1. **Problem:** Show that $AB = I$ where $$A = \begin{bmatrix}1 & 0 & 2 \\ 2 & -1 & 3 \\ 4 & 1 & 8\end{bmatrix}, \quad B = \begin{bmatrix}-11 & 2 & 2 \\ -4 & 0 & 1 \\ 6 & -1 & -1\end{bmatrix}$$ 2. **Formula:** Matrix multiplication is defined as $$ (AB)_{ij} = \sum_k A_{ik} B_{kj} $$ 3. **Step-by-step multiplication:** - Calculate each element of $AB$ by multiplying rows of $A$ by columns of $B$. $$AB = \begin{bmatrix} 1\cdot(-11) + 0\cdot(-4) + 2\cdot6 & 1\cdot2 + 0\cdot0 + 2\cdot(-1) & 1\cdot2 + 0\cdot1 + 2\cdot(-1) \\ 2\cdot(-11) + (-1)\cdot(-4) + 3\cdot6 & 2\cdot2 + (-1)\cdot0 + 3\cdot(-1) & 2\cdot2 + (-1)\cdot1 + 3\cdot(-1) \\ 4\cdot(-11) + 1\cdot(-4) + 8\cdot6 & 4\cdot2 + 1\cdot0 + 8\cdot(-1) & 4\cdot2 + 1\cdot1 + 8\cdot(-1) \end{bmatrix}$$ 4. **Calculate each element:** - First row: $-11 + 0 + 12 = 1$, $2 + 0 - 2 = 0$, $2 + 0 - 2 = 0$ - Second row: $-22 + 4 + 18 = 0$, $4 + 0 - 3 = 1$, $4 - 1 - 3 = 0$ - Third row: $-44 - 4 + 48 = 0$, $8 + 0 - 8 = 0$, $8 + 1 - 8 = 1$ 5. **Result:** $$AB = \begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix} = I$$ **Therefore, $AB = I$ as required.**