$\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1&{ - 1}&2\\0&2&{ - 3}\\3&{ - 2}&4\end{array}} \right]$

Let $A = \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1&{ - 1}&2\\0&2&{ - 3}\\3&{ - 2}&4\end{array}} \right]$
$\therefore$ $|A| = 1(8 - 6) + 1(0 + 9) + 2(0 - 6)$

$= 2 + 9 - 12 = - 1 \ne 0$
${A_{11}} = {( - 1)^{1 + 1}}\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}2&{ - 3}\\{ - 2}&4\end{array}} \right] = 8 - 6 = 2$

${A_{12}} = {( - 1)^{1 + 2}}\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}0&{ - 3}\\3&4\end{array}} \right] = - (0 + 9) = - 9$

${A_{13}} = {( - 1)^{1 + 3}}\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}0&2\\3&{ - 2}\end{array}} \right] = 0 - 6 = - 6$

${A_{21}} = {( - 1)^{2 + 1}}\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{ - 1}&2\\{ - 2}&4\end{array}} \right] = - ( - 4 + 4) = 0$

${A_{22}} = {( - 1)^{2 + 2}}\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1&2\\3&4\end{array}} \right] = 4 - 6 = - 2$

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

${A_{31}} = {( - 1)^{3 + 1}}\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{ - 1}&2\\2&{ - 3}\end{array}} \right] = 3 - 4 = - 1$

${A_{32}} = {( - 1)^{3 + 2}}\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1&2\\0&{ - 3}\end{array}} \right] = - ( - 3 - 0) = 3$
${A_{33}} = {( - 1)^{3 + 3}}\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1&{ - 1}\\0&2\end{array}} \right] = 2 + 0 = 2$

% $\therefore$ $adj\;A = {\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}2&{ - 9}&{ - 6}\\0&{ - 2}&{ - 1}\\{ - 1}&3&2\end{array}} \right]^{\prime}} = \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}2&0&{ - 1}\\{ - 9}&{ - 2}&3\\{ - 6}&{ - 1}&2\end{array}} \right]$

Hence, ${A^{ - 1}} = \cfrac{{adjA}}{{|A|}} = \cfrac{1}{{ - 1}}\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}2&0&{ - 1}\\{ - 9}&{ - 2}&3\\{ - 6}&{ - 1}&2\end{array}} \right] = \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{ - 2}&0&1\\9&2&{ - 3}\\6&1&{ - 2}\end{array}} \right]$