$\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}2&1&3\\4&{ - 1}&0\\{ - 7}&2&1\end{array}} \right]$

Let $A = \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}2&1&3\\4&{ - 1}&0\\{ - 7}&2&1\end{array}} \right]$

Then, $|A| = \left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}2&1&3\\4&{ - 1}&0\\{ - 7}&2&1\end{array}} \right|$

$= 2\left( { - 1 - 0} \right) - 1\left( {4 - 0} \right) + 3\left( {8 - 7} \right) = - 2 - 4 + 3 = - 3 \ne 0.$

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

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

Then, the cofactors of elements of A are given by

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

${A_{12}} = {( - 1)^{1 + 2}}\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}4&0\\{ - 7}&1\end{array}} \right] = - 4$

${A_{13}} = {( - 1)^{1 + 3}}\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}4&{ - 1}\\{ - 7}&2\end{array}} \right] = 8 - 7 = 1$

${A_{21}} = {( - 1)^{2 + 1}}\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1&3\\2&1\end{array}} \right] = - (1 - 6) = 5$

${A_{22}} = {( - 1)^{2 + 2}}\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}2&3\\{ - 7}&1\end{array}} \right] = 2 + 21 = 23$

${A_{23}} = {( - 1)^{2 + 3}}\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}2&1\\{ - 7}&2\end{array}} \right] = - (4 + 7) = - 11$

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

${A_{32}} = {( - 1)^{3 + 2}}\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}2&3\\4&0\end{array}} \right] = 12$

${A_{33}} = {( - 1)^{3 + 3}}\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}2&1\\4&{ - 1}\end{array}} \right] = - 2 - 4 = - 6$

$\therefore$ $adjA = \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{ - 1}&{ - 4}&1\\5&{23}&{ - 11}\\3&{12}&{ - 6}\end{array}} \right] = \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{ - 1}&5&3\\{ - 4}&{23}&{12}\\1&{ - 11}&{ - 6}\end{array}} \right]$

Hence, ${A^{ - 1}} = \cfrac{1}{{|A|}}(adjA) = - \cfrac{1}{3}\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{ - 1}&5&3\\{ - 4}&{23}&{12}\\1&{ - 11}&{ - 6}\end{array}} \right]$