1. **Problem:** Calculate $A^2 + 4A - 5I$ for matrix $A = \begin{bmatrix}1 & 2 & 1 \\ 2 & 1 & 2 \\ 2 & 2 & 1\end{bmatrix}$ and $I$ is the $3 \times 3$ identity matrix. 2. **Formula and rules:** - Matrix multiplication: $(AB)_{ij} = \sum_k A_{ik}B_{kj}$. - Scalar multiplication and addition are element-wise. - Identity matrix $I$ has 1's on diagonal, 0's elsewhere. 3. **Calculate $A^2$:** $$A^2 = A \times A = \begin{bmatrix}1 & 2 & 1 \\ 2 & 1 & 2 \\ 2 & 2 & 1\end{bmatrix} \times \begin{bmatrix}1 & 2 & 1 \\ 2 & 1 & 2 \\ 2 & 2 & 1\end{bmatrix}$$ Calculate each element: - $(1,1): 1\times1 + 2\times2 + 1\times2 = 1 + 4 + 2 = 7$ - $(1,2): 1\times2 + 2\times1 + 1\times2 = 2 + 2 + 2 = 6$ - $(1,3): 1\times1 + 2\times2 + 1\times1 = 1 + 4 + 1 = 6$ - $(2,1): 2\times1 + 1\times2 + 2\times2 = 2 + 2 + 4 = 8$ - $(2,2): 2\times2 + 1\times1 + 2\times2 = 4 + 1 + 4 = 9$ - $(2,3): 2\times1 + 1\times2 + 2\times1 = 2 + 2 + 2 = 6$ - $(3,1): 2\times1 + 2\times2 + 1\times2 = 2 + 4 + 2 = 8$ - $(3,2): 2\times2 + 2\times1 + 1\times2 = 4 + 2 + 2 = 8$ - $(3,3): 2\times1 + 2\times2 + 1\times1 = 2 + 4 + 1 = 7$ So, $$A^2 = \begin{bmatrix}7 & 6 & 6 \\ 8 & 9 & 6 \\ 8 & 8 & 7\end{bmatrix}$$ 4. **Calculate $4A$:** $$4A = 4 \times \begin{bmatrix}1 & 2 & 1 \\ 2 & 1 & 2 \\ 2 & 2 & 1\end{bmatrix} = \begin{bmatrix}4 & 8 & 4 \\ 8 & 4 & 8 \\ 8 & 8 & 4\end{bmatrix}$$ 5. **Calculate $5I$:** $$5I = 5 \times \begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix} = \begin{bmatrix}5 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 5\end{bmatrix}$$ 6. **Sum $A^2 + 4A - 5I$:** $$\begin{bmatrix}7 & 6 & 6 \\ 8 & 9 & 6 \\ 8 & 8 & 7\end{bmatrix} + \begin{bmatrix}4 & 8 & 4 \\ 8 & 4 & 8 \\ 8 & 8 & 4\end{bmatrix} - \begin{bmatrix}5 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 5\end{bmatrix} = \begin{bmatrix}7+4-5 & 6+8-0 & 6+4-0 \\ 8+8-0 & 9+4-5 & 6+8-0 \\ 8+8-0 & 8+8-0 & 7+4-5\end{bmatrix}$$ $$= \begin{bmatrix}6 & 14 & 10 \\ 16 & 8 & 14 \\ 16 & 16 & 6\end{bmatrix}$$ **Final answer:** $$A^2 + 4A - 5I = \begin{bmatrix}6 & 14 & 10 \\ 16 & 8 & 14 \\ 16 & 16 & 6\end{bmatrix}$$