If \(A = \begin{bmatrix}0&2&-3\\-2&0&1\\3&-1&0\end{bmatrix}\) and \(X = \begin{bmatrix}1\\2\\3\end{bmatrix}\), then \(X'AX\) is equal to
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(\begin{bmatrix}0\end{bmatrix}\)
X' is of order \(1 \times 3\), A of order \(3 \times 3\) and X of order \(3 \times 1\), so \(X'AX\) is of order \(1 \times 1\).
\( \displaystyle AX = \begin{bmatrix}0 + 4 - 9\\-2 + 0 + 3\\3 - 2 + 0\end{bmatrix} = \begin{bmatrix}-5\\1\\1\end{bmatrix} \)
\(X'(AX) = \begin{bmatrix}1(-5) + 2(1) + 3(1)\end{bmatrix} = \begin{bmatrix}0\end{bmatrix}\)
This is no accident: A is skew-symmetric (\(A' = -A\)). A \(1 \times 1\) matrix equals its own transpose, so
\(X'AX = (X'AX)' = X'A'X = -X'AX\),
which forces \(X'AX = \begin{bmatrix}0\end{bmatrix}\) for every column X of order \(3 \times 1\).