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JEE Main 2026 · 6 Apr · Shift 1, Mathematics question 21

Let A=[−11−1101001]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))))=[2−22−20−100−1]A^2+\alpha\left(\operatorname{adj}\left(\operatorname{adj}(A)\right)\right)+\beta\left(\operatorname{adj}(A)\left(\operatorname{adj}\left(\operatorname{adj}(A)\right)\right)\right)=\begin{bmatrix}2&-2&2\\-2&0&-1\\0&0&-1\end{bmatrix}
for some α,β∈R\alpha,\beta\in\mathbb R. Then (α−β)2(\alpha-\beta)^2 is equal to ________.

Answer and worked solution

Official answer: 4

Adjugate identities (3×33\times3, det⁡A=−1\det A=-1).

  • adj⁡(adj⁡A)=(det⁡A)A=−A\operatorname{adj}(\operatorname{adj}A)=(\det A)A=-A.
  • adj⁡(A) (−A)=−(det⁡A)I=I\operatorname{adj}(A)\,(-A)=-(\det A)I=I.

Equation. It becomes A2−αA+βI=RA^2-\alpha A+\beta I=R. With A2=[2−11−110001]A^2=\begin{bmatrix}2&-1&1\\-1&1&0\\0&0&1\end{bmatrix}:

  • An off-diagonal entry, −α(1)=−1−0-\alpha(1)=-1-0, gives α=1\alpha=1.
  • The (2,2) entry gives β=−1\beta=-1.
  • All other entries agree.

Value: (α−β)2=4(\alpha-\beta)^2=4.

Answer: 4

Topics:Adjoint and inverseMatrix algebra
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