In Problems 41– 60, use the inverses found in Problems 31–40 to solve each system of equations

$\left\{\begin{array}{r}x-y+z=2\\ -2y+z=2\\ -2x-3y=\frac{1}{2}\end{array}\right.$

A matrix is given as,

$\left\{\begin{array}{r}x-y+z=2\\ -2y+z=2\\ -2x-3y=\frac{1}{2}\end{array}\right.$

We need to find the solution of the matrix using inverse of the matrix.

The inverse of the matrix can be found by :

$$\begin{gathered} \left[A\mid I_3\right]=\left[\begin{array}{cccccc}1& -1& 1& 1& 0& 0\\ 0& -2& 1& 0& 1& 0\\ -2& -3& 0& 0& 0& 1\end{array}\right] \\ {r}_3-r_3+2r_1 \\ \left[\begin{array}{cccccc}1& -1& 1& 1& 0& 0\\ 0& -2& 1& 0& 1& 0\\ -2& -3& 0& 0& 0& 1\end{array}\right]\to \left[\begin{array}{cccccc}1& -1& 1& 1& 0& 0\\ 0& -2& 1& 0& 1& 0\\ -2+2& -3-2& 0+2& 0+2& 0+0& 1+0\end{array}\right]\to \left[\begin{array}{cccccc}1& -1& 1& 1& 0& 0\\ 0& -2& 1& 0& 1& 0\\ 0& -5& 2& 2& 0& 1\end{array}\right] \\ {R}_2=\frac{r_2}{-2} \\ \left[\begin{array}{cccccc}1& -1& 1& 1& 0& 0\\ 0& -2& 1& 0& 1& 0\\ 0& -5& 2& 2& 0& 1\end{array}\right]\to \left[\begin{array}{cccccc}1& -1& 1& 1& 0& 0\\ \frac{0}{-2}& \frac{-2}{-2}& \frac{1}{-2}& \frac{0}{-2}& \frac{1}{-2}& \frac{0}{-2}\\ 0& -5& 2& 2& 0& 1\end{array}\right]\to \left[\begin{array}{cccccc}1& -1& 1& 1& 0& 0\\ 0& 1& -\frac{1}{2}& 0& -\frac{1}{2}& 0\\ 0& -5& 2& 2& 0& 1\end{array}\right] \\ {R}_1=r_1+r_2 \\ \left[\begin{array}{cccccc}1& -1& 1& 1& 0& 0\\ 0& 1& -\frac{1}{2}& 0& -\frac{1}{2}& 0\\ 0& -5& 2& 2& 0& 1\end{array}\right]\to \left[\begin{array}{cccccc}1+0& -1+1& 1-\frac{1}{2}& 1+0& 0-\frac{1}{2}& 0+0\\ 0& 1& -\frac{1}{2}& 0& -\frac{1}{2}& 0\\ 0& -5& 2& 2& 0& 1\end{array}\right]\to \left[\begin{array}{cccccc}1& 0& \frac{1}{2}& 1& -\frac{1}{2}& 0\\ 0& 1& -\frac{1}{2}& 0& -\frac{1}{2}& 0\\ 0& -5& 2& 2& 0& 1\end{array}\right] \end{gathered}$$

Further simplifying,

$$\begin{gathered} R_3=r_3+5r_2 \\ \left[\begin{array}{cccccc}1& 0& \frac{1}{2}& 1& -\frac{1}{2}& 0\\ 0& 1& -\frac{1}{2}& 0& -\frac{1}{2}& 0\\ 0& -5& 2& 2& 0& 1\end{array}\right]\to \left[\begin{array}{cccccc}1& 0& \frac{1}{2}& 1& -\frac{1}{2}& 0\\ 0& 1& -\frac{1}{2}& 0& -\frac{1}{2}& 0\\ 0+0& -5+5& 2-\frac{5}{2}& 2+0& 0-\frac{5}{2}& 1+0\end{array}\right] \\ \to \left[\begin{array}{rrrrrr}1& 0& \frac{1}{2}& 1& -\frac{1}{2}& 0\\ 0& 1& -\frac{1}{2}& 0& -\frac{1}{2}& 0\\ 0& 0& -\frac{1}{2}& 2& -\frac{5}{2}& 1\end{array}\right] \\ {R}_3=-2r_3 \\ \left[\begin{array}{cccccc}1& 0& \frac{1}{2}& 1& -\frac{1}{2}& 0\\ 0& 1& -\frac{1}{2}& 0& -\frac{1}{2}& 0\\ 0& 0& -\frac{1}{2}& 2& -\frac{5}{2}& 1\end{array}\right]\to \left[\begin{array}{cccccc}1& 0& \frac{1}{2}& 1& -\frac{1}{2}& 0\\ 0& 1& -\frac{1}{2}& 0& -\frac{1}{2}& 0\\ 0(-2)& 0(-2)& -\frac{1}{2}(-2)& 2(-2)& -\frac{5}{2}(-2)& 1(-2)\end{array}\right] \\ \to \left[\begin{array}{cccccc}1& 0& \frac{1}{2}& 1& -\frac{1}{2}& 0\\ 0& 1& -\frac{1}{2}& 0& -\frac{1}{2}& 0\\ 0& 0& 1& -4& 5& -2\end{array}\right] \\ {R}_1=r_1-\frac{r_3}{2} \\ \left[\begin{array}{cccccc}1& 0& \frac{1}{2}& 1& -\frac{1}{2}& 0\\ 0& 1& -\frac{1}{2}& 0& -\frac{1}{2}& 0\\ 0& 0& 1& -4& 5& -2\end{array}\right]\to \left[\begin{array}{cccccc}1-0& 0-0& \frac{1}{2}-\frac{1}{2}& 1+2& -\frac{1}{2}-\frac{5}{2}& 0+1\\ 0& 1& -\frac{1}{2}& 0& -\frac{1}{2}& 0\\ 0& 0& 1& -4& 5& -2\end{array}\right]\to \left[\begin{array}{cccccc}1& 0& 0& 3& -3& 1\\ 0& 1& -\frac{1}{2}& 0& -\frac{1}{2}& 0\\ 0& 0& 1& -4& 5& -2\end{array}\right] \\ {R}_2=r_2+\frac{r_3}{2} \\ \left[\begin{array}{cccccc}1& 0& 0& 3& -3& 1\\ 0& 1& -\frac{1}{2}& 0& -\frac{1}{2}& 0\\ 0& 0& 1& -4& 5& -2\end{array}\right]\to \left[\begin{array}{cccccc}1& 0& 0& 3& -3& 1\\ 0+0& 1+0& -\frac{1}{2}+\frac{1}{2}& 0-2& -\frac{1}{2}+\frac{5}{2}& 0-1\\ 0& 0& 1& -4& 5& -2\end{array}\right]\to \left[\begin{array}{cccccc}1& 0& 0& 3& -3& 1\\ 0& 1& 0& -2& 2& -1\\ 0& 0& 1& -4& 5& -2\end{array}\right] \\ Therefore, \\ {A}^{-1}=\left[\begin{array}{ccc}3& -3& 1\\ -2& 2& -1\\ -4& 5& -2\end{array}\right] \end{gathered}$$

We know,

$X=A^{-1}B$

$$\begin{gathered} X=\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=A^{-1}B \\ =\left[\begin{array}{ccc}3& -3& 1\\ -2& 2& -1\\ -4& 5& -2\end{array}\right]·\left[\begin{array}{c}2\\ 2\\ \frac{1}{2}\end{array}\right] \\ X=\left[\begin{array}{c}3\left(2\right)+(-3)\left(2\right)+1\left(\frac{1}{2}\right)\\ -2\left(2\right)+2\left(2\right)+(-1)\left(\frac{1}{2}\right)\\ -4\left(2\right)+5\left(2\right)+(-2)\frac{1}{2}\end{array}\right] \\ =\left[\begin{array}{c}6-6+\frac{1}{2}\\ -4+4-\frac{1}{2}\\ -8+10-1\end{array}\right] \\ =\left[\begin{array}{c}\frac{1}{2}\\ -\frac{1}{2}\\ 1\end{array}\right] \end{gathered}$$