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`

Step-by-step 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
`-(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
`-(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`