Question

( matrix 1&3\\ 3&10 matrix )^ -1 =​

Answer 1

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

Let assume that,

$\rm :\longmapsto\:A = \bigg[ \begin{matrix}1&3 \\ 3&10 \end{matrix} \bigg]$

So, We know that,

$\boxed{ \bf{ \: {A}^{ - 1} = \frac{1}{ |A| } \: adjA \: }}$

Let evaluate,

$\red{\rm :\longmapsto\: |A| }$

$\red{\rm \:  =  \: \begin{array}{|cc|}\sf 1 &\sf 3 \\ \sf 3 &\sf 10 \\\end{array}}$

$\red{\rm \:  =  \: 10 - 9}$

$\red{\rm \:  =  \: 1}$

$\red{\bf\implies \: |A| = 1 \: \ne \: 0}$

Hence,

$\red{\boxed{ \bf{ \: {A}^{ - 1} \: exist \: }}}$

Now, Let find cofactors.

$\rm :\longmapsto\:c_{11} = {( - 1)}^{1 + 1}(10) = 10$

$\rm :\longmapsto\:c_{12} = {( - 1)}^{1 + 2}(3) = - 3$

$\rm :\longmapsto\:c_{21} = {( - 1)}^{2+ 1}(3) = - 3$

$\rm :\longmapsto\:c_{22} = {( - 1)}^{2+ 2}(1) = 1$

So,

$\rm :\longmapsto\:adjA = \bigg[ \begin{matrix}10& - 3 \\ - 3&1 \end{matrix} \bigg]'$

$\bf :\longmapsto\:adjA = \bigg[ \begin{matrix}10& - 3 \\ - 3&1 \end{matrix} \bigg]$

Hence,

$\bf :\longmapsto\: {A}^{ - 1} = \dfrac{1}{1} \bigg[ \begin{matrix}10& - 3 \\ - 3&1 \end{matrix} \bigg]$

$\bf :\longmapsto\: {A}^{ - 1} = \bigg[ \begin{matrix}10& - 3 \\ - 3&1 \end{matrix} \bigg]$

Additional Information :-

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

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

$\rm :\longmapsto\:\boxed{ \bf{ \: |adj \: A| = { |A| }^{n - 1}}}$

$\rm :\longmapsto\:\boxed{ \bf{ \: {(AB)}^{ - 1} = {B}^{ - 1} {A}^{ - 1}}}$

Answer 2

Answer:

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

Let assume that,

$\rm :\longmapsto\:A = \bigg[ \begin{matrix}1&3 \\ 3&10 \end{matrix} \bigg]$

So, We know that,

$\boxed{ \bf{ \: {A}^{ - 1} = \frac{1}{ |A| } \: adjA \: }}$

Let evaluate,

$\red{\rm :\longmapsto\: |A| }$

$\red{\rm \:  =  \: \begin{array}{|cc|}\sf 1 &\sf 3 \\ \sf 3 &\sf 10 \\\end{array}}$

$\red{\rm \:  =  \: 10 - 9}$

$\red{\rm \:  =  \: 1}$

$\red{\bf\implies \: |A| = 1 \: \ne \: 0}$

Hence,

$\red{\boxed{ \bf{ \: {A}^{ - 1} \: exist \: }}}$

Now, Let find cofactors.

$\rm :\longmapsto\:c_{11} = {( - 1)}^{1 + 1}(10) = 10$

$\rm :\longmapsto\:c_{12} = {( - 1)}^{1 + 2}(3) = - 3$

$\rm :\longmapsto\:c_{21} = {( - 1)}^{2+ 1}(3) = - 3$

$\rm :\longmapsto\:c_{22} = {( - 1)}^{2+ 2}(1) = 1$

So,

$\rm :\longmapsto\:adjA = \bigg[ \begin{matrix}10& - 3 \\ - 3&1 \end{matrix} \bigg]'$

$\bf :\longmapsto\:adjA = \bigg[ \begin{matrix}10& - 3 \\ - 3&1 \end{matrix} \bigg]$

Hence,

$\bf :\longmapsto\: {A}^{ - 1} = \dfrac{1}{1} \bigg[ \begin{matrix}10& - 3 \\ - 3&1 \end{matrix} \bigg]$

$\bf :\longmapsto\: {A}^{ - 1} = \bigg[ \begin{matrix}10& - 3 \\ - 3&1 \end{matrix} \bigg]$

Additional Information :-

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

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

$\rm :\longmapsto\:\boxed{ \bf{ \: |adj \: A| = { |A| }^{n - 1}}}$

$\rm :\longmapsto\:\boxed{ \bf{ \: {(AB)}^{ - 1} = {B}^{ - 1} {A}^{ - 1}}}$