Let A=pmatrix4&-2&pmatrix.

Question

Let $A=\begin{pmatrix}4&-2\\alpha&\beta\end{pmatrix}$. If $A^2+\gamma A+18I=O$, then $\det(A)$ is equal to \_\_\_\_\_\_\_\_.

Accepted Answer

The characteristic equation of $A$ is $\lambda^2-(4+\beta)\lambda+(4\beta+2\alpha)=0$, so by Cayley--Hamilton, $A^2-(4+\beta)A+(4\beta+2\alpha)I=0$.

Comparing with $A^2+\gamma A+18I=0$: the constant term gives $\det(A)=4\beta+2\alpha=18$.

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