Question

Let $a$ denote the element of the $i^{th}$ row and $j^{th}$ column in a $3\times 3$ matrix and let $a_{ij}=-a_{ji}$ for every i and j then this matrix is an -

• A. Orthogonal matrix
• B. singular matrix
• C. matrix whose principal diagonal elements are all zero
• D. skew-symmetric matrix

Answer 1

Answer: (D)

skew-symmetric matrix

If $a_{ij}$ is the element of $i^{th}$ row and $j^{th}$ column and $a_{ij}=-a_{ji}$

then lets assume

$A=\left[\begin{array}{ccc}{a}_{11}& a_{12}& a_{13}\\ a_{21}& a_{22}& a_{23}\\ a_{31}& a_{32}& a_{33}\end{array}\right]$

Where, $a_{ij}=-a_{ji}$

$a_{11}=-a_{11}⟹a_{11}=0$, similarly $a_{22}=a_{33}=0$

and $a_{21}=-a_{12},a_{31}=-a_{13},a_{32}=-a_{23}$

Putting the values

$A=\left[\begin{array}{ccc}0& a_{12}& a_{13}\\ {-a}_{12}& 0& a_{23}\\ {-a}_{13}& {-a}_{23}& 0\end{array}\right]$

$A^T=\left[\begin{array}{ccc}0& {-a}_{12}& {-a}_{13}\\ a_{12}& 0& {-a}_{23}\\ a_{13}& a_{23}& 0\end{array}\right]$
and
$-A=\left[\begin{array}{ccc}0& -a_{12}& -a_{13}\\ a_{12}& 0& -a_{23}\\ a_{13}& a_{23}& 0\end{array}\right]$
thus
$A^T=-A$

$\therefore$ The matrix is a skew-symmetric matrix.