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Given that $A=\begin{bmatrix}-5 & -3\\ 2 & 0\end{bmatrix}$ and $I=\begin{bmatrix}1 & 0\\ 0 & 1\end{bmatrix}$, the value of $A^3$ is

1. $15A+12I$
2. $19A+30I$
3. $17A+15I$
4. $17A+21I$

Correct answer: $19A+30I$

Solution

The characteristic polynomial of $A$ is $\lambda^2+5\lambda+6=(\lambda+2)(\lambda+3)$, so by Cayley-Hamilton, $A^2+5A+6I=0$. Multiplying by $A$ gives $A^3+5A^2+6A=0$, and substituting $A^2=-5A-6I$ yields $A^3=19A+30I$.

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