ISRO EC 2025 Q.32: solution and answer

Official answer: DSource: ISRO's final answer key of 15 May 2026

1. Similar matrices means there is an invertible $P$ with $B=P^{-1}AP$.

2. Determinant: $\det (B)=\det (P^{-1})\det (A)\det (P)=\frac{1}{\det (P)}\det (A)\det (P)=\det (A)$, so option (a) is true.

3. Characteristic polynomial: $\det (\lambda I-B)=\det \left(P^{-1}(\lambda I-A)P\right)=\det (\lambda I-A)$. So $A$ and $B$ have the same characteristic polynomial, hence the same eigenvalues, so option (c) is true.

4. Trace: using $\mathrm{tr}(XY)=\mathrm{tr}(YX)$, $\mathrm{tr}(P^{-1}AP)=\mathrm{tr}(AP{P}^{-1})=\mathrm{tr}(A)$, so option (b) is true. (Equivalently, the trace is the sum of the eigenvalues, which are the same.)

5. Options (a), (b) and (c) are all true, so the answer is (d), All of the above.