If \(A=\begin{bmatrix}x&-3&4\\3&y&-5\\-4&z&0\end{bmatrix}\) is a Skew-Symmetric matrix and \(\mathrm{adj}\,A = [a_{ij}]_{3\times3}\), then \(a_{11} + a_{22} + a_{33}\) is equal to
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 50
In a skew-symmetric matrix \(a_{ji} = -a_{ij}\), so every diagonal entry is 0. Hence
\(x = 0,\quad y = 0,\quad z = -(-5) = 5\)
\(\displaystyle A = \begin{bmatrix}0&-3&4\\3&0&-5\\-4&5&0\end{bmatrix}\)
adj A is the transpose of the matrix of cofactors. Transposing does not move the diagonal, so the diagonal entries of adj A are the cofactors \(A_{11}, A_{22}, A_{33}\) of A:
\(\displaystyle a_{11} = A_{11} = \begin{vmatrix}0&-5\\5&0\end{vmatrix} = 0 + 25 = 25\)
\(\displaystyle a_{22} = A_{22} = \begin{vmatrix}0&4\\-4&0\end{vmatrix} = 0 + 16 = 16\)
\(\displaystyle a_{33} = A_{33} = \begin{vmatrix}0&-3\\3&0\end{vmatrix} = 0 + 9 = 9\)
\(a_{11} + a_{22} + a_{33} = 25 + 16 + 9 = 50\)