Question

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Answer 1

If 3A= A(x)=(x -2 )

If= x=2 A(2)=(2-2) A(2)=0 answer hai or Achcha Lage to comment karna ok see you

Answer 2

Answer:

Given :

$\mathrm{3A = \left[\begin{array}{ccc}1&2&2\\2&1&-2\\-2&2&-1\end{array}\right] }$

To Prove :

$\mathrm{A^{-1} = A'}$

Solution :

$\longmapsto \mathrm{3A = \left[\begin{array}{ccc}1&2&2\\2&1&-2\\-2&2&-1\end{array}\right] }$

$\implies \mathrm{A = \dfrac{1}{3}\left[\begin{array}{ccc}1&2&2\\2&1&-2\\-2&2&-1\end{array}\right] }$

• Transverse of Matrix A,

Transverse of a Matrix is to replace rows by columns or vice versa.

$\implies\mathrm{A' = \dfrac{1}{3}\left[\begin{array}{ccc}1&2&-2\\2&1&2\\2&-2&-1\end{array}\right]}$ ______[1]

• Det of A is :

$\implies\mathrm{|A| = \bigg(\dfrac{1}{3}\bigg)^3\left|\begin{array}{ccc}1&2&2\\2&1&-2\\-2&2&-1\end{array}\right|}$

$\implies\mathrm{|A| = \bigg(\dfrac{1}{27}\bigg) \Big(1(-1+4) - 2(-2-4)+2(4+2)\Big)}$

$\implies\mathrm{|A| = \bigg(\dfrac{1}{27}\bigg) \Big(3 +12+12\Big)}$

$\implies\mathrm{|A| = \bigg(\dfrac{1}{\cancel{27}}\bigg) \Big(\cancel{27}\Big)}$

$\implies\mathrm{|A| = 1}$

• Cofactor matrix of A :

$\implies \mathrm{ \dfrac{1}{9}\left[\begin{array}{ccc}3&6&6\\6&3&-6\\-6&6&-3\end{array}\right] }$

$\implies \mathrm{ \dfrac{1}{3}\left[\begin{array}{ccc}1&2&2\\2&1&-2\\-2&2&-1\end{array}\right] }$

• Adjoint of A :

Adjoint of A is transverse of cofactor matrix of A

$\implies \mathrm{\dfrac{1}{3} \left[\begin{array}{ccc}1&2&-2\\2&1&2\\2&-2&-1\end{array}\right] }$

• Inverse of A :

We know that,

$\boxed{\mathrm{A^{-1} = \dfrac{AdjA}{|A|}}}$

$\implies \mathrm{A^{-1} = \dfrac{\dfrac{1}{3} \left[\begin{array}{ccc}1&2&-2\\2&1&2\\2&-2&-1\end{array}\right]}{1}}$

$\implies \mathrm{A^{-1} = \dfrac{1}{3} \left[\begin{array}{ccc}1&2&-2\\2&1&2\\2&-2&-1\end{array}\right]}$_____[2]

So, from [1] and [2], we can conclude that,

$\blue \bigstar \ \underline{\boxed{\mathbf{A^{-1} = A'}}}$

$\underline{\Large{\textsf{Hence Proved!!}}}$

Hope it helps!!