Chapter 8.3, Problem 59E

To determine

The system of linear equations by using inverse matrix.

Answer to Problem 59E

  $x=-7,y=3,$ and $z=-2$ .

Explanation of Solution

Given information:

The given system of equation is:

  $\{\begin{array}{l}3x-2y+z=-29\\ -4x+y-3z=37\\ x-5y+z=-24\end{array}$

Calculation:

The given system of equation is:

  $\{\begin{array}{l}3x-2y+z=-29\\ -4x+y-3z=37\\ x-5y+z=-24\end{array}$

To find the system of equations by writing the system in the matrix form $AX=B$ .

  $\left[\begin{array}{ccc}3& -2& 1\\ -4& 1& -3\\ 1& -5& 1\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}-29\\ 37\\ -24\end{array}\right]$

Find the determinant of $A$ :

  $\begin{array}{l}\text{det }A=\left|\begin{array}{ccc}3& -2& 1\\ -4& 1& -3\\ 1& -5& 1\end{array}\right|\\ =3\cdot 1\cdot 1+(-2)\cdot -3\cdot 1+1\cdot (-4)\cdot (-5)-1\cdot 1\cdot 1-3\cdot (-3)\cdot (-5)-(-2)\cdot (-4)\cdot 1\end{array}$

  $\begin{array}{l}=-45-8\\ =-25\end{array}$

The determinant of $A$ is not zero,therefore the inverse matrix $A^{-1}$ exist. To calculate the inverse matrix find additional minors and cofactors of matrix $A$ :

  $A^{-1}=\frac{C^T}{\det A}=\left(\begin{array}{ccc}\frac{14}{25}& \frac{3}{25}& \frac{-1}{5}\\ -\frac{1}{25}& -\frac{2}{25}& -\frac{1}{5}\\ -\frac{19}{25}& -\frac{13}{25}& \frac{1}{5}\end{array}\right)$

Now find the solution of system of equation by usinf an inverse of matrix:

  $\begin{array}{l}X=A^{-1}\cdot B\\ =\left(\begin{array}{ccc}\frac{14}{25}& \frac{3}{25}& \frac{-1}{5}\\ -\frac{1}{25}& -\frac{2}{25}& -\frac{1}{5}\\ -\frac{19}{25}& -\frac{13}{25}& \frac{1}{5}\end{array}\right)\left(\begin{array}{c}-2\\ 3\\ -2\end{array}\right)\\ =\left(\begin{array}{ccc}\frac{14}{25}(-29)& \frac{3}{25}\cdot 37& \left(\frac{-1}{5}\right)\cdot (-24)\\ \frac{-1}{25}\cdot (-29)& \frac{-2}{25}\cdot 37& \left(\frac{-1}{5}\right)\cdot (-24)\\ \left(\frac{-19}{25}\right)\cdot (-29)& \frac{-13}{25}\cdot 37& \frac{1}{5}\cdot (-24)\end{array}\right)\\ =\left(\begin{array}{ccc}-\frac{406}{25}& \frac{111}{25}& \frac{24}{5}\\ \frac{29}{25}& \frac{-74}{25}& \frac{24}{5}\\ \frac{551}{25}& -\frac{481}{25}& -\frac{24}{5}\end{array}\right)\\ =\left(\begin{array}{c}-7\\ \begin{array}{l}3\\ -2\end{array}\end{array}\right)\end{array}$

Therefore, the solutions are $x=-7,y=3,$ and $z=-2$ .