## Algebra and Trigonometry 10th Edition

When there is an inverse of a matrix $A A^{-1} = I$ (the Identity Matrix) $AB = \begin{bmatrix} -1 &1 & 0 & -1 \\1 & -1 & 1 & 0 \\ -1 &1 & -2 &0 \\ 0 &-1 &1 & 1 \end{bmatrix} \dfrac{1}{3}\begin{bmatrix} -3 & 1 & 1 & -3\\-3 & -1 & 2 &-3 \\ 0&1&1&0 \\ -3 & -2 & 1 & 0\end{bmatrix} =\begin{bmatrix} 1 & 0 & 0 & 0\\0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 &1 \end{bmatrix}$ Now, multiplying by columns, we have: $BA = \dfrac{1}{3} \begin{bmatrix} -3 & 1 & 1 & -3\\-3 & -1 & 2 &-3 \\ 0&1&1&0 \\ -3 & -2 & 1 & 0\end{bmatrix} \begin{bmatrix} -1 &1 & 0 & -1 \\1 & -1 & 1 & 0 \\ -1 &1 & -2 &0 \\ 0 &-1 &1 & 1 \end{bmatrix} =\begin{bmatrix} 1 & 0 & 0 & 0\\0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 &1 \end{bmatrix}$ Since $AB=BA$ equals the identity matrix, B is the inverse of A.