$\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1&2&3\\0&2&4\\0&0&5\end{array}} \right]$

Let $A = \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1&2&3\\0&2&4\\0&0&5\end{array}} \right]$

Then, $|A| = 10 \ne 0$ .

${A_{11}} = {( - 1)^{1 + 1}}\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}2&4\\0&5\end{array}} \right] = 10$

${A_{12}} = {( - 1)^{1 + 2}}\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}0&4\\0&5\end{array}} \right] = 0$

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

${A_{21}} = {( - 1)^{2 + 1}}\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}2&3\\0&5\end{array}} \right] = - 10$

${A_{22}} = {( - 1)^{2 + 2}}\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1&3\\0&5\end{array}} \right] = 5$

${A_{23}} = {( - 1)^{2 + 3}}\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1&2\\0&0\end{array}} \right] = 0$

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

${A_{32}} = {( - 1)^{3 + 2}}\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1&3\\0&4\end{array}} \right] = - 4$

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

% $\therefore$ adj $A = {\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{10}&0&0\\{ - 10}&5&0\\2&{ - 4}&2\end{array}} \right]^{\prime}} = \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{10}&{ - 10}&2\\0&5&{ - 4}\\0&0&2\end{array}} \right]$

Hence, ${A^{ - 1}} = \cfrac{1}{{|A|}}(adj\,A) = \cfrac{1}{{10}}\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{10}&{ - 10}&2\\0&5&{ - 4}\\0&0&2\end{array}} \right]$