1. ## Inverse

Set up a system belw as a matrix multiplication of the form AB=C. Find the inverse A^-1 of A and use it to solve the system.

3x + -6y + -17z = 288

11x + -17y + 110z = 79

-5x + 7y + 8z = 20.

I am able to get the first part of the matrix but I get lost when it comes to working on the second row.

2. Hello,

$\displaystyle \begin{pmatrix} 3&-6&-17 \\ 11&-17&110 \\ -5&7&8 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 288 \\ 79 \\ 20 \end{pmatrix}$

After that, I don't know which methods you've used to calculate the inverse of a 3x3 matrix.
Here is the most common one :

$\displaystyle \left( \begin{array}{ccc|ccc} 3&-6&-17&1&0&0 \\ 11&-17&110&0&1&0 \\ -5&7&8&0&0&1 \end{array} \right)$

And make row operations to find the ?'s in this matrix :
$\displaystyle \left( \begin{array}{ccc|ccc} 1&0&0&?&?&? \\ 0&1&0&?&?&? \\ 0&0&1&?&?&? \end{array} \right)$

The ?'s will form $\displaystyle A^{-1}$

After that, get back to the equation :
$\displaystyle AB=C$

multiply on the left by $\displaystyle A^{-1}$ :

$\displaystyle \underbrace{A^{-1}A}_{=I}B=A^{-1}C$

$\displaystyle B=\begin{pmatrix} x \\ y \\ z \end{pmatrix}=A^{-1}C=A^{-1}\begin{pmatrix} 288 \\ 79 \\ 20 \end{pmatrix}$