Let $A =\begin{bmatrix}0&2α+1\\1&β\end{bmatrix}$ and $B = [b_{ij}]$ be a skew symmetric matrix of order 2 such that $b_{12} = 1$. If $AB = I_2$, where $I_2$ is identity matrix of order 2, then

$β-α=1$

The correct answer is Option (2) → $β-α=1$

Given: $A = \begin{bmatrix} 0 & 2\alpha+1 \\ 1 & \beta \end{bmatrix}$, $B = [b_{ij}]$ skew-symmetric of order 2 with $b_{12} = 1$, and $AB = I_2$

For a 2×2 skew-symmetric matrix $B$:

$B = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}$

Compute AB = I:

$AB = \begin{bmatrix} 0 & 2\alpha+1 \\ 1 & \beta \end{bmatrix} \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix} = \begin{bmatrix} (0*0 + (2\alpha+1)*(-1)) & (0*1 + (2\alpha+1)*0) \\ (1*0 + \beta*(-1)) & (1*1 + \beta*0) \end{bmatrix} = \begin{bmatrix} -(2\alpha+1) & 0 \\ -\beta & 1 \end{bmatrix}$

Set AB = I₂

$-(2\alpha +1) = 1 \Rightarrow 2\alpha +1 = -1 \Rightarrow \alpha = -1$

$0 = 0$ ✅

$-\beta = 0 \Rightarrow \beta = 0$

$1 = 1$ ✅

Values: α = -1, β = 0