If $A=\begin{bmatrix}x&-3&4\\3&y&-5\\-4&z&0\end{bmatrix}$ is a Skew-Symmetric matrix and $adj\, A = [a_{ij}]_{3×3}$, then $A_{11} + A_{22} + A_{33}$ is equal to

50

The correct answer is Option (3) → 50

Since $A$ is skew–symmetric,

$A^T = -A$

So all diagonal entries must be zero:

$x=0,\; y=0,\; A_{33}=0$ already.

Also off–diagonal conditions give:

$A_{23}=-5 = -A_{32}$ ⟹ $z=5$

Thus

$A=\begin{bmatrix} 0 & -3 & 4\\ 3 & 0 & -5\\ -4 & 5 & 0 \end{bmatrix}$

$A_{11} = \det\begin{bmatrix}0 & -5\\ 5 & 0\end{bmatrix} = 25$

$A_{22} = \det\begin{bmatrix}0 & 4\\ -4 & 0\end{bmatrix} = 16$

$A_{33} = \det\begin{bmatrix}0 & -3\\ 3 & 0\end{bmatrix} = 9$

Sum:

$A_{11}+A_{22}+A_{33} = 25+16+9 = 50$