Let $A = \begin{bmatrix} 1 & -1 \\ 2 & \alpha \end{bmatrix}$ and $B = \begin{bmatrix} \beta & 1 \\ 1 & 0 \end{bmatrix}$, $\alpha, \beta \in \mathbb{R}$. Let $\alpha_1$ be the value of $\alpha$ which satisfies $(A + B)^2 = A^2 + \begin{bmatrix} 2 & 2 \\ 2 & 2 \end{bmatrix}$ and $\alpha_2$ be the value of $\alpha$ which satisfies $(A + B)^2 = B^2$. Then $|\alpha_1 - \alpha_2|$ is equal to
Solution
Given $A=\begin{bmatrix} 1 & -1 \\ 2 & \alpha \end{bmatrix}$ and $B=\begin{bmatrix} \beta & 1 \\ 1 & 0 \end{bmatrix}$, $\alpha, \beta \in \mathbb{R}$,
So, $A+B=\begin{bmatrix} \beta+1 & 0 \\ 3 & \alpha \end{bmatrix}$
Now $(A+B)^2=\begin{bmatrix} \beta+1 & 0 \\ 3 & \alpha \end{bmatrix}\begin{bmatrix} \beta+1 & 0 \\ 3 & \alpha \end{bmatrix}$
$=\begin{bmatrix} (\beta+1)^2 & 0 \\ 3(\beta+1)+3\alpha & \alpha^2 \end{bmatrix}$
Also, $A^2=\begin{bmatrix} 1 & -1 \\ 2 & \alpha \end{bmatrix}\begin{bmatrix} 1 & -1 \\ 2 & \alpha \end{bmatrix}$
$=\begin{bmatrix} -1 & -1-\alpha \\ 2+2\alpha & \alpha^2-2 \end{bmatrix}$
Now solving $(A+B)^2=A^2+\begin{bmatrix} 2 & 2 \\ 2 & 2 \end{bmatrix}$
$\Rightarrow \begin{bmatrix} (\beta+1)^2 & 0 \\ 3\alpha+\beta+1 & \alpha^2 \end{bmatrix}=\begin{bmatrix} 1 & -\alpha+1 \\ 2\alpha+4 & \alpha^2 \end{bmatrix}$
Now on comparing both side we get, $\alpha=1=\alpha_1$
And $B^2=\begin{bmatrix} \beta & 1 \\ 1 & 0 \end{bmatrix}\begin{bmatrix} \beta & 1 \\ 1 & 0 \end{bmatrix}=\begin{bmatrix} \beta^2+1 & \beta \\ \beta & 1 \end{bmatrix}$
Now using $(A+B)^2=B^2$
$\Rightarrow \begin{bmatrix} \beta^2+1 & \beta \\ \beta & 1 \end{bmatrix}=\begin{bmatrix} (\beta+1)^2 & 0 \\ 3\beta+1+3\alpha & \alpha^2 \end{bmatrix}$
Again on comparing both side we get, $\beta=0$, $\alpha=-1=\alpha_2$
So, $|\alpha_1-\alpha_2|=|1-(-1)|=2$