Why is the JEE Advanced 2020 adjoint of adjoint question rated hard?
The JEE Advanced 2020 Paper 2 question carries difficulty tag 4 and an expected solving time of 180 seconds. It requires the exact identity adj(adj A)=(det A)^{n-2}A for n=3. The multi-correct format demands verifying each of A, B, C, D independently after finding det M.
Students know the definitions of adjoint and inverse plus det(AB)=det(A)det(B). The gap is deriving all consequences of M^{-1}=adj(adj M) for 3×3 matrices. You must first force det M=1 from the determinant step, then chain equalities to prove B, C and D are always true.
What does the JEE Advanced 2020 question on matrices and determinants actually ask?
M is any 3×3 invertible matrix with real entries. Given that the inverse of M equals the adjoint of the adjoint of M, decide which of the four statements must hold for every such M:
B, C and D must hold. A does not.
How do you solve the official JEE Advanced 2020 adjoint and inverse question step by step?
Let d=det(M)≠0. For 3×3 matrix adj(adj M)=d^{3-2}M=dM.
The given condition becomes
Take determinant:
Left side is 1/d. Right side is d^3·d=d^4. Thus
which gives
Since d is real and nonzero, d=1. This proves B.
Substitute d=1:
Multiply both sides on the left by M:
This is C. Now adj M=(det M)M^{-1}=1·M^{-1}=M. Hence
This is D. Option A is not necessarily true because any non-identity involutory matrix with det=1 satisfies the condition.
Why do students wrongly pick option A in this matrices and determinants JEE 2020 question?
Students stop at M^{-1}=M after finding d=1 and declare M must equal the identity matrix. They skip the left-multiplication by M that produces M^2=I.
This error skips the recognition that M^2=I defines involutory matrices of which the identity is only one example. The correct chain is d→1 then M^{-1}=M then M^2=I. The slip jumps straight from M^{-1}=M to M=I and wrongly marks A.
Which practice questions test the same adjoint and inverse ideas?
Question 1 (2×2 case): If adj(adj A)=kA for a 2×2 invertible matrix A, find possible values of det(A) and compare the exponent used versus the 3×3 case.
For n=2 the identity gives adj(adj A)=(det A)^0 A=A. Thus k=1 and det(A) can be any nonzero real. The exponent is 0 here versus 1 in the 3×3 case.
Question 2: For a 3×3 matrix with det M=−1, simplify (adj M)^2 in terms of M and M^{-1}.
adj M=(det M)M^{-1}=−M^{-1}. Squaring gives
Question 3 (JEE-style multi-correct): Given M^3=I and det M=1, which of adj(M)=M^2, M^{-1}=M^2, det(adj M)=1 must be true.
From M^3=I multiply on the left by M^{-1} to get M^{-1}=M^2. Then adj(M)=(det M)M^{-1}=M^2 and det(adj M)=(det M)^2=1. All three hold.
Solve these on paper, then check against the past-paper archive to search every JEE Main paper from 2002 and every Advanced paper from 2007 by chapter for worked solutions.
What key identities and resources help after this matrices and determinants JEE 2020 question?
The general formula adj(adj A)=(det A)^{n-2}A must be memorized for n=2,3,4. Taking determinant at the earliest step usually pins down possible values of det M.
See the solved paper on consistency in match-list format: Matrices and Determinants JEE 2022.
What practical exam tips improve speed on adjoint-inverse questions in JEE Advanced?
Always work with d=det M as a scalar variable rather than assuming its value. In multi-correct questions prove or disprove each option after fixing d. A real determinant forces d=1 as the only real fifth root of unity.
When stuck on a similar adjoint question, photograph a doubt and get a step-by-step solution with a free-body diagram when the question needs one (20 free).
Next step: the past-paper archive on JEEnius and search every JEE Main paper from 2002 and every Advanced paper from 2007, by year, subject or chapter, each with a worked solution (free).
Frequently asked questions
How do you solve the adjoint of adjoint question in matrices and determinants jee 2020?
Let d = det(M). For 3x3 matrices adj(adj M) = d M. The condition M^{-1} = adj(adj M) gives M^{-1} = d M. Taking determinants yields 1/d = d^4 so d^5 = 1. Since d is real, d=1. Then M^{-1}=M so M^2=I. Also adj(M)=M so (adj M)^2=I.
Why is option A incorrect in the JEE Advanced 2020 adjoint question?
Option A claims M equals the identity. After finding M^{-1}=M students often stop here. However M^{-1}=M only implies M^2=I. Any involutory matrix with det=1 satisfies the condition, not just the identity. Thus A is not necessarily true.
What is adj(adj M) for a 3x3 matrix in JEE Advanced?
For an nxn matrix the identity is adj(adj A)=(det A)^{n-2}A. When n=3 this simplifies to adj(adj M)=(det M)M. This formula is essential to solve the given condition M^{-1}=adj(adj M) and derive the required results.
Why do many students get the JEE 2020 matrices and determinants question wrong?
Students correctly find d=1 and reach M^{-1}=M but then wrongly conclude M=I. They miss multiplying both sides by M to obtain M^2=I. This skips the fact that there are many involutory matrices with det=1 besides the identity matrix.