4x-3y=3 3x-5y=7

Matrix form of given equations is AX = B

{tex}\Rightarrow \left[ {\begin{array}{*{20}{c}} 4&{ - 3} \\ 3&{ - 5} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} x \\ y \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 3 \\ 7 \end{array}} \right]{/tex}

Here {tex}A = \left[ {\begin{array}{*{20}{c}} 4&{ - 3} \\ 3&{ - 5} \end{array}} \right],X = \left[ {\begin{array}{*{20}{c}} x \\ y \end{array}} \right]{/tex} and {tex}B = \left[ {\begin{array}{*{20}{c}} 3 \\ 7 \end{array}} \right]{/tex}

{tex}\therefore \left| A \right| = \left| {\begin{array}{*{20}{c}} 4&{ - 3} \\ 3&{ - 5} \end{array}} \right| {/tex} = - 20 - (-9) = - 20 + 9 = - 11 {tex} \ne {/tex} 0

Therefore, solution is unique and {tex}X = {A^{ - 1}}B = \frac{1}{{\left| A \right|}}\left( {adj.A} \right)B{/tex}

{tex}\Rightarrow \left[ {\begin{array}{*{20}{c}} x \\ y \end{array}} \right] = \frac{1}{{ - 11}}\left[ {\begin{array}{*{20}{c}} { - 5}&3 \\ { - 3}&4 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 3 \\ 7 \end{array}} \right]{/tex}

{tex}= \frac{1}{{ - 11}}\left[ {\begin{array}{*{20}{c}} { - 15 + 21} \\ { - 9 + 28} \end{array}} \right] = \frac{1}{{ - 11}}\left[ {\begin{array}{*{20}{c}} 6 \\ 9 \end{array}} \right]{/tex}

{tex}= \left[ {\begin{array}{*{20}{c}} {\frac{{ - 6}}{{11}}} \\ {\frac{{ - 19}}{{11}}} \end{array}} \right]{/tex}

Therefore, {tex}x = \frac{{ - 6}}{{11}}{/tex} and {tex}y = \frac{{ - 19}}{{11}}{/tex}