Discussion
Suppose the eigen values of a 3 × 3 matrix A are 0, 1 and 2. Then, the determinant of \((I + A^{2})^{-1}\).
Answer: B
The determinant is 1/10.
If A has eigenvalues 0, 1, 2, then \(A^{2}\) has eigenvalues 0, 1, 4 and \(I + A^{2}\) has eigenvalues \(1 + 0 = 1\), \(1 + 1 = 2\), \(1 + 4 = 5\). This holds for any matrix, diagonalizable or not, because eigenvalues of a polynomial in A are the polynomial evaluated at the eigenvalues of A.
\[ \begin{aligned} \det(I + A^{2}) &= 1 \times 2 \times 5 = 10 \\ \det(I + A^{2})^{-1} &= \frac{1}{10} \end{aligned} \]
So the information is enough, and the answer is not 0: \(I + A^{2}\) is invertible because none of its eigenvalues is zero. The value 1/8 comes from wrongly taking the eigenvalues of \(I + A^{2}\) as 1, 2, 4 (forgetting to square the 2).
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