1. ## Matrix Inversion

I'm trying to solve the following system of equations by matrix inversion, however when I substitute back my final solutions for x,y,z into the equations, one of them doesn't work. I checked everything a couple of times but I can't find the mistake.

Q: Solve the system of equations by matrix inversion if possible:

3x-2y+z=1
x+y+z=0
2x-y+2z=1

and here is the solution: ImageShack Album - 6 images
)))))))

2. You inverse for the matrix is correct, but I think there are simpler and more efficient ways to invert a matrix. With that said I think you just made multiplication error
when you multiplied the column vector by the inverse. It should give
$\begin{bmatrix} \frac{1}{2} & \frac{1}{2} & -\frac{1}{2} \\
0 & \frac{2}{3} & -\frac{1}{3} \\ -\frac{1}{2} & -\frac{1}{6} & \frac{5}{6}
\end{bmatrix} \begin{bmatrix} 1 \\ 0 \\1\end{bmatrix}=\begin{bmatrix} 0 \\ -\frac{1}{3} \\ \frac{1}{3}\end{bmatrix}$

3. Originally Posted by TheEmptySet
You inverse for the matrix is correct, but I think there are simpler and more efficient ways to invert a matrix. With that said I think you just made multiplication error
when you multiplied the column vector by the inverse. It should give
$\begin{bmatrix} \frac{1}{2} & \frac{1}{2} & -\frac{1}{2} \\
0 & \frac{2}{3} & -\frac{1}{3} \\ -\frac{1}{2} & -\frac{1}{6} & \frac{5}{6}
\end{bmatrix} \begin{bmatrix} 1 \\ 0 \\1\end{bmatrix}=\begin{bmatrix} 0 \\ -\frac{1}{3} \\ \frac{1}{3}\end{bmatrix}$
Cheers