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

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

Then, $|A| = \left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1&0&0\\3&3&0\\5&2&{ - 1}\end{array}} \right| = 1\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}3&0\\2&{ - 1}\end{array}} \right| = - 3 - 0 = - 3 \ne 0$

So, A is non-singular matrix and therefore, it is invertible.

Let ${A_{ij}}$ be cofactors of ${a_{ij}}$ in A.

Then, the cofactor of elements of A are given by
${A_{11}} = {( - 1)^{1 + 1}}\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}3&0\\2&{ - 1}\end{array}} \right] = ( - 3 - 0) = - 3$

${A_{12}} = {( - 1)^{1 + 2}}\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}3&0\\5&{ - 1}\end{array}} \right] = - ( - 3 - 0) = 3$

${A_{13}} = {( - 1)^{1 + 3}}\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}3&3\\5&2\end{array}} \right] = 6 - 15 = - 9$

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

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

${A_{23}} = {( - 1)^{2 + 3}}\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1&0\\5&2\end{array}} \right] = - 2$

${A_{31}} = {( - 1)^{3 + 1}}\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}0&0\\3&0\end{array}} \right] = 0$

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

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

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

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