If C is a 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

Hint: First of all, assume that \[C=\left[ \begin{align}
  & \begin{matrix}
    0 & a & b \\
   -a & 0 & c \\
   -b & -c & 0 \\
\end{matrix} \\
\end{align} \right]\] and \[X=\left[ \begin{matrix}
   p \\
   q \\
   r \\
\end{matrix} \right]\] . Now, get the transpose of the matrix X, \[{{X}^{T}}=\left[ \begin{matrix}
   p & q & r \\
\end{matrix} \right]\] . Now, multiply the matrix C and X, and get the value of \[CX\] . We know the property that the transpose of a matrix is the interchange of its rows by columns. Use this property and get the transpose of the matrix X, \[{{X}^{T}}\] . Now, multiply the matrix \[{{X}^{T}}\] and \[CX\] , and get the result. We also know the property that the determinant value of a non-invertible matrix is equal to zero. At last, conclude the answer.

Complete step-by-step solution:
According to the question, it is given that we have a skew-symmetric matrix C of order n and a matrix X of order \[n\times 1\] column matrix.
We know that the diagonal elements of a skew-symmetric matrix are zero and also the transpose of the skew-symmetric its negative.
First of all, let us assume that,
A skew-symmetric matrix C of order n = \[C=\left[ \begin{align}
  & \begin{matrix}
   \,\,\,0 & \,\,a & \,b \\
\end{matrix} \\
 & \begin{matrix}
   -a & \,0 & \,c \\
\end{matrix} \\
 & \begin{matrix}
   -b & -c & 0 \\
\end{matrix} \\
\end{align} \right]\] ……………………………………….(1)
The matrix X of order \[n\times 1\] column matrix = \[X=\left[ \begin{matrix}
   p \\
   q \\
   r \\
\end{matrix} \right]\] ……………………………………….(2)
We know the property that the transpose of a matrix is simply the interchange of the rows and columns of the matrix.
Now, the transpose of the matrix X = \[{{X}^{T}}=\left[ \begin{matrix}
   p & q & r \\
\end{matrix} \right]\] ………………………………………..(3)
From equation (1) and equation (2), we have the matrix C and X.
Now, on multiplying the matrix C and X, we get
\[CX=\left[ \begin{align}
  & \begin{matrix}
   \,\,\,0 & \,\,a & \,b \\
\end{matrix} \\
 & \begin{matrix}
   -a & \,0 & \,\,c \\
\end{matrix} \\
 & \begin{matrix}
   -b & -c & 0 \\
\end{matrix} \\
\end{align} \right]\left[ \begin{matrix}
   p \\
   q \\
   r \\
\end{matrix} \right]=\left[ \begin{matrix}
   0\times p+a\times q+b\times r \\
   \left( -a \right)\times p+0\times q+c\times r \\
   \left( -b \right)\times p+\left( -c \right)\times q+0\times r \\
\end{matrix} \right]=\left[ \begin{matrix}
   aq+br \\
   -ap+cr \\
   -bp-cq \\
\end{matrix} \right]\] ……………………………………….(4)
From equation (3) and equation (4), we have the matrix \[{{X}^{T}}\] and \[CX\] .
Now, on multiplying the matrix \[{{X}^{T}}\] and \[CX\] , we get
\[\begin{align}
  & \Rightarrow {{X}^{T}}CX=\left[ \begin{matrix}
   p & q & r \\
\end{matrix} \right]\left[ \begin{matrix}
   aq+br \\
   -ap+cr \\
   -bp-cq \\
\end{matrix} \right] \\
 & \Rightarrow {{X}^{T}}CX=\left[ paq+pbr-qap+cqr-rbp-rcq \right] \\
 & \Rightarrow {{X}^{T}}CX=\left[ paq+pbr-paq+cqr-pbr-cqr \right] \\
\end{align}\]
\[\Rightarrow {{X}^{T}}CX=\left[ 0 \right]\] ……………………………………..(5)
From equation (5), we have the value of \[{{X}^{T}}CX=\left[ 0 \right]\] .
Now, we can see that the determinant value of \[{{X}^{T}}CX\] is zero and we know that the determinant value of a singular matrix is zero ……………………………. (6)
So, the matrix \[{{X}^{T}}CX\] is singular ………………………………………(7)
We know the property that the determinant value of a non-invertible matrix is equal to zero.
From equation (6), we have the determinant value of matrix \[{{X}^{T}}CX\] .
So, the matrix \[{{X}^{T}}CX\] is non-invertible …………………………………………….(8)
Now, from equation (7) and equation (8), we can say that the matrix \[{{X}^{T}}CX\] is singular and also non-invertible.
Therefore, the correct option is (A) and (D).

Note: In this question, one might assume matrix C as \[\left[ \begin{align}
  & \begin{matrix}
   d & a & b \\
\end{matrix} \\
 & \begin{matrix}
   -a & e & c \\
\end{matrix} \\
 & \begin{matrix}
   -b & -c & f \\
\end{matrix} \\
\end{align} \right]\] . If we do so then our calculation will get complex. Since the matrix C is skew-symmetric iso, its diagonal matrix is zero, and the transpose is negative of the skew-symmetric matrix.