Let $A = \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1&{ - 2}&1\\{ - 2}&3&1\\1&1&5\end{array}} \right]$ . Verify that,

(i) ${[adj\;A]^{ - 1}} = adj({A^{ - 1}})$

(ii) ${({A^{ - 1}})^{ - 1}} = A$

$A = \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1&{ - 2}&1\\{ - 2}&3&1\\1&1&5\end{array}} \right]$

Now, $|A| = 1(15 - 1) + 2( - 10 - 1) + 1( - 2 - 3)$

$= 14 - 22 - 5 = - 13 \ne 0$

$\therefore$ ${A^{ - 1}}$ exists.

Cofactors of elements of A are :

${A_{11}} = {( - 1)^{1 + 1}}\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}3&1\\1&5\end{array}} \right| = 15 - 1 = 14,$

${A_{12}} = {( - 1)^{1 + 2}}\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{ - 2}&1\\1&5\end{array}} \right| = - ( - 10 - 1) = 11,$

${A_{13}} = {( - 1)^{1 + 3}}\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{ - 2}&3\\1&1\end{array}} \right| = ( - 3 - 2) = - 5,$

${A_{21}} = {( - 1)^{2 + 1}}\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{ - 2}&1\\1&5\end{array}} \right| = - ( - 10 - 1) = 11,$

${A_{22}} = {( - 1)^{2 + 2}}\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1&1\\1&5\end{array}} \right| = 5 - 1 = 4,$

${A_{23}} = {( - 1)^{2 + 3}}\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1&{ - 2}\\1&1\end{array}} \right| = - (1 + 2) = - 3,$,

${A_{31}} = {( - 1)^{3 + 1}}\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{ - 2}&1\\3&1\end{array}} \right| = - 2 - 3 = - 5,$

${A_{32}} = {( - 1)^{3 + 2}}\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1&1\\{ - 2}&1\end{array}} \right| = - (1 + 2) = - 3,$

${A_{33}} = {( - 1)^{3 + 3}}\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1&{ - 2}\\{ - 2}&3\end{array}} \right| = 3 - 4 = - 1$

%$\therefore$ $adj\;A = {\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{14}&{11}&{ - 5}\\{11}&4&{ - 3}\\{ - 5}&{ - 3}&{ - 1}\end{array}} \right]^{\prime}} = \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{14}&{11}&{ - 5}\\{11}&4&{ - 3}\\{ - 5}&{ - 3}&{ - 1}\end{array}} \right]$

So, ${A^{ - 1}} = \cfrac{1}{{|A|}}(adj\;A) = \cfrac{1}{{ - 13}}\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{14}&{11}&{ - 5}\\{11}&4&{ - 3}\\{ - 5}&{ - 3}&{ - 1}\end{array}} \right] = \cfrac{1}{{13}}\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{ - 14}&{ - 11}&5\\{ - 11}&{ - 4}&3\\5&3&1\end{array}} \right]$

(i) Now, $|adj\;A| = \left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{14}&{11}&{ - 5}\\{11}&4&{ - 3}\\{ - 5}&{ - 3}&{ - 1}\end{array}} \right|$

$= 14( - 4 - 9) - 11( - 11 - 15) - 5( - 33 + 20)$

$= - 182 + 286 + 65 = 269 \ne 0$

$\therefore$ adj A is invertible and ${(adj\;A)^{ - 1}} = \cfrac{1}{{|adj\;A|}}\{ adj\;(adj\;A)\}$

Now, to obtain $adj(adj\;A)$,

the cofactors of adj A are :
$adj{A_{11}} = {( - 1)^{1 + 1}}\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}4&{ - 3}\\{ - 3}&{ - 1}\end{array}} \right| = - 4 - 9 = - 13$

$adj{A_{12}} = {( - 1)^{1 + 2}}\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{11}&{ - 3}\\{ - 5}&{ - 1}\end{array}} \right| = - ( - 11 - 15) = 26,$

$adj{A_{13}} = {( - 1)^{1 + 3}}\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{11}&4\\{ - 5}&{ - 3}\end{array}} \right| = - 33 + 20 = - 13,$

$adj{A_{21}} = {( - 1)^{2 + 1}}\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{11}&{ - 5}\\{ - 3}&{ - 1}\end{array}} \right| = - ( - 11 - 15) = 26,$

$adj{A_{22}} = {( - 1)^{2 + 2}}\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{14}&{ - 5}\\{ - 5}&{ - 1}\end{array}} \right| = - 14 - 25 = - 39,$

$adj{A_{23}} = {( - 1)^{2 + 3}}\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{14}&{11}\\{ - 5}&{ - 3}\end{array}} \right| = - ( - 42 + 55) = - 13,$,

$adj{A_{32}} = {( - 1)^{3 + 2}}\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{14}&{ - 5}\\{11}&{ - 3}\end{array}} \right| = - ( - 42 + 55) = - 13,$

$adj{A_{33}} = {( - 1)^{3 + 3}}\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{14}&{11}\\{11}&4\end{array}} \right| = (56 - 121) = - 65$

%$\therefore$ $adj(adjA) = {\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{ - 13}&{26}&{ - 13}\\{26}&{ - 39}&{ - 13}\\{ - 13}&{ - 13}&{ - 65}\end{array}} \right]^{\prime}} = \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{ - 13}&{26}&{ - 13}\\{26}&{ - 39}&{ - 13}\\{ - 13}&{ - 13}&{ - 65}\end{array}} \right]$

So, ${(adj\;A)^{ - 1}} = \cfrac{1}{{|adj\;A|}}\{ adj(adjA)\} = \cfrac{1}{{169}}\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{ - 13}&{26}&{ - 13}\\{26}&{ - 39}&{ - 13}\\{ - 13}&{ - 13}&{ - 65}\end{array}} \right]$

$= \cfrac{1}{{13}}\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{ - 1}&2&{ - 1}\\2&{ - 3}&{ - 1}\\{ - 1}&{ - 1}&{ - 5}\end{array}} \right]$

..(1)
Also, $adj({A^{ - 1}}) = adj\left( {\cfrac{1}{{13}}\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{ - 14}&{ - 11}&5\\{ - 11}&{ - 4}&3\\5&3&1\end{array}} \right]} \right)$

Similarly,
$adj\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{ - \cfrac{{14}}{{13}}}&{ - \cfrac{{11}}{{13}}}&{\cfrac{5}{{13}}}\\{ - \cfrac{{11}}{{13}}}&{ - \cfrac{4}{{13}}}&{\cfrac{3}{{13}}}\\{\cfrac{5}{{13}}}&{\cfrac{3}{{13}}}&{\cfrac{1}{{13}}}\end{array}} \right]$

$= \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{ - \cfrac{{13}}{{169}}}&{\cfrac{{26}}{{169}}}&{ - \cfrac{{13}}{{169}}}\\{\cfrac{{26}}{{169}}}&{ - \cfrac{{39}}{{169}}}&{ - \cfrac{{13}}{{169}}}\\{ - \cfrac{{13}}{{169}}}&{ - \cfrac{{13}}{{169}}}&{ - \cfrac{{65}}{{169}}}\end{array}} \right]$

$\Rightarrow$ $(adj\;{A^{ - 1}}) = \cfrac{1}{{13}}\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{ - 1}&2&{ - 1}\\2&{ - 3}&{ - 1}\\{ - 1}&{ - 1}&{ - 5}\end{array}} \right]$

….(2)

From (1) and (2),

we find that, $adj({A^{ - 1}}) = {(adjA)^{ - 1}}.$

(ii) $|{A^{ - 1}}| = \left| {\cfrac{1}{{13}}\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{ - 14}&{ - 11}&5\\{ - 11}&{ - 4}&3\\5&3&1\end{array}} \right]} \right|$

$= \cfrac{1}{{{{(13)}^2}}}\{ - 14( - 4 - 9) + 11( - 11 - 15) + 5( - 33 + 20)\}$

$= \cfrac{1}{{13 \times 13 \times 13}}( - 169) = - \cfrac{1}{{13}} \ne 0.$

$\therefore$ ${({A^{ - 1}})^{ - 1}}$

exists and ${({A^{ - 1}})^{ - 1}} = \cfrac{1}{{|{A^{ - 1}}|}}(adj\;{A^{ - 1}})$

$= \cfrac{1}{{ - \cfrac{1}{{13}}}} \cdot \cfrac{1}{{13}}\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{ - 1}&2&{ - 1}\\2&{ - 3}&{ - 1}\\{ - 1}&{ - 1}&{ - 5}\end{array}} \right] = \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1&{ - 2}&1\\{ - 2}&3&1\\1&1&5\end{array}} \right]$

(using (2))
= A.