Question

Use a matrix to find the solution to the systems of equations

Answer 1

The solutions of the system are:  $x=-6$ and $y=8$

Explanation

Given system of equations......

$-8x-8y=-16\\ \\ 6x-9y=-108$

First we need to make the augmented matrix using the given equations....

$\left[\begin{array}{cccc}-8&-8&|&-16\\6&-9&|&-108\end{array}\right]$

Now, we need to transform the augmented matrix to the reduced row echelon form using the row operations.

Row operation 1 :   Multiply the 1st row by $-\frac{1}{8}$. So, we will get...

$\left[\begin{array}{cccc}1&1&|&2\\6&-9&|&-108\end{array}\right]$

Row operation 2 :   Add -6 times the 1st row to the 2nd row. So, we will get...

$\left[\begin{array}{cccc}1&1&|&2\\0&-15&|&-120\end{array}\right]$

Row operation 3 :   Multiply the 2nd row by $-\frac{1}{15}$. So, we will get...

$\left[\begin{array}{cccc}1&1&|&2\\0&1&|&8\end{array}\right]$

Row operation 4 :   Add -1 times the 2nd row to the 1st row. So, we will get....

$\left[\begin{array}{cccc}1&0&|&-6\\0&1&|&8\end{array}\right]$

So, this is the reduced row echelon form.

We can get the equations from the above reduced row echelon form as.....

$1x+0y=-6\\ x=-6 \\ \\ and \\ \\ 0x+1y=8\\ y=8$

So, the solutions of the system are:  $x=-6$ and $y=8$