Matrices & DeterminantshardPYQ · JEE Main · 6 Apr 2026 · Shift 1 (Morning)Free

Adjugate Matrix Equation: (α−β)² = 4 | JEE Main 2026

JEE Maths question with a full step-by-step solution.

Question
Let A=[111101001]A=\begin{bmatrix}-1&1&-1\\1&0&1\\0&0&1\end{bmatrix} satisfy A2+α(adj(adj(A)))+β(adj(A)adj(adj(A)))=[222201001]A^2+\alpha\big(\mathrm{adj}(\mathrm{adj}(A))\big)+\beta\big(\mathrm{adj}(A)\cdot\mathrm{adj}(\mathrm{adj}(A))\big)=\begin{bmatrix}2&-2&2\\-2&0&-1\\0&0&-1\end{bmatrix} for some α,βR\alpha,\beta\in\mathbb{R}. Then (αβ)2(\alpha-\beta)^2 is equal to
Solution
Answer: 4 (± 0.01)
Step 1: Expand A=[111101001]A=\begin{bmatrix}-1&1&-1\\1&0&1\\0&0&1\end{bmatrix} along the third row:
A=11110=(1)(0)(1)(1)=1.|A|=1\cdot\begin{vmatrix}-1&1\\1&0\end{vmatrix}=(-1)(0)-(1)(1)=-1.
Step 2: For n=3n=3: \bullet adj(adj(A))=An2A=AA=(1)A=A.\mathrm{adj}(\mathrm{adj}(A))=|A|^{n-2}A=|A|A=(-1)A=-A. \bullet adj(A)adj(adj(A))=adj(A)(AA)=A(adj(A)A)=AAI=(1)2I=I.\mathrm{adj}(A)\cdot\mathrm{adj}(\mathrm{adj}(A))=\mathrm{adj}(A)\cdot\big(|A|A\big)=|A|\big(\mathrm{adj}(A)\,A\big)=|A|\cdot|A|I=(-1)^2I=I. Step 3: Substitute:
A2+α(A)+β(I)=M,M=[222201001].A^2+\alpha(-A)+\beta(I)=M,\qquad M=\begin{bmatrix}2&-2&2\\-2&0&-1\\0&0&-1\end{bmatrix}.
A2αA+βI=M.\Rightarrow A^2-\alpha A+\beta I=M.
Step 4:
A2=[111101001][111101001]=[211110001].A^2=\begin{bmatrix}-1&1&-1\\1&0&1\\0&0&1\end{bmatrix}\begin{bmatrix}-1&1&-1\\1&0&1\\0&0&1\end{bmatrix}=\begin{bmatrix}2&-1&1\\-1&1&0\\0&0&1\end{bmatrix}.
(1,1)=(1)(1)+(1)(1)+(1)(0)=2(1,1)=(-1)(-1)+(1)(1)+(-1)(0)=2; (1,2)=(1)(1)+(1)(0)+(1)(0)=1(1,2)=(-1)(1)+(1)(0)+(-1)(0)=-1; (2,1)=(1)(1)+(0)(1)+(1)(0)=1(2,1)=(1)(-1)+(0)(1)+(1)(0)=-1. Step 5: Compare A2αA+βI=MA^2-\alpha A+\beta I=M: \bullet (2,2)(2,2): 1α(0)+β=1+β=M22=0β=1.1-\alpha(0)+\beta=1+\beta=M_{22}=0\Rightarrow\beta=-1. \bullet (1,2)(1,2): 1α(1)=1α=M12=2α=1.-1-\alpha(1)=-1-\alpha=M_{12}=-2\Rightarrow\alpha=1. Step 6:
(αβ)2=(1(1))2=22=4.\therefore(\alpha-\beta)^2=(1-(-1))^2=2^2=4.
Correct answer: 4
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