Question

If \(A=\begin{pmatrix}1 & 2 \\ 3 & 5\end{pmatrix}\), the value of \(\left|A^{4}+3A^{2}-5A+6I\right|\) is ________________.

Hint:

Use Cayley–Hamilton to reduce powers and \(\det(A+\alpha I)=\alpha^2+\alpha\,\mathrm{tr}(A)+\det(A)\) for \(2\times2\) matrices.

Answer 1

Correct Answer: 10551

Step 1: For \(A\), \(\mathrm{tr}(A)=6\), \(\det(A)=-1\). By Cayley–Hamilton: \(A^2-6A-I=0\Rightarrow A^2=6A+I\).
Step 2: Compute: \(A^3=A(A^2)=A(6A+I)=6A^2+A=36A+6I+A=37A+6I\).
\(A^4=A(A^3)=A(37A+6I)=37A^2+6A=37(6A+I)+6A=228A+37I\).
Step 3: Then \[ A^{4}+3A^{2}-5A+6I=(228A+37I)+3(6A+I)-5A+6I=241A+46I. \] Step 4: For any \(2\times2\) matrix, \(\det(A+\alpha I)=\alpha^2+\alpha\,\mathrm{tr}(A)+\det(A)\). With \(\alpha=\tfrac{46}{241}\): \[ \det(241A+46I)=241^2\det\!\left(A+\tfrac{46}{241}I\right) =241^2\left(\alpha^2+6\alpha-1\right)=\boxed{10551}. \]