$x - y + 2z = 7,3x + 4y - 5z = - 5,2x - y + 3z = 12$

The given system of equations can be written in the form $AX = B$,

where
$A = \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1&{ - 1}&2\\3&4&{ - 5}\\2&{ - 1}&3\end{array}} \right],X = \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}x\\y\\z\end{array}} \right]$ and $B = \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}7\\{ - 5}\\{12}\end{array}} \right]$

Now, $|A| = \left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1&{ - 1}&2\\3&4&{ - 5}\\2&{ - 1}&3\end{array}} \right| = 1\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}4&{ - 5}\\{ - 1}&3\end{array}} \right| + 1\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}3&{ - 5}\\2&3\end{array}} \right| + 2\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}3&4\\2&{ - 1}\end{array}} \right|$

$\Rightarrow$ $|A| = 1(12 - 5) + 1(9 + 10) + 2( - 3 - 8)$

$\Rightarrow$ $|A| = 7 + 19 - 22 = 4 \ne 0$

$\Rightarrow$ A is a non-singular matrix and so the given equations have a unique solution.

Here, cofactors of elements of A are :
${A_{11}} = {( - 1)^{1 + 1}}\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}4&{ - 5}\\{ - 1}&3\end{array}} \right| = (12 - 5) = 7,$

${A_{12}} = {( - 1)^{1 + 2}}\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}3&{ - 5}\\2&3\end{array}} \right| = - (9 + 10) = - 19,$

${A_{13}} = {( - 1)^{1 + 3}}\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}3&4\\2&{ - 1}\end{array}} \right| = ( - 3 - 8) = - 11,$

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

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

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

${A_{31}} = {( - 1)^{3 + 1}}\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{ - 1}&2\\4&{ - 5}\end{array}} \right| = (5 - 8) = - 3,$

${A_{32}} = {( - 1)^{3 + 2}}\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1&2\\3&{ - 5}\end{array}} \right| = - (5 - 6) = 11,$

${A_{33}} = {( - 1)^{3 + 3}}\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1&{ - 1}\\3&4\end{array}} \right| = (4 + 3) = 7,$

%$\therefore$ $adj\;A = {\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}7&{ - 19}&{ - 11}\\1&{ - 1}&{ - 1}\\{ - 3}&{11}&7\end{array}} \right]^{\prime}} = \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}7&1&{ - 3}\\{ - 19}&{ - 1}&{11}\\{ - 11}&{ - 1}&7\end{array}} \right]$

Hence, ${A^{ - 1}} = \cfrac{1}{{|A|}}(adj\;A) = \cfrac{1}{4}\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}7&1&{ - 3}\\{ - 19}&{ - 1}&{11}\\{ - 11}&{ - 1}&7\end{array}} \right]$

Solution of given system is given by $X = {A^{ - 1}}B.$

$\Rightarrow$ $\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}x\\y\\z\end{array}} \right] = \cfrac{1}{4}\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}7&1&{ - 3}\\{ - 19}&{ - 1}&{11}\\{ - 11}&{ - 1}&7\end{array}} \right]\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}7\\{ - 5}\\{12}\end{array}} \right]$

$= \cfrac{1}{4}\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{49 - 5 - 36}\\{ - 133 + 5 + 132}\\{ - 77 + 5 + 84}\end{array}} \right] = \cfrac{1}{4}\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}8\\4\\{12}\end{array}} \right] = \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}2\\1\\3\end{array}} \right]$

$\Rightarrow$ $x = 2,y = 1$ and $z = 3$.