Q. 53 - Precalculus Enhanced with Graphing Utilities

Question

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

$\left\{\begin{array}{rr}x-y+z=& 0\\ -2y+z=& -1\\ -2x-3y=& -5\end{array}\right.$

Step-by-Step Solution

Answer

The solution of the matrix is $x=-2,y=3,\text{ and }z=5$

Step 1. Given information

A matrix is given as, 

$\left\{\begin{array}{rr}x-y+z=& 0\\ -2y+z=& -1\\ -2x-3y=& -5\end{array}\right.$

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

Step 2. Finding Inverse of the matrix.

The inverse of the matrix can be found by : 

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

Step 3. Inverse of the matrix

Further simplifying,

$$\begin{gathered} R_2=r_2+\frac{r_3}{2} \\ \left[\begin{array}{cccccc}1& 0& 0& 3& -3& 1\\ 0& 1& -\frac{1}{2}& 0& -\frac{1}{2}& 0\\ 0& 0& 1& -4& 5& -2\end{array}\right]\to \left[\begin{array}{ccccccc}1& 0& 0& 3& -3& 1& \\ 0+0& 1+0& -\frac{1}{2}+\frac{1}{2}& 0-2& -\frac{1}{2}+\frac{5}{2}& 0& -1\\ 0& 0& 1& -4& 5& -2& \end{array}\right]\to \left[\begin{array}{cccccc}1& 0& 0& 3& -3& 1\\ 0& 1& 0& -2& 2& -1\\ 0& 0& 1& -4& 5& -2\end{array}\right] \\ Therefore, \\ {A}^{-1}=\left[\begin{array}{ccc}3& -3& 1\\ -2& 2& -1\\ -4& 5& -2\end{array}\right] \end{gathered}$$

Step 4. Solution of the matrix

We know,

$X=A^{-1}B$

Therefore,

$$\begin{gathered} X=\left[\begin{array}{l}x\\ y\\ z\end{array}\right] \\ =A^{-1}B \\ =\left[\begin{array}{ccc}3& -3& 1\\ -2& 2& -1\\ -4& 5& -2\end{array}\right]·\left[\begin{array}{c}0\\ -1\\ -5\end{array}\right] \\ X=\left[\begin{array}{c}3\left(0\right)+(-3)(-1)+1(-5)\\ -2\left(0\right)+2(-1)+(-1)(-5)\\ -4\left(0\right)+5(-1)+10\end{array}\right] \\ =\left|\begin{array}{c}0+3-5\\ 0-2+5\\ 0-5+10\end{array}\right| \\ =\left[\begin{array}{c}-2\\ 3\\ 5\end{array}\right] \\ x=-2,y=3,\text{ and }z=5 \end{gathered}$$