Question

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

Answer 1

First of all, assume that $C=\left[ \begin{aligned} & \begin{matrix} 0 & a & b \\\ -a & 0 & c \\\ -b & -c & 0 \\\ \end{matrix} \\\ \end{aligned} \right]$ 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 = $C=\left[ \begin{aligned} & \begin{matrix} \,\,\,0 & \,\,a & \,b \\\ \end{matrix} \\\ & \begin{matrix} -a & \,0 & \,c \\\ \end{matrix} \\\ & \begin{matrix} -b & -c & 0 \\\ \end{matrix} \\\ \end{aligned} \right]$ ……………………………………….(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

& \begin{matrix} \,\,\,0 & \,\,a & \,b \\\ \end{matrix} \\\ & \begin{matrix} -a & \,0 & \,\,c \\\ \end{matrix} \\\ & \begin{matrix} -b & -c & 0 \\\ \end{matrix} \\\ \end{aligned} \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{aligned} & \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{aligned}$$ $$\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{aligned} & \begin{matrix} d & a & b \\\ \end{matrix} \\\ & \begin{matrix} -a & e & c \\\ \end{matrix} \\\ & \begin{matrix} -b & -c & f \\\ \end{matrix} \\\ \end{aligned} \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.