Let \(A = \begin{bmatrix}0 & 2\alpha + 1\\ 1 & \beta\end{bmatrix}\) 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
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(\beta - \alpha = 1\)
A skew-symmetric matrix has zero diagonal and \(b_{21} = -b_{12} = -1\):
\( \displaystyle B = \begin{bmatrix}0 & 1\\ -1 & 0\end{bmatrix} \)
\( \displaystyle AB = \begin{bmatrix}0 & 2\alpha + 1\\ 1 & \beta\end{bmatrix}\begin{bmatrix}0 & 1\\ -1 & 0\end{bmatrix} \)
\( \displaystyle = \begin{bmatrix}-(2\alpha + 1) & 0\\ -\beta & 1\end{bmatrix} \)
Comparing with \(I_2\), entry by entry:
\(-(2\alpha + 1) = 1,\quad \alpha = -1\)
\(-\beta = 0,\quad \beta = 0\)
So \(\beta - \alpha = 0 - (-1) = 1\).