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

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

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

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

$= 2\left( {4 + 1} \right) - 3\left( { - 2 - 3} \right) + 3\left( { - 1 + 6} \right) = 10 + 15 + 15 = 40 \ne 0$

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

Here, cofactors of elements of A are :
${A_{11}} = {( - 1)^{1 + 1}}\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{ - 2}&1\\{ - 1}&{ - 2}\end{array}} \right| = (4 + 1) = 5,$

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

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

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

${A_{22}} = {( - 1)^{2 + 2}}\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}2&3\\3&{ - 2}\end{array}} \right| = ( - 4 - 9) = - 13,$

${A_{23}} = {( - 1)^{2 + 3}}\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}2&3\\3&{ - 1}\end{array}} \right| = - ( - 2 - 9) = 11,$,

${A_{31}} = {( - 1)^{3 + 1}}\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}3&3\\{ - 2}&1\end{array}} \right| = (3 + 6) = 9,$

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

${A_{33}} = {( - 1)^{3 + 3}}\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}2&3\\1&{ - 2}\end{array}} \right| = ( - 4 - 3) = - 7,$

% $\therefore$ $adj\;\;A = {\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}5&5&5\\3&{ - 13}&{11}\\9&1&{ - 7}\end{array}} \right]^{\prime}} = \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}5&3&9\\5&{ - 13}&1\\5&{11}&{ - 7}\end{array}} \right]$

Hence, ${A^{ - 1}} = \cfrac{1}{{|A|}}(adj\;A) = \cfrac{1}{{40}}\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}5&3&9\\5&{ - 13}&1\\5&{11}&{ - 7}\end{array}} \right]$

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

$\Rightarrow$ $\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}x\\y\\z\end{array}} \right] = \cfrac{1}{{40}}\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}5&3&9\\5&{ - 13}&1\\5&{11}&{ - 7}\end{array}} \right]\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}5\\{ - 4}\\3\end{array}} \right]$

$= \cfrac{1}{{40}}\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{25 - 12 + 27}\\{25 + 52 + 3}\\{25 - 44 - 21}\end{array}} \right] = \cfrac{1}{{40}}\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{40}\\{80}\\{ - 40}\end{array}} \right] = \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1\\2\\{ - 1}\end{array}} \right]$

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