Using matrix method, solve the system of equations $3x + 2y - 2z = 3$, $x + 2y + 3z = 6$ and $2x - y + z = 2$.

Given system of equations is
$3x + 2y - 2z = 3$

$x + 2y + 3z = 6$

and $2x - y + z = 2$

In the form of $AX = B$

$\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}3&2&{ - 2}\\1&2&3\\2&{ - 1}&1\end{array}} \right]\left[ {\begin{array}{llllllllllllllllllll}x\\y\\z\end{array}} \right] = \left[ {\begin{array}{llllllllllllllllllll}3\\6\\2\end{array}} \right]$

For ${A^{ - 1}}$, $|A| = |3(5) - 2(1 - 6) + ( - 2)( - 5)|$

$= |15 + 10 + 10| = |35| \ne 0$

$\therefore$ ${A_{11}} = 5,{A_{12}} = 5,{A_{13}} = - 5,{A_{21}} = 0,{A_{22}} = 7,{A_{23}} = 7,{A_{31}} = 10,{A_{32}} = - 11$

and ${A_{33}} = 4$

$\therefore$ ${\mathop{\rm adj}\nolimits} A = \left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}5&5&{ - 5}\\0&7&7\\{10}&{ - 11}&4\end{array}} \right| = \left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}5&0&{10}\\5&7&{ - 11}\\{ - 5}&7&4\end{array}} \right|$

Now, ${A^{ - 1}} = \frac{{{\mathop{\rm adj}\nolimits} A}}{{|A|}} = \frac{1}{{35}}\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}5&0&{10}\\5&7&{ - 11}\\{ - 5}&7&4\end{array}} \right|$

For $X = {A^{ - 1}}B$,
$\left[ {\begin{array}{llllllllllllllllllll}x\\y\\z\end{array}} \right] = \frac{1}{{35}}\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}5&0&{10}\\5&7&{ - 11}\\{ - 5}&7&4\end{array}} \right]\left[ {\begin{array}{llllllllllllllllllll}3\\6\\2\end{array}} \right]$

$= \frac{1}{{35}}\left[ {\begin{array}{cccccccccccccccccccc}{15 + 20}\\{15 + 42 - 22}\\{ - 15 + 42 + 8}\end{array}} \right]$

$= \frac{1}{{35}}\left[ {\begin{array}{llllllllllllllllllll}{35}\\{35}\\{35}\end{array}} \right] = \left[ {\begin{array}{llllllllllllllllllll}1\\1\\1\end{array}} \right]$

$\therefore$ $x = 1,y = 1$ and $z = 1$