Question

☆BRAINLIEAST QUESTION☆

Answer the question which is in attachment

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Answer 1

$\large\underline{\sf{Solution-}}$

Given matrix is

$\red{\rm :\longmapsto\:A = \begin{gathered}\sf\left[\begin{array}{ccc}1&2& -3\\6&1&4\\3&5& - 3\end{array}\right]\end{gathered}}$

To find Adjoint of matrix A, we have to find co - factors.

We know,

$\boxed{ \bf \: { c_{ij} \: = \: {( - 1)}^{i \: + \: j} \: m_{ij} \: }}$

Now,

$\rm :\longmapsto\:c_{11} = {( - 1)}^{1 + 1}\begin{array}{|cc|}\sf 1 &\sf 4 \\ \sf 5 &\sf - 3 \\\end{array} = ( - 3 - 20) = - 23$

$\rm :\longmapsto\:c_{12} = {( - 1)}^{1 + 2}\begin{array}{|cc|}\sf 6 &\sf 4 \\ \sf 3 &\sf - 3 \\\end{array} = - ( - 18 - 12) = 30$

$\rm :\longmapsto\:c_{13} = {( - 1)}^{1 + 3}\begin{array}{|cc|}\sf 6 &\sf 1 \\ \sf 3 &\sf 5 \\\end{array} = (30 - 3) = 27$

$\rm :\longmapsto\:c_{21} = {( - 1)}^{2 + 1}\begin{array}{|cc|}\sf 2 &\sf - 3 \\ \sf 5 &\sf - 3 \\\end{array} = - ( - 6 + 15) = - 9$

$\rm :\longmapsto\:c_{22} = {( - 1)}^{2 + 2}\begin{array}{|cc|}\sf 1 &\sf - 3 \\ \sf 3 &\sf - 3 \\\end{array} = ( - 3 + 9) = 6$

$\rm :\longmapsto\:c_{23} = {( - 1)}^{2 + 3}\begin{array}{|cc|}\sf 1 &\sf 2 \\ \sf 3 &\sf 5 \\\end{array} = - (5 - 6) =1$

$\rm :\longmapsto\:c_{31} = {( - 1)}^{3 + 1}\begin{array}{|cc|}\sf 2 &\sf - 3 \\ \sf 1 &\sf 4 \\\end{array} =(8 + 3) =11$

$\rm :\longmapsto\:c_{32} = {( - 1)}^{3 + 2}\begin{array}{|cc|}\sf 1 &\sf - 3 \\ \sf 6 &\sf 4 \\\end{array} = - (4 + 18) = - 22$

$\rm :\longmapsto\:c_{33} = {( - 1)}^{3 + 3}\begin{array}{|cc|}\sf 2 &\sf - 3 \\ \sf 1 &\sf 4 \\\end{array} =(8 + 3) = 11$

So,

$\rm :\longmapsto\:adjA = \begin{gathered}\sf\left[\begin{array}{ccc} - 23&30&27\\ - 9&6&1\\11& - 22& 11\end{array}\right]\end{gathered}'$

$\rm :\longmapsto\:adjA = \begin{gathered}\sf\left[\begin{array}{ccc} - 23& - 9&11\\ 30&6& - 22\\27&1& 11\end{array}\right]\end{gathered}$

Additional Information : -

If A is a square matrix of order n, then

$\boxed{ \rm{ |adjA| = { |A| }^{n - 1} }}$

$\boxed{ \rm{ A \: adjA \: = \: (adjA) \: A \: = \: |A| I}}$

$\boxed{ \rm{ |A'| = |A|}}$

$\boxed{ \rm{ | {A}^{ - 1} | = \frac{1}{ |A| }}}$