JEE Main 2022 — Matrices Question with Solution

Given $A=\left[\begin{array}{cc}1& -1\\ 2& \alpha \end{array}\right]$ and $B=\left[\begin{array}{ll}\beta & 1\\ 1& 0\end{array}\right],\alpha ,\beta \in R$,

So, $A+B=\left[\begin{array}{cc}\beta +1& 0\\ 3& \alpha \end{array}\right]$

Now ${\left(A+B\right)}^2=\left[\begin{array}{cc}\beta +1& 0\\ 3& \alpha \end{array}\right]\left[\begin{array}{cc}\beta +1& 0\\ 3& \alpha \end{array}\right]$

$=\left[\begin{array}{cc}{\left(\beta +1\right)}^2& 0\\ 3\left(\beta +1\right)+3\alpha & {\alpha }^2\end{array}\right]$

Also, $A^2=\left[\begin{array}{cc}1& -1\\ 2& \alpha \end{array}\right]\left[\begin{array}{cc}1& -1\\ 2& \alpha \end{array}\right]$$=\left[\begin{array}{cc}-1& -1-\alpha \\ 2+2\alpha & {\alpha }^2-2\end{array}\right]$

Now solving ${\left(A+B\right)}^2=A^2+\left[\begin{array}{ll}2& 2\\ 2& 2\end{array}\right]$

$\Rightarrow \left[\begin{array}{cc}{\left(\beta +1\right)}^2& 0\\ 3\left(\alpha +\beta +1\right)& {\alpha }^2\end{array}\right]=\left[\begin{array}{cc}1& -\alpha +1\\ 2\alpha +4& {\alpha }^2\end{array}\right]$

Now on comparing both side we get, $\alpha =1={\alpha }_1$

And $B^2=\left[\begin{array}{ll}\beta & 1\\ 1& 0\end{array}\right]\left[\begin{array}{ll}\beta & 1\\ 1& 0\end{array}\right]=\left[\begin{array}{cc}{\beta }^2+1& \beta \\ \beta & 1\end{array}\right]$

Now using ${\left(A+B\right)}^2=B^2$

$\Rightarrow \left[\begin{array}{cc}{\beta }^2+1& \beta \\ \beta & 1\end{array}\right]=\left[\begin{array}{cc}{\left(\beta +1\right)}^2& 0\\ 3\left(\beta +1\right)+3\alpha & {\alpha }^2\end{array}\right]$

Again on comparing both side we get, $\beta =0,\alpha =-1={\alpha }_2$

So, $\left|{\alpha }_1-{\alpha }_2\right|=\left|1-\left(-1\right)\right|=2$