1. **State the problem:** Find the eigenvalues of the matrix $$A = \begin{bmatrix} 4 & 0 & 0 \\ 0 & 6 & 0 \\ 8 & 0 & 5 \end{bmatrix}$$ given the expression $$3 + 8A - 5I$$ where $$I$$ is the identity matrix. 2. **Recall eigenvalue properties:** If $$\lambda$$ is an eigenvalue of $$A$$ with eigenvector $$v$$, then for any scalar constants $$a$$ and $$b$$, the eigenvalues of $$aA + bI$$ are $$a\lambda + b$$. 3. **Find eigenvalues of $$A$$:** To find eigenvalues $$\lambda$$ of $$A$$, solve $$\det(A - \lambda I) = 0$$. Calculate: $$A - \lambda I = \begin{bmatrix} 4-\lambda & 0 & 0 \\ 0 & 6-\lambda & 0 \\ 8 & 0 & 5-\lambda \end{bmatrix}$$ The determinant is: $$\det(A - \lambda I) = (4-\lambda) \cdot \det \begin{bmatrix} 6-\lambda & 0 \\ 0 & 5-\lambda \end{bmatrix} - 0 + 0 = (4-\lambda)(6-\lambda)(5-\lambda)$$ Note: The third row first column element 8 does not affect the determinant because the matrix is upper block triangular with zeros in the first row's off-diagonal. 4. **Solve characteristic equation:** $$ (4-\lambda)(6-\lambda)(5-\lambda) = 0 $$ So eigenvalues of $$A$$ are: $$ \lambda_1 = 4, \quad \lambda_2 = 6, \quad \lambda_3 = 5 $$ 5. **Find eigenvalues of $$3 + 8A - 5I$$:** Using the property from step 2, eigenvalues of $$3 + 8A - 5I = 8A + (3 - 5)I = 8A - 2I$$ are: $$ 8\lambda - 2 $$ for each eigenvalue $$\lambda$$ of $$A$$. Calculate: $$ 8 \times 4 - 2 = 32 - 2 = 30 $$ $$ 8 \times 6 - 2 = 48 - 2 = 46 $$ $$ 8 \times 5 - 2 = 40 - 2 = 38 $$ **Final answer:** The eigenvalues of $$3 + 8A - 5I$$ are $$30, 46, 38$$.