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
Answer & explanation
Correct answer: option 3
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:
$A_{11}+A_{22}+A_{33} = 25+16+9 = 50$