Let $A = \begin{bmatrix} 0 & 2\alpha+1 \\ \ 1& \beta \end{bmatrix}$ and $B = \begin{bmatrix} b_{ij}\end{bmatrix}$ 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

A skew-symmetric matrix satisfies $B^T = -B$.

For a $2 \times 2$ skew-symmetric matrix $B = \begin{bmatrix} b_{11} & b_{12} \\ b_{21} & b_{22} \end{bmatrix}$, this property gives us:

• All diagonal elements must be zero: $b_{11} = 0$ and $b_{22} = 0$
• Off-diagonal elements are opposites: $b_{21} = -b_{12}$

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Given that $b_{12} = 1$:

$b_{11} = 0$

$b_{12} = 1$

$b_{21} = -b_{12} = -1$

$b_{22} = 0$

Therefore:

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

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Since $AB = I_2$, we have:

$\begin{bmatrix} 0 & 2\alpha+1 \\ 1 & \beta \end{bmatrix} \times \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$

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Multiplying the matrices:

Row 1, Column 1: $(0)(0) + (2\alpha+1)(-1) = -2\alpha - 1$

Row 1, Column 2: $(0)(1) + (2\alpha+1)(0) = 0$

Row 2, Column 1: $(1)(0) + (\beta)(-1) = -\beta$

Row 2, Column 2: $(1)(1) + (\beta)(0) = 1$

This gives:

$\begin{bmatrix} -2\alpha-1 & 0 \\ -\beta & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$

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Comparing corresponding elements:

From position $(1,1)$:

$-2\alpha - 1 = 1$

$-2\alpha = 2$

$\alpha = -1$

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From position $(2,1)$:

$-\beta = 0$

$\beta = 0$

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Therefore, $\alpha = -1$ and $\beta = 0$.