Solution The system of equations can be written in the form $A X=B$, where

$A=\left[\begin{array}{ll}
2 & 5 \\
3 & 2
\end{array}\right], X=\left[\begin{array}{l}
x \\
y
\end{array}\right] \text { and } \mathrm{B}=\left[\begin{array}{l}
1 \\
7
\end{array}\right]$

Now, $|\mathrm{A}|=-11 \neq 0,$ Hence, $\mathrm{A}$ is nonsingular matrix and so has a unique solution.

Note that $A^{-1}=-\frac{1}{11}\left[\begin{array}{cc}2 & -5 \\ -3 & 2\end{array}\right]$

${{\text{ Therefore}}}$ $X=A^{-1} B=-\frac{1}{11}\left[\begin{array}{cc}2 & -5 \\ -3 & 2\end{array}\right]\left[\begin{array}{l}1 \\ 7\end{array}\right]$

${\text{i}}{\text{.e}}{\text{.}}$ $\left[ {\begin{array}{*{20}{l}}
  x \\ 
  y 
\end{array}} \right] =  – \frac{1}{{11}}\left[ {\begin{array}{*{20}{l}}
  { – 33} \\ 
  {11} 
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
  3 \\ 
  { – 1} 
\end{array}} \right]$

Hence $x=3, y=-1$