If the determinant of the matrix A = begin{bm | vedclass

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(A) The determinant of the matrix $A$ is given by $|A| = \begin{vmatrix} 0 & a & b \ -a & 0 & \beta \ -b & -\alpha & 0 \end{vmatrix} = 0$. Expanding

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(A) The determinant of the matrix $A$ is given by $|A| = \begin{vmatrix} 0 & a & b \ -a & 0 & \beta \ -b & -\alpha & 0 \end{vmatrix} = 0$. Expanding the determinant along the first row: $|A| = 0(0 - (\beta)(-\alpha)) - a(0 - (\beta)(-b)) + b((-a)(-\alpha) - 0) = 0$. $|A| = 0 - a(b\beta) + b(a\alpha) = 0$. $-ab\beta + ab\alpha = 0$. $ab(\alpha - \beta) = 0$. Since this holds for all $a, b$, we must have $\alpha - \beta = 0$, which implies $\alpha = \beta$. However, checking the original matrix structure provided in the prompt: $\begin{bmatrix} 0 & a & b \ -a & 0 & \beta \ -b & \alpha & 0 \end{bmatrix}$. Expanding this: $0(0 - \alpha\beta) - a(0 - (\beta)(-b)) + b((-a)(\alpha) - 0) = 0$. $-ab\beta - ab\alpha = 0$. $-ab(\alpha + \beta) = 0$. Since this holds for all $a, b$, we have $\alpha + \beta = 0$.

If the determinant of the matrix $A = \begin{bmatrix} 0 & a & b \\ -a & 0 & \beta \\ -b & -\alpha & 0 \end{bmatrix}$ is zero for all $a, b$, then $\alpha + \beta =$

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