Question

If A is a skew-symmetric matrix of order n and $C$ is a column matrix of the order $n\times 1$, Then ${{C}^{T}}AC$ is
(1) an identity matrix of order n
(2) an identity matrix of order 1
(3) a zero matrix of order 1
(4) None of the above

Answer 1

In this type of question we have to use the concept of matrices. We know that, if $A$ is a skew symmetric matrix then ${{A}^{T}}=-A$. Also we have by the property of transpose of a matrix, ${{\left( {{A}^{T}} \right)}^{T}}=A$ and ${{\left( AB \right)}^{T}}={{B}^{T}}{{A}^{T}}$. We know that the transpose of a column matrix of order $n\times 1$ is always a row matrix of the order $1\times n$.

Complete step-by-step solution:
Now here we have to find the nature of ${{C}^{T}}AC$ where $A$ is a skew-symmetric matrix of order n and $C$ is a column matrix of the order $n\times 1$
As we know, the transpose of a column matrix of order $n\times 1$ is always a row matrix of the order $1\times n$.
Hence, as $C$ is a column matrix of the order $n\times 1$, ${{C}^{T}}$ is a row matrix of the order $1\times n$
Thus we have, $A$ is a skew-symmetric matrix of order n i.e. $n\times n$, $C$ is a column matrix of the order $n\times 1$ and ${{C}^{T}}$ is a row matrix of the order $1\times n$

& \Rightarrow \text{The order of }{{C}^{T}}AC=\left( 1\times n \right)\times \left( n\times n \right)\times \left( n\times 1 \right) \\\ & \Rightarrow \text{The order of }{{C}^{T}}AC=\left( 1\times 1 \right) \\\ \end{aligned}$$ Now as the order of $${{C}^{T}}AC$$ is $$\left( 1\times 1 \right)$$ we can write $$\Rightarrow {{\left( {{C}^{T}}AC \right)}^{T}}={{C}^{T}}AC\cdots \cdots \cdots \left( i \right)$$ Let us assume that, $$\Rightarrow {{C}^{T}}AC=B$$ Thus from the equation $$\left( i \right)$$ we can write $$\Rightarrow {{B}^{T}}=B\cdots \cdots \cdots \left( ii \right)$$ Now we have given that $$A$$ is a skew-symmetric matrix $$\Rightarrow {{A}^{T}}=-A\cdots \cdots \cdots \left( iii \right)$$ Let us consider, $$\begin{aligned} & \Rightarrow {{\left( {{C}^{T}}AC \right)}^{T}}={{B}^{T}} \\\ & \Rightarrow \left( {{C}^{T}} \right){{A}^{T}}{{\left( {{C}^{T}} \right)}^{T}}={{B}^{T}}\cdots \cdots \cdots \left\\{ {{\left( AB \right)}^{T}}={{B}^{T}}{{A}^{T}} \right\\} \\\ & \Rightarrow {{C}^{T}}\left( -A \right){{\left( {{C}^{T}} \right)}^{T}}={{B}^{T}}\cdots \cdots \cdots \left\\{ \text{from e}{{\text{q}}^{n}}\left( iii \right) \right\\} \\\ & \Rightarrow {{C}^{T}}\left( -A \right)C={{B}^{T}}\cdots \cdots \cdots \left\\{ {{\left( {{A}^{T}} \right)}^{T}}=A \right\\} \\\ & \Rightarrow -\left( {{C}^{T}}AC \right)=B\cdots \cdots \ \cdots \left\\{ \text{from e}{{\text{q}}^{n}}\left( ii \right) \right\\} \\\ & \Rightarrow -B=B \\\ & \Rightarrow 2B=0 \\\ & \Rightarrow B=\left[ 0 \right] \\\ \end{aligned}$$ Hence, we get $${{C}^{T}}AC$$ as a zero matrix of order 1. **Thus, option (3) is the correct option.** **Note:** In this type of question students have to remember the different properties of transpose of a matrix. Also students have to remember the definitions of different types of matrices. Students have to remember that in case of matrix multiplication, the number of columns in the first matrix must be equal to the number of rows in the second matrix; the resultant matrix has the number of rows of the first and the number of columns of the second matrix.