Given the matrix $A = \begin{pmatrix} 1 & 0 & 3 \\ -2 & 4 & 5 \\ 2 & 0 & 1 \end{pmatrix}$, we need to find $A^2 - 2A + 5I$, where $I$ is the identity matrix.

Let's break it down step by step:

1. Compute $A^2$

First, we need to multiply $A$ by itself:

$A^2 = A \times A = \begin{pmatrix} 1 & 0 & 3 \\ -2 & 4 & 5 \\ 2 & 0 & 1 \end{pmatrix} \times \begin{pmatrix} 1 & 0 & 3 \\ -2 & 4 & 5 \\ 2 & 0 & 1 \end{pmatrix}$

(1 \times 1) + (0 \times -2) + (3 \times 2) & (1 \times 0) + (0 \times 4) + (3 \times 0) & (1 \times 3) + (0 \times 5) + (3 \times 1) \\ (-2 \times 1) + (4 \times -2) + (5 \times 2) & (-2 \times 0) + (4 \times 4) + (5 \times 0) & (-2 \times 3) + (4 \times 5) + (5 \times 1) \\ (2 \times 1) + (0 \times -2) + (1 \times 2) & (2 \times 0) + (0 \times 4) + (1 \times 0) & (2 \times 3) + (0 \times 5) + (1 \times 1) \end{pmatrix}$$ $$= \begin{pmatrix} 1 + 0 + 6 & 0 + 0 + 0 & 3 + 0 + 3 \\ -2 - 8 + 10 & 0 + 16 + 0 & -6 + 20 + 5 \\ 2 + 0 + 2 & 0 + 0 + 0 & 6 + 0 + 1 \end{pmatrix}$$ $$= \begin{pmatrix} 7 & 0 & 6 \\ 0 & 16 & 19 \\ 4 & 0 & 7 \end{pmatrix}$$ ### 2. Compute $$2A$$ Now, multiply $$A$$ by 2: $$2A = 2 \times \begin{pmatrix} 1 & 0 & 3 \\ -2 & 4 & 5 \\ 2 & 0 & 1 \end{pmatrix} = \begin{pmatrix} 2 & 0 & 6 \\ -4 & 8 & 10 \\ 4 & 0 & 2 \end{pmatrix}$$ ### 3. Compute $$5I$$ Here, $$I$$ is the identity matrix: $$I = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}$$ Now multiply it by 5: $$5I = 5 \times \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} = \begin{pmatrix} 5 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 5 \end{pmatrix}$$ ### 4. Compute $$A^2 - 2A + 5I$$ Finally, subtract $$2A$$ from $$A^2$$ and add $$5I$$: $$A^2 - 2A + 5I = \begin{pmatrix} 7 & 0 & 6 \\ 0 & 16 & 19 \\ 4 & 0 & 7 \end{pmatrix} - \begin{pmatrix} 2 & 0 & 6 \\ -4 & 8 & 10 \\ 4 & 0 & 2 \end{pmatrix} + \begin{pmatrix} 5 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 5 \end{pmatrix}$$ $$= \begin{pmatrix} (7 - 2 + 5) & (0 - 0 + 0) & (6 - 6 + 0) \\ (0 + 4 + 0) & (16 - 8 + 5) & (19 - 10 + 0) \\ (4 - 4 + 0) & (0 - 0 + 0) & (7 - 2 + 5) \end{pmatrix}$$ $$= \begin{pmatrix} 10 & 0 & 0 \\ 4 & 13 & 9 \\ 0 & 0 & 10 \end{pmatrix}$$ Thus, $$A^2 - 2A + 5I = \begin{pmatrix} 10 & 0 & 0 \\ 4 & 13 & 9 \\ 0 & 0 & 10 \end{pmatrix}$$. Would you like further details or have any questions about this solution? ### Related Questions: 1. How can you verify that the calculation of $$A^2$$ is correct? 2. What would change if the scalar multiplication was by another number instead of 2 in $$2A$$? 3. How is the identity matrix $$I$$ important in linear algebra? 4. How would the result change if the matrix $$A$$ had different dimensions? 5. What are the properties of matrix multiplication that apply to this problem? ### Tip: Always double-check matrix multiplications by verifying each element calculation to avoid small errors that could lead to incorrect results.