct choice.
If $$A=\begin{pmatrix} 3 & 1 \\ -9 & -3 \end{pmatrix}$$ then $${ \left( 1+2A+3{ A }^{ 2 }+....\infty \right) }^{ -1 }$$ equals
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$$\begin{pmatrix} -5 & -2 \\ 18 & 7 \end{pmatrix}$$
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$$\begin{pmatrix} -5 & 18 \\ -2 & 7 \end{pmatrix}$$
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$$\begin{pmatrix} 7 & -2 \\ 18 & -5 \end{pmatrix}$$
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None of these
Explanation
$$A=\begin{pmatrix} 3 & 1 \\ -9 & -3 \end{pmatrix}$$
$$\quad \therefore { A }^{ 2 }=\begin{pmatrix} 3 & 1 \\ -9 & -3 \end{pmatrix}\begin{pmatrix} 3 & 1 \\ -9 & -3 \end{pmatrix}=\begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}$$
$$\therefore 1+2A+3{ A }^{ 2 }+....\infty =1+2A$$
$$=\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}+\begin{pmatrix} 6 & 2 \\ -18 & -6 \end{pmatrix}=\begin{pmatrix} 7 & 2 \\ -18 & -5 \end{pmatrix}$$
$$\therefore (1+2A+3{ A }^{ 2 }+....\infty)^{-1}=(1+2A)^{-1} $$
$$\quad ={ \begin{pmatrix} 7 & 2 \\ -18 & -5 \end{pmatrix} }^{ -1 }=\begin{pmatrix} -5 & -2 \\ 18 & 7 \end{pmatrix}$$
If $$\begin{bmatrix} \alpha & \beta \\ \gamma & -\alpha \end{bmatrix}$$ to the square is two rowed unit matrix, then $$\alpha ,\beta ,\gamma $$ should satisfy the relation
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$$1+{ \alpha }^{ 2 }+\beta \gamma =0$$
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$$1-{ \alpha }^{ 2 }-\beta \gamma =0$$
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$$1-{ \alpha }^{ 2 }+\beta \gamma =0$$
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$${ \alpha }^{ 2 }+\beta \gamma -1=0$$
Explanation
since $$\begin{bmatrix} \alpha & \beta \\ \gamma & -\alpha \end{bmatrix}$$ is a square root of $${I}_{2}$$ i.e., two rowed unit matrix
$$\therefore { \begin{bmatrix} \alpha & \beta \\ \gamma & -\alpha \end{bmatrix} }^{ 2 }=\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$$
$$\Rightarrow \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}=\begin{bmatrix} \alpha & \beta \\ \gamma & -\alpha \end{bmatrix}\begin{bmatrix} \alpha & \beta \\ \gamma & -\alpha \end{bmatrix}$$
$$\Rightarrow \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}=\begin{bmatrix} { \alpha }^{ 2 }+\beta \gamma & 0 \\ 0 & { \alpha }^{ 2 }+\beta \gamma \end{bmatrix}$$
$$\therefore { \alpha }^{ 2 }+\beta \gamma =1$$
$$\Rightarrow 1-{ \alpha }^{ 2 }-\beta \gamma =0$$
If $$A=\begin{bmatrix} \alpha & 0 \\ 1 & 1 \end{bmatrix}$$ and $$B=\begin{bmatrix} 1 & 0 \\ 5 & 1 \end{bmatrix}$$, then value of $$\alpha$$ for which $${A}^{2}=B$$, is
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$$1$$
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$$-1$$
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$$4$$
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No real value
Explanation
Given :
$$A=\begin{bmatrix} \alpha & 0 \\ 1 & 1 \end{bmatrix}$$ and $$B=\begin{bmatrix} 1 & 0 \\ 5 & 1 \end{bmatrix}$$
$${ A }^{ 2 }=\begin{bmatrix} \alpha & 0 \\ 1 & 1 \end{bmatrix}\begin{bmatrix} \alpha & 0 \\ 1 & 1 \end{bmatrix}=\begin{bmatrix} { \alpha }^{ 2 } & 0 \\ \alpha +1 & 1 \end{bmatrix}$$
Now $${ A }^{ 2 }=B\quad $$
$$\begin{bmatrix} { \alpha }^{ 2 } & 0 \\ \alpha +1 & 1 \end{bmatrix}=\begin{bmatrix} 1 & 0 \\ 5 & 1 \end{bmatrix}$$
$$\Rightarrow { \alpha }^{ 2 }=1$$ and $$\alpha +1=5$$
Clearly, no real value of $$\alpha$$ exist.
If $$A = \begin{bmatrix}1 &3 \\ 3 & 4\end{bmatrix}$$ and $$A^{2} - kA - 5I_{2} = 0$$, then the value of $$k$$ is
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$$3$$
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$$5$$
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$$7$$
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$$-7$$
Explanation
Given, $$A = \begin{bmatrix}1 &3 \\ 3 & 4\end{bmatrix}$$
Now, $$A^{2} - 5I_{2} = \begin{bmatrix}1 &3 \\ 3 & 4\end{bmatrix}\begin{bmatrix}1 &3 \\ 3 & 4\end{bmatrix} - 5\begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}$$
$$= \begin{bmatrix}1 + 9 & 3 + 12 \\ 3 + 12 & 9 + 16\end{bmatrix} - \begin{bmatrix}5 &0 \\ 0 & 5\end{bmatrix}$$
$$= \begin{bmatrix}5 &15 \\ 15 & 20\end{bmatrix} = 5\begin{bmatrix}1 &3 \\ 3 & 4\end{bmatrix}$$
$$\Rightarrow A^{2} - 5I_{2} = 5A$$
But it is given that $$A^{2} - 5I_{2} = kA$$
$$k = 5$$.
If $$A=\begin{bmatrix} 1 & -3 \\ 2 & k \end{bmatrix}$$ and $${ A }^{ 2 }-4A+10I=A$$, then $$k$$ is equal to
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$$0$$
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$$-4$$
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$$4$$ and not $$1$$
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$$1$$ or $$4$$
Explanation
Given, $$A=\begin{bmatrix} 1 & -3 \\ 2 & k \end{bmatrix}$$ and
$${ A }^{ 2 }-4A+10I=A$$
$$\Rightarrow \begin{bmatrix} 1 & -3 \\ 2 & k \end{bmatrix} \begin{bmatrix} 1 & -3 \\ 2 & k \end{bmatrix} -4 \begin{bmatrix} 1 & -3 \\ 2 & k \end{bmatrix} + 10 \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & -3 \\ 2 & k \end{bmatrix}$$
$$\Rightarrow \begin{bmatrix} -5 & -3-3k \\ 2+2k & -6+{ k }^{ 2 } \end{bmatrix} - \begin{bmatrix} 4 & -12 \\ 8 & 4k \end{bmatrix} + \begin{bmatrix} 10 & 0 \\ 0 & 10 \end{bmatrix} = \begin{bmatrix} 1 & -3 \\ 2 & k \end{bmatrix}$$
$$\Rightarrow \begin{bmatrix} 1 & 9-3k \\ -6+2k & 4+{ k }^{ 2 }-4k \end{bmatrix} = \begin{bmatrix} 1 & -3 \\ 2 & k \end{bmatrix}$$
$$\Rightarrow 9-3k=-3, -6+2k=2$$ ....(i)
and $$4+{ k }^{ 2 }-4k=k$$
$${ k }^{ 2 }-5k+4=0$$
$$\Rightarrow \left( k-4 \right) \left( k-1 \right) =0 \Rightarrow k=4,1$$
But $$k=1$$ is not satisfied the equation (i).
If $$A = \begin{vmatrix} 5 & x-2 \\ 2x+3 & x+1 \end{vmatrix}$$ is symmetric, then $$x = $$_____
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$$4$$
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$$5$$
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$$-5$$
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$$-4$$
Explanation
A=$$ \begin {vmatrix} 5&x-2\\2x-3&x+1\end{vmatrix}$$
$$\implies A^T=$$
$$ \begin {vmatrix} 5&2x-3\\x-2&x+1\end{vmatrix}$$
Since $$A$$ is symmetric, $$\implies A=A^T$$
$$\implies x-2=2x+3$$
$$\implies x=-5$$
If $$A$$ is a non zero square matrix of order $$n$$ with $$det\left( I+A \right) \neq 0$$, and $${A}^{3}=0$$, where $$I,O$$ are unit and null matrices of order $$n\times n$$ respectively, then $${ \left( I+A \right) }^{ -1 }=$$
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$$I-A+{ A }^{ 2 }$$
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$$I+A+{ A }^{ 2 }$$
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$$I+{ A }^{ 2 }$$
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$$I+A$$
Explanation
$$det(I+A)\neq 0$$
$$A^3=0$$ where $$0$$ is null matrix, $$I$$ is the identity matrix
$$A^3+I=I$$ [adding $$I$$ on both sides]
$$(A+I)(A^2-IA+I^2)=I$$ [by the formula of $$a^3+b^3$$]
$$(A+I)(A^2-A+I)=I$$
$$(I+A)(I+A)^{-1}=I$$ [by the rule of inverse matrix]
hence $$(I+A)^{-1}=(A^2-A+I)$$
Ans: $$I-A+A^2$$
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