I want to solve the following equation $$AX=\left( \begin{matrix} 9 \\ 3 \\ -3 \\ \end{matrix} \right) $$
where $$A=\left( \begin{matrix} 1 & -2 & -2 \\ -2 & 1 & -2 \\ -2 & -2 & 1 \\ \end{matrix} \right)$$
How can I do this? I know the first step is to calculate the inverse of $A$.
There are many ways of solving this problem. One way is to reduce the matrix A to reduced echelon form. If you do the same for the vector [9, 3, -3] then you can easily read the solution of.
Other way is indeed to calculate the inverse of A. Then we can multiply both sides of the equation with $A^{-1}$ and you will see that the lhs reads $IX$ or just $X$. Note that in this case the unity matrix I is the reduced echelon form.
You can always obtain the inverse matrix (if it exists) by putting matrix A next to I and reducing A to echelon while doing the same type reduction steps on I.