A 2×2 real matrix A has trace 5 and determinant 6. Then det (A^2+I) equals

By the Cayley-Hamilton theorem $A^2-(\operatorname{tr}A)A+(\det A)I=0$, so $A^2=5A-6I$. Then $A^2+I=5A-5I=5(A-I)$ and, for a $2\times2$ matrix, $\det(5(A-I))=25\det(A-I)$. Also $\det(A-I)=\det A-\operatorname{tr}A+1=6-5+1=2$. Hence $\det(A^2+I)=25\cdot2=50$. Check with eigenvalues $2$ and $3$ (roots of $\lambda^2-5\lambda+6$): $(2^2+1)(3^2+1)=5\cdot10=50$. A frequent error is to forget the factor $5^2$ when taking the determinant of $5(A-I)$. Any matrix with these invariants, such as $\operatorname{diag}(2,3)$, can be used to test the result quickly.