Step 1

In matrix form, given system of equations cab be written as

\(x+2y = -1\)

\(x-2y=3\)

\(AX=B\)

\(\displaystyle{A}={\left[\begin{array}{cc} {1}&{2}\\{1}&-{2}\end{array}\right]},{X}={\left[\begin{array}{c} {x}\\{y}\end{array}\right]},{B}={\left[\begin{array}{c} -{1}\\{3}\end{array}\right]}\)

We know that

\(\displaystyle{\left[\begin{array}{cc} {a}&{b}\\{c}&{d}\end{array}\right]}^{{-{1}}}=\frac{{1}}{{{a}{d}-{b}{c}}}{\left[\begin{array}{cc} {d}&-{b}\\-{c}&{a}\end{array}\right]}\)

Step 2

Therefore,

\(\displaystyle{A}^{{-{1}}}=\frac{{1}}{{{1}{\left(-{2}\right)}-{2}{\left({1}\right)}}}{\left[\begin{array}{cc} -{2}&-{2}\\-{1}&{1}\end{array}\right]}\)

\(\displaystyle{A}^{{-{1}}}=-\frac{{1}}{{4}}{\left[\begin{array}{cc} -{2}&-{2}\\-{1}&{1}\end{array}\right]}\)

Therefore, solution of system of equations is

\(\displaystyle{X}={A}^{{-{1}}}{B}\)

\(\displaystyle{X}=-\frac{{1}}{{4}}{\left[\begin{array}{cc} -{2}&-{2}\\-{1}&{1}\end{array}\right]}{\left[\begin{array}{c} -{1}\\{3}\end{array}\right]}\)

\(\displaystyle{X}=-\frac{{1}}{{4}}{\left[\begin{array}{cc} {2}&-{6}\\{1}&{3}\end{array}\right]}{\left[\begin{array}{c} {1}\\-{1}\end{array}\right]}\)

\(x=1,y=-1\)

Result:\(x=1,y=-1\)