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

$\displaystyle \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}$