. $x - y + z = 4,2x + y - 3z = 0,x + y + z = 2.$

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

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

and $B = \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}4\\0\\2\end{array}} \right]$

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

$= 1\left( {1 + 3} \right) + 1\left( {2 + 3} \right) + 1\left( {2 - 1} \right) = 4 + 5 + 1 = 10 \ne 0$

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

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

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

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

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

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

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

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

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

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

% $\therefore$ $adj\;A = {\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}4&{ - 5}&1\\2&0&{ - 2}\\2&5&3\end{array}} \right]^{\prime}} = \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}4&2&2\\{ - 5}&0&5\\1&{ - 2}&3\end{array}} \right]$

and ${A^{ - 1}} = \cfrac{1}{{|A|}}(adj\;A) = \cfrac{1}{{10}}\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}4&2&2\\{ - 5}&0&5\\1&{ - 2}&3\end{array}} \right]$

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

$\Rightarrow$ $\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}x\\y\\z\end{array}} \right] = \cfrac{1}{{10}}\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}4&2&2\\{ - 5}&0&5\\1&{ - 2}&3\end{array}} \right]\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}4\\0\\2\end{array}} \right] = \cfrac{1}{{10}}\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{16 + 0 + 4}\\{ - 20 + 0 + 10}\\{4 - 0 + 6}\end{array}} \right]$

$\Rightarrow$ $\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}x\\y\\z\end{array}} \right] = \cfrac{1}{{10}}\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{20}\\{ - 10}\\{10}\end{array}} \right] = \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}2\\{ - 1}\\1\end{array}} \right] \Rightarrow x = 2,y = - 1$ and $z = 1$.