# How do you solve 9a + 7b = -30, 8b + 5c = 11, -3a + 10c = 73?

Oct 13, 2015

I have shown how to find $a = - 1$ using Kramer's rule in linear matrix algebra. The other 2 unknowns may be found in a similar fashion and I leave it as an exercise. :)

#### Explanation:

This is a system of 3 linear equations in 3 unknowns, so the easiest will be to use linear matrix algebra to solve it. (either Gaussian elimination, or the inverse matrix method, or Kramer's Rule).

I will use Kramer's Rule as it is probably the quickest.

First write the system of linear equations in coefficient matrix form :

$\left[\begin{matrix}9 & 7 & 0 \\ 0 & 8 & 5 \\ - 3 & 0 & 10\end{matrix}\right]$

Now use co-factor expansion along any row or column of your choice to find the determinant of this matrix.
Proceeding along row 1, I get
$\Delta = 9 \left(80 - 0\right) - 7 \left(0 + 15\right) + 0 = 615$

We now form the 3 matrices by replacing each column vector in the coefficient matrix by the column vector of solutions and find the corresponding determinants in each case.

${\Delta}_{a} = | \left(- 30 , 7 , 0\right) , \left(11 , 8 , 5\right) , \left(73 , 0 , 10\right) | = \left(- 30\right) \left(80\right) - 7 \left(110 - 365\right) + 0 = - 615$

$\therefore$ by Kramer's Rule, $a = \frac{{\Delta}_{a}}{\Delta} = \frac{- 615}{615} = - 1$

Proceeding in this fashion, we eventually find b and c. I leave the details as an exercise.