$2x + y + z = 1,x - 2y - z = 3/2,3y - 5z = 9.$

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

where
$A = \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}2&1&1\\1&{ - 2}&{ - 1}\\0&3&{ - 5}\end{array}} \right],X = \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}X\\Y\\Z\end{array}} \right]$ and $B = \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1\\{\cfrac{3}{2}}\\9\end{array}} \right]$

Now, $|A| = \left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}2&1&1\\1&{ - 2}&{ - 1}\\0&3&{ - 5}\end{array}} \right| = 2\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{ - 2}&{ - 1}\\3&{ - 5}\end{array}} \right| - 1\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1&{ - 1}\\0&{ - 5}\end{array}} \right| + 1\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1&{ - 2}\\0&3\end{array}} \right|$

$|A| = 2(10 + 3) - 1( - 5) + 1(3) \Rightarrow |A| = 26 + 8 = 34 \ne 0.$

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

Here, cofactors of elements of A are :
${A_{11}} = {( - 1)^{1 + 1}}\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{ - 2}&{ - 1}\\3&{ - 5}\end{array}} \right| = (10 + 3) = 13,$

${A_{12}} = {( - 1)^{1 + 2}}\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1&{ - 1}\\0&{ - 5}\end{array}} \right| = ( - 5 + 0) = 5,$

${A_{13}} = {( - 1)^{1 + 3}}\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1&{ - 2}\\0&3\end{array}} \right| = (3 - 0) = 3,$

${A_{21}} = {( - 1)^{2 + 1}}\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1&1\\3&{ - 5}\end{array}} \right| = - ( - 5 - 3) = 8,$

${A_{22}} = {( - 1)^{2 + 2}}\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}2&1\\0&{ - 5}\end{array}} \right| = ( - 10 - 0) = - 10,$

${A_{23}} = {( - 1)^{2 + 3}}\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}2&1\\0&3\end{array}} \right| = - (6 - 0) = - 6,$,

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

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

${A_{33}} = {( - 1)^{3 + 3}}\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}2&1\\1&{ - 2}\end{array}} \right| = - 4 - 1 = - 5$

% $\therefore$ $adj\;A = {\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{13}&5&3\\8&{ - 10}&{ - 6}\\1&3&{ - 5}\end{array}} \right]^{\prime}} = \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{13}&8&1\\5&{ - 10}&3\\3&{ - 6}&{ - 5}\end{array}} \right]$

and ${A^{ - 1}} = \cfrac{1}{{|A|}}(adj\;A) = \cfrac{1}{{34}}\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{13}&8&1\\5&{ - 10}&3\\3&{ - 6}&{ - 5}\end{array}} \right]$

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

$\Rightarrow$ $\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}x\\y\\z\end{array}} \right] = \cfrac{1}{{34}}\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{13}&8&1\\5&{ - 10}&3\\3&{ - 6}&{ - 5}\end{array}} \right]\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1\\{3/2}\\9\end{array}} \right]$

$= \cfrac{1}{{34}}\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{13 + 12 + 9}\\{5 - 15 + 27}\\{3 - 9 - 45}\end{array}} \right] = \cfrac{1}{{34}}\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{34}\\{17}\\{ - 51}\end{array}} \right] = \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1\\{1/2}\\{ - 3/2}\end{array}} \right]$

Hence, $x = 1,y = \cfrac{1}{2}$ and $z = \cfrac{{ - 3}}{2}$ .