Chapter 8.3, Problem 60E

To determine

The system of linear equations by using inverse matrix.

Answer to Problem 60E

  $x=10,y=-3,$ and $z=5$ .

Explanation of Solution

Given information:

The given system of equation is:

  $\{\begin{array}{l}-8x+7y-10z=-151\\ 12x+3y-5z=86\\ 15x-9y+2z=187\end{array}$

Calculation:

The given system of equation is:

  $\{\begin{array}{l}-8x+7y-10z=-151\\ 12x+3y-5z=86\\ 15x-9y+2z=187\end{array}$

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

  $\left[\begin{array}{ccc}-8& 7& -10\\ 12& 3& -5\\ 15& -9& 2\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}-151\\ 86\\ 187\end{array}\right]$

Find the determinant of $A$ :

  $\begin{array}{l}\text{det }A=\left|\begin{array}{ccc}-8& 7& -10\\ 12& 3& -5\\ 15& -9& 2\end{array}\right|\\ =(-8)\cdot 3\cdot 2+7\cdot (-5)\cdot (15)+(-10)\cdot 12\cdot (-9)-(-10)\cdot 3\cdot 15-(-8)\cdot (-5)\cdot (-9)-7\cdot 12\cdot 2\end{array}$

  $\begin{array}{l}=-48-525+1080+450+360-168\\ =1149\end{array}$

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 $A$ :

  $A^{-1}=\frac{C^T}{\det A}=\left(\begin{array}{ccc}-\frac{13}{383}& \frac{76}{1149}& \frac{-5}{1149}\\ -\frac{33}{383}& \frac{134}{1149}& -\frac{160}{1149}\\ -\frac{51}{383}& \frac{11}{383}& -\frac{36}{383}\end{array}\right)$

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

  $\begin{array}{l}X=A^{-1}\cdot B\\ =\left(\begin{array}{ccc}-\frac{13}{383}& \frac{76}{1149}& \frac{-5}{1149}\\ -\frac{33}{383}& \frac{134}{1149}& -\frac{160}{1149}\\ -\frac{51}{383}& \frac{11}{383}& -\frac{36}{383}\end{array}\right)\left(\begin{array}{c}-151\\ 86\\ 187\end{array}\right)\\ =\left(\begin{array}{ccc}-\frac{13}{383}(-151)& \frac{76}{1149}(86)& \frac{-5}{1149}(187)\\ -\frac{33}{383}(-151)& \frac{134}{1149}(86)& -\frac{160}{1149}(187)\\ -\frac{51}{383}(-151)& \frac{11}{383}(86)& -\frac{36}{383}(187)\end{array}\right)\\ =\left(\begin{array}{ccc}\frac{1963}{383}& \frac{6536}{1149}& -\frac{935}{1149}\\ \frac{4983}{383}& \frac{11524}{1149}& -\frac{29920}{1149}\\ \frac{7701}{383}& \frac{946}{383}& -\frac{6732}{383}\end{array}\right)\\ =\left(\begin{array}{c}10\\ \begin{array}{l}-3\\ 5\end{array}\end{array}\right)\end{array}$

Therefore, the solutions are $x=10,y=-3,$ and $z=5$ .