Answer
\[x_1=1\;,\; x_2=0\]
Work Step by Step
\[x_1-x_2=1\\
2x_1+x_2=2\]
Here $A=\left[\begin{array}{cc}
1&-1\\
2&1\end{array}\right]\;,\;b=\left[\begin{array}{c}
1\\
2\end{array}\right]$
and $X=\left[\begin{array}{c}
x_1\\
x_2\end{array}\right]$
Consider $\;A=\left[\begin{array}{cc}
1&-1\\
2&1\end{array}\right]$
Here $|A|=1(1)-(-1)2=3\neq 0$
$\Rightarrow A$ is invertible
Using Theorem 3.8
$\;\Rightarrow A^{-1}=\frac{1}{1-(-2)}\left[\begin{array}{cc}
1&1\\
-2&1\end{array}\right]$
$\;\Rightarrow A^{-1}=\frac{1}{3}\left[\begin{array}{cc}
1&1\\
-2&1\end{array}\right]$
$X=A^{-1}b$
$\Rightarrow X=\frac{1}{3}\left[\begin{array}{cc}
1&1\\
-2&1\end{array}\right]\left[\begin{array}{c}
1\\
2\end{array}\right]$
$\Rightarrow X=\frac{1}{3}\left[\begin{array}{c}
1+2\\
-2+2\end{array}\right]$
$\Rightarrow X=\frac{1}{3}\left[\begin{array}{c}
3\\
0\end{array}\right]$
$\Rightarrow X=\left[\begin{array}{c}
1\\
0\end{array}\right]$
$\Rightarrow \left[\begin{array}{c}
x_1\\
x_2\end{array}\right] =\left[\begin{array}{c}
1\\
0\end{array}\right]$
$\Rightarrow x_1=1\;,\; x_2=0$
Hence $x_1=1\;,\; x_2=0$.