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← JEE Main 2022 · 28 Jul · Shift 1
MathematicsInteger+4 / -1Question 25 of 90

JEE Main 2022 · 28 Jul · Shift 1, Mathematics question 25

Let A=[1−12α]A=\begin{bmatrix}1&-1\\2&\alpha\end{bmatrix} and B=[β110]B=\begin{bmatrix}\beta&1\\1&0\end{bmatrix}, α,β∈R\alpha,\beta\in\mathbf R. Let α1\alpha_1 be the value of α\alpha which satisfies (A+B)2=A2+[2222](A+B)^2=A^2+\begin{bmatrix}2&2\\2&2\end{bmatrix} and α2\alpha_2 be the value of α\alpha which satisfies (A+B)2=B2(A+B)^2=B^2. Then ∣α1−α2∣|\alpha_1-\alpha_2| is equal to ______.

Answer and worked solution

Official answer: 2

First equation. AB+BA+B2=[2222]AB+BA+B^2=\begin{bmatrix}2&2\\2&2\end{bmatrix} forces β=0\beta=0, α1=1\alpha_1=1.

Second. A2+AB+BA=OA^2+AB+BA=O forces β=0\beta=0, α2=−1\alpha_2=-1.

∣α1−α2∣=2|\alpha_1-\alpha_2|=2.

Answer: 2