Exercise 3.3.4

With the technique used before we can calculate $A^{-1}$ first and then the solution of $\mathit{Ax}=b$ would be $A^{-1}b$ if $A$ is invertible.

1.

Calculate $A^{-1}=\left(\begin{array}{cc}\hfill -5\hfill & \hfill 3\hfill \\ \hfill 2\hfill & \hfill -1\hfill \end{array}\right)$ and solution is $x_1=-11$, $x_2=5$.

2.

Calculate $A^{-1}=\left(\begin{array}{ccc}\hfill \frac{1}{3}\hfill & \hfill 0\hfill & \hfill \frac{1}{3}\hfill \\ \hfill \frac{1}{9}\hfill & \hfill \frac{1}{3}\hfill & \hfill -\frac{2}{9}\hfill \\ \hfill -\frac{4}{9}\hfill & \hfill \frac{2}{3}\hfill & \hfill -\frac{1}{9}\hfill \end{array}\right)$ and solution is $x_1=3$, $x_2=0$, $x_3=-2$.