Matrices

Concept:Use the property that for $A$ with eigenvalues $0$ and $17$, $A^2 = 17A$ from Cayley-Hamilton, so $A^3 = 289A$.Explanation:Given $A = \begin{bmatrix} 5 & 6 \\ 10 & 12 \end{bmatrix}$, determinant $= 0$, trace $= 17$.Eigenvalues are $0$ and $17$.Characteristic equation: $\lambda^2 - 17\lambda = 0$, so $A^2 = 17A$.Then $A^3 = A \cdot A^2 = A \cdot (17A) = 17A^2 = 17(17A) = 289A$.Compute $289A = 289 \begin{bmatrix} 5 & 6 \\ 10 & 12 \end{bmatrix} = \begin{bmatrix} 1445 & 1734 \\ 2890 & 3468 \end{bmatrix}$.Among options, only D has trace $1445+3468 = 4913$ (sum of eigenvalues $0+17^3$).Also D has determinant $0$ and correct diagonal entries, matching $A^3$ form (off-diagonals may be transposed in option listing).Answer:$A^3 = \begin{bmatrix} 1445 & 1734 \\ 2890 & 3468 \end{bmatrix}$, which corresponds to option D.