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JEE Advanced 2022 Paper 1 · Mathematics · Matrices and Determinants · medium

List-I (P) det (adj(adjA)) (Q) det (2A^-1) (R) trace of adjA (S) det (A^TA^2 List-II (1) 3 (2) 4 (3) 6 (4) 8 (5) 16

Single correct option

Let A=(110011101)A=\begin{pmatrix}1&1&0\\0&1&1\\1&0&1\end{pmatrix}. Match the quantities in List-I with the values in List-II.

List-I
(P) det⁡(adj⁡(adj⁡A))\det\big(\operatorname{adj}(\operatorname{adj}A)\big)
(Q) det⁡(2A−1)\det\big(2A^{-1}\big)
(R) trace of adj⁡A\operatorname{adj}A
(S) det⁡(ATA2)\det\big(A^{T}A^{2}\big)

List-II
(1) 33
(2) 44
(3) 66
(4) 88
(5) 1616
  1. A. P→5, Q→2, R→3, S→4
  2. B. P→5, Q→2, R→1, S→4
  3. C. P→4, Q→2, R→1, S→5
  4. D. P→5, Q→4, R→1, S→2
Show answer and solution

Answer: (B) P→5, Q→2, R→1, S→4

Expanding, det⁡A=1(1)−1(0−1)+0=2\det A=1(1)-1(0-1)+0=2. For a 3×33\times3 matrix det⁡(adj⁡B)=(det⁡B)2\det(\operatorname{adj}B)=(\det B)^2 and adj⁡(adj⁡A)=(det⁡A)A\operatorname{adj}(\operatorname{adj}A)=(\det A)A. (P) det⁡((det⁡A)A)=23⋅2=16\det((\det A)A)=2^3\cdot2=16 (5). (Q) det⁡(2A−1)=8/det⁡A=4\det(2A^{-1})=8/\det A=4 (2). (R) tr⁡(adj⁡A)\operatorname{tr}(\operatorname{adj}A) equals the sum of principal 2×22\times2 minors, i.e. the coefficient of λ\lambda in det⁡(λI−A)=λ3−3λ2+3λ−2\det(\lambda I-A)=\lambda^3-3\lambda^2+3\lambda-2, which is 33 (1). (S) det⁡(AT)det⁡(A)2=2⋅4=8\det(A^T)\det(A)^2=2\cdot4=8 (4). Hence P→5, Q→2, R→1, S→4.
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