Question

Q.1) find à by using adjoint Method ​

Answer 1

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

Given matrix is

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

Let first evaluate |A|.

So,

$\rm :\longmapsto\: |A| = 3(9 - 4) - 2(3 - 8) + 1(1 - 6)$

$\rm :\longmapsto\: |A| = 3(5) - 2( -5 ) + 1( - 5)$

$\rm :\longmapsto\: |A| = 15 + 10 - 5$

$\rm :\longmapsto\: |A| = 20 \: \ne \: 0$

$\rm\implies \: {A}^{ - 1} \: exist$

So, Let evaluate Cofactors of matrix A

$\rm :\longmapsto\:c_{11} = {( - 1)}^{1 + 1}\begin{array}{|cc|}\sf 3 &\sf 4 \\ \sf 1 &\sf 3 \\\end{array} = 9 - 4 = 5$

$\rm :\longmapsto\:c_{12} = {( - 1)}^{1 + 2}\begin{array}{|cc|}\sf 1 &\sf 4 \\ \sf 2 &\sf 3 \\\end{array} = - (3 - 8) = 5$

$\rm :\longmapsto\:c_{13} = {( - 1)}^{1 + 3}\begin{array}{|cc|}\sf 1 &\sf 3 \\ \sf 2 &\sf 1 \\\end{array} = 1 - 6 = - 5$

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

$\rm :\longmapsto\:c_{22} = {( - 1)}^{2 + 2}\begin{array}{|cc|}\sf 3 &\sf 1 \\ \sf 2 &\sf 3 \\\end{array} = 9 - 2 = 7$

$\rm :\longmapsto\:c_{23} = {( - 1)}^{2 + 3}\begin{array}{|cc|}\sf 3 &\sf 2\\ \sf 2 &\sf 1 \\\end{array} = - (3 - 4)= 1$

$\rm :\longmapsto\:c_{31} = {( - 1)}^{3+ 1}\begin{array}{|cc|}\sf 2 &\sf 1\\ \sf 3 &\sf 4\\\end{array} = 8 - 3 = 5$

$\rm :\longmapsto\:c_{32} = {( - 1)}^{3+ 2}\begin{array}{|cc|}\sf 3 &\sf 1\\ \sf 1 &\sf 4\\\end{array} = - (12 - 1) = - 11$

$\rm :\longmapsto\:c_{33} = {( - 1)}^{3+ 3}\begin{array}{|cc|}\sf 3 &\sf 2\\ \sf 1 &\sf 3\\\end{array} = 9 - 2 = 7$

So,

$\rm\implies \:adjA = \begin{gathered}\sf \left[\begin{array}{ccc}5& - 5&5\\5& 7& - 11\\ - 5&1&7\end{array}\right]\end{gathered}$

We know,

$\purple{\rm :\longmapsto\: {A}^{ - 1} = \dfrac{1}{ |A| } adjA}$

$\rm\implies \: {A}^{ - 1} = \dfrac{1}{20} \begin{gathered}\sf \left[\begin{array}{ccc}5& - 5&5\\5& 7& - 11\\ - 5&1&7\end{array}\right]\end{gathered}$

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More to know

$\purple{\rm :\longmapsto\:\boxed{\tt{ A \: adjA = (adjA) \: A \: = \: |A| }}}$

$\purple{\rm :\longmapsto\:\boxed{\tt{ {AA}^{ - 1} = {A}^{ - 1}A = I}}}$

$\purple{\rm :\longmapsto\:\boxed{\tt{ | {A}^{ - 1} | = \frac{1}{ |A| } }}}$

$\purple{\rm :\longmapsto\:\boxed{\tt{ |A| = 0\rm\implies \: {A}^{ - 1} \: doesnot \: exist}}}$