1. **State the problem:** Given matrix $A = \begin{bmatrix}1 & -1 & 2 \\ 0 & 2 & 1 \\ 0 & 0 & 4\end{bmatrix}$, find matrix $B = A^2 - 2A + 3I$, where $I$ is the $3 \times 3$ identity matrix. 2. **Calculate $A^2$:** Multiply $A$ by itself: $$A^2 = A \times A = \begin{bmatrix}1 & -1 & 2 \\ 0 & 2 & 1 \\ 0 & 0 & 4\end{bmatrix} \times \begin{bmatrix}1 & -1 & 2 \\ 0 & 2 & 1 \\ 0 & 0 & 4\end{bmatrix}$$ Calculate each element: - First row: - $(1)(1) + (-1)(0) + (2)(0) = 1$ - $(1)(-1) + (-1)(2) + (2)(0) = -1 - 2 = -3$ - $(1)(2) + (-1)(1) + (2)(4) = 2 - 1 + 8 = 9$ - Second row: - $(0)(1) + (2)(0) + (1)(0) = 0$ - $(0)(-1) + (2)(2) + (1)(0) = 4$ - $(0)(2) + (2)(1) + (1)(4) = 2 + 4 = 6$ - Third row: - $(0)(1) + (0)(0) + (4)(0) = 0$ - $(0)(-1) + (0)(2) + (4)(0) = 0$ - $(0)(2) + (0)(1) + (4)(4) = 16$ So, $$A^2 = \begin{bmatrix}1 & -3 & 9 \\ 0 & 4 & 6 \\ 0 & 0 & 16\end{bmatrix}$$ 3. **Calculate $-2A$:** Multiply each element of $A$ by $-2$: $$-2A = \begin{bmatrix}-2 & 2 & -4 \\ 0 & -4 & -2 \\ 0 & 0 & -8\end{bmatrix}$$ 4. **Calculate $3I$:** The identity matrix $I$ is: $$I = \begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}$$ Multiply by 3: $$3I = \begin{bmatrix}3 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3\end{bmatrix}$$ 5. **Sum $A^2 - 2A + 3I$:** Add the matrices element-wise: $$B = \begin{bmatrix}1 & -3 & 9 \\ 0 & 4 & 6 \\ 0 & 0 & 16\end{bmatrix} + \begin{bmatrix}-2 & 2 & -4 \\ 0 & -4 & -2 \\ 0 & 0 & -8\end{bmatrix} + \begin{bmatrix}3 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3\end{bmatrix} = \begin{bmatrix}1 - 2 + 3 & -3 + 2 + 0 & 9 - 4 + 0 \\ 0 + 0 + 0 & 4 - 4 + 3 & 6 - 2 + 0 \\ 0 + 0 + 0 & 0 + 0 + 0 & 16 - 8 + 3\end{bmatrix}$$ Simplify each element: $$B = \begin{bmatrix}2 & -1 & 5 \\ 0 & 3 & 4 \\ 0 & 0 & 11\end{bmatrix}$$ **Final answer:** $$B = \begin{bmatrix}2 & -1 & 5 \\ 0 & 3 & 4 \\ 0 & 0 & 11\end{bmatrix}$$