Question

What is the inverse of a skew symmetric matrix of odd order?
(a) a symmetric matrix.
(b) a skew symmetric matrix.
(c) diagonal matrix.
(d) does not exist.

Answer 1

Hint: We start solving by assuming the matrix and recalling the definitions of skew symmetric and transpose of the matrix. We use the condition of the skew symmetric matrix and find the elements in the matrix. We then find the determinant of the matrix and we use the fact that the determinant of the matrix should not be zero in order to have an inverse to get the desired result.

Complete step-by-step answer:
According to the problem, we need to find the inverse of a skew symmetric matrix of odd order.
Let us check for the $3\times 3$ matrix and let us assume the matrix be $A=\left[ \begin{matrix}
   a & b & c \\
   d & e & f \\
   g & h & i \\
\end{matrix} \right]$. We know that a square matrix is defined as a skew symmetric matrix if the transpose of the matrix is equal to the negative of the matrix i.e., ${{A}^{T}}=-A$.
So, let us first transpose of the given matrix A.
We know that the transpose of a matrix is formed by interchanging the rows with columns of given matrix. We use this to find the transpose of the matrix $A=\left[ \begin{matrix}
   a & b & c \\
   d & e & f \\
   g & h & i \\
\end{matrix} \right]$.
So, we have ${{A}^{T}}=\left[ \begin{matrix}
   a & d & g \\
   b & e & h \\
   c & f & i \\
\end{matrix} \right]$.
We have ${{A}^{T}}=-A$,
\[\Rightarrow \left[ \begin{matrix}
   a & d & g \\
   b & e & h \\
   c & f & i \\
\end{matrix} \right]=-\left[ \begin{matrix}
   a & b & c \\
   d & e & f \\
   g & h & i \\
\end{matrix} \right]\].
\[\Rightarrow \left[ \begin{matrix}
   a & d & g \\
   b & e & h \\
   c & f & i \\
\end{matrix} \right]=\left[ \begin{matrix}
   -a & -b & -c \\
   -d & -e & -f \\
   -g & -h & -i \\
\end{matrix} \right]\].
We know that if two matrices are equal, then the elements in the corresponding places are also equal.
So, we get $a=-a$, $d=-b$, $g=-c$, $e=-e$, $h=-f$, $i=-i$.
Let us solve for one element in principal diagonal as the other elements are following same pattern.
So, $a+a=0$.
$\Rightarrow 2a=0$.
$\Rightarrow a=0$.
We have got $a=0$, $d=-b$, $g=-c$, $e=0$, $h=-f$, $i=0$. Using these values, we get the matrix A as $\left[ \begin{matrix}
   0 & b & c \\
   -b & 0 & f \\
   -c & -f & 0 \\
\end{matrix} \right]$.
We know that the determinant of a matrix should not be zero in order to have an inverse. So, let us find the determinant of matrix A.
$\Rightarrow \left| \begin{matrix}
   0 & b & c \\
   -b & 0 & f \\
   -c & -f & 0 \\
\end{matrix} \right|=0\left| \begin{matrix}
   0 & f \\
   -f & 0 \\
\end{matrix} \right|-b\left| \begin{matrix}
   -b & f \\
   -c & 0 \\
\end{matrix} \right|+c\left| \begin{matrix}
   -b & 0 \\
   -c & -f \\
\end{matrix} \right|$.
$\Rightarrow \left| \begin{matrix}
   0 & b & c \\
   -b & 0 & f \\
   -c & -f & 0 \\
\end{matrix} \right|=0-b\left( \left( -b\times 0 \right)-\left( -c\times f \right) \right)+c\left( \left( -b\times -f \right)-\left( -c\times 0 \right) \right)$.
$\Rightarrow \left| \begin{matrix}
   0 & b & c \\
   -b & 0 & f \\
   -c & -f & 0 \\
\end{matrix} \right|=0-b\left( \left( 0 \right)-\left( -cf \right) \right)+c\left( \left( bf \right)-\left( 0 \right) \right)$.
$\Rightarrow \left| \begin{matrix}
   0 & b & c \\
   -b & 0 & f \\
   -c & -f & 0 \\
\end{matrix} \right|=0-b\left( cf \right)+c\left( bf \right)$.
$\Rightarrow \left| \begin{matrix}
   0 & b & c \\
   -b & 0 & f \\
   -c & -f & 0 \\
\end{matrix} \right|=0-b\left( cf \right)+c\left( bf \right)$.
$\Rightarrow \left| \begin{matrix}
   0 & b & c \\
   -b & 0 & f \\
   -c & -f & 0 \\
\end{matrix} \right|=-bcf+bcf$.
$\Rightarrow \left| \begin{matrix}
   0 & b & c \\
   -b & 0 & f \\
   -c & -f & 0 \\
\end{matrix} \right|=0$.
We have got the determinant of skew symmetric as 0.
So, the inverse doesn’t exist.
∴ We have found that the inverse of the skew symmetric matrix of odd order doesn’t exist.
The correct option for the given problem is (d).

Note: Alternatively, we can solve the problem as follows,
We know that for a symmetric matrix ${{A}^{T}}=-A$.
Let us apply determinant on both sides.
$\det \left( {{A}^{T}} \right)=\det \left( -A \right)$.
We know that $\det \left( {{A}^{T}} \right)=\det \left( A \right)$ and $\det \left( kA \right)={{k}^{n}}\det \left( A \right)$, where n is the order of the matrix.
$\det \left( A \right)={{\left( -1 \right)}^{n}}\det \left( A \right)$.
According to the problem, we have the order of the matrix as negative.
So, $\det \left( A \right)=\left( -1 \right)\det \left( A \right)$.
$\det \left( A \right)=-\det \left( A \right)$.
$\det \left( A \right)+\det \left( A \right)=0$.
$2\det \left( A \right)=0$.
$\det \left( A \right)=0$. This makes sure that the determinant of matrix A doesn’t exist.