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JEE Advanced 2023 Paper 2 · Mathematics · Matrices and Determinants · medium

Let A be a 3×3 real matrix with A^2=A+2I and tr(A)=3. The value of tr(A^5+det (adjA) is

Numerical answer

Let AA be a 3×33\times3 real matrix with A2=A+2IA^2=A+2I and tr⁡(A)=3\operatorname{tr}(A)=3. The value of tr⁡(A5)+det⁡(adj⁡A)\operatorname{tr}(A^5)+\det(\operatorname{adj}A) is
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Answer: 79

(A−2I)(A+I)=0(A-2I)(A+I)=0 with distinct roots, so AA is diagonalisable with eigenvalues in {2,−1}\{2,-1\}. If 22 occurs pp times and −1-1 occurs qq times, p+q=3p+q=3 and 2p−q=32p-q=3 give p=2p=2, q=1q=1. Then tr⁡(A5)=2⋅32+(−1)=63\operatorname{tr}(A^5)=2\cdot32+(-1)=63, det⁡A=2⋅2⋅(−1)=−4\det A=2\cdot2\cdot(-1)=-4 and det⁡(adj⁡A)=(det⁡A)2=16\det(\operatorname{adj}A)=(\det A)^{2}=16. Total =79=79.
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