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For the matrix $A = \begin{bmatrix} 1 & 5 \\ 6 & 7 \end{bmatrix}$ , verify that $(ii) (A - A')$ is a skew symmetric matrix.

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Toolbox:
• A square matrix A=[a$_{ij}$] is said to be symmetric if A'=A that is $[a_{ij}]=[a_{ji}]$ for all possible value of i and j.
• A square matrix A=[a$_{ij}$] is said to be skew symmetric if A'=-A that is $[a_{ij}]= -[a_{ji}]$ for all possible value of i and j.
• The diagonal elements of a skew symmetric matrix is always equal to zero
(ii)(A-A') is a skew symmetric matrix:
Given
$A = \begin{bmatrix} 1 & 5 \\ 6 & 7 \end{bmatrix}$
$A' = \begin{bmatrix} 1 & 6 \\ 5 & 7 \end{bmatrix}$
A-A'=A+(-1)(A')
$A-A'=\begin{bmatrix} 1 & 5 \\ 6 & 7 \end{bmatrix}+(-)\begin{bmatrix} 1 & 6 \\ 5 & 7 \end{bmatrix}$
$\;\;\;=\begin{bmatrix} 1 & 5 \\ 6 & 7 \end{bmatrix}+\begin{bmatrix} -1 & -6 \\ -5 & -7 \end{bmatrix}$
$\;\;\;=\begin{bmatrix} 1-1 & 5-6 \\ 6-5 & 7-7 \end{bmatrix}$
$\;\;\;=\begin{bmatrix} 0 & -1 \\ 1 & 0\end{bmatrix}$
Here $a_{21}=1\Rightarrow a_{12}=-1.$
$\Rightarrow a_{21}=-a_{12}$
$a_{11}=a_{22}=0.$
The diagonal elements of the above obtained matrix is equal to zero hence its a skew symmetric matrix
A-A' is a skew symmetric matrix.