Question

If C is skew-symmetric matrix of order n and X in n $\times$ 1 column matrix, then X${}^T$CX is

• A. singular
• B. non-singular
• C. invertible
• D. non-invertible

Answer 1

Answer: (A), (D)

singular
non-invertible

Skew Symmetric matrix means if $A$ represents a matrix then $A^T=-A$

Now,

Let $C=\left[\begin{array}{ccc}0& 1& 2\\ -1& 0& 3\\ -2& -3& 0\end{array}\right]$ and $X=\left[\begin{array}{c}1\\ 2\\ 3\end{array}\right]$ ($\because$ values are assumed randomly keeping in mind, the order of the matrices, as given in the question)

$\therefore CX=\left[\begin{array}{ccc}0& 1& 2\\ -1& 0& 3\\ -2& -3& 0\end{array}\right]\left[\begin{array}{c}1\\ 2\\ 3\end{array}\right]=\left[\begin{array}{c}1\times 2+2\times 3\\ -1\times 1+3\times 3\\ -2\times 1-3\times 2\end{array}\right]=\left[\begin{array}{c}8\\ 8\\ -8\end{array}\right]\therefore X^{T}CX=\left[\begin{array}{ccc}1& 2& 3\end{array}\right]\cdot \left[\begin{array}{c}8\\ 8\\ -8\end{array}\right]=[8+16-24]=\left[0\right]\therefore \left|X^{T}CX\right|=0$

$\therefore X^{T}CX$ is singular and non-invertible