Home  >>  CBSE XII  >>  Math  >>  Matrices

# Express the following matrices as the sum of a symmetric and a skew symmetric matrix: $(ii) \begin{bmatrix} 6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3 \end{bmatrix}$

This question has multiple parts. Therefore each part has been answered as a separate question on Clay6.com

Toolbox:
• Any square matrix can be expressed as the sum of a symmetric and a skew symmetric matrix by A=$\frac{1}{2}(A+A') +\frac{1}{2}(A-A')$ Where A+A' --> symmetric matrix A-A' --> Skew symmetric matrix
• If A_{i,j} be a matrix m*n matrix , then the matrix obtained by interchanging the rows and column of A is called as transpose of A.
Step1:
(ii)Let B=1/2(A+A') [Symmetric matrix]
Given:$A=\begin{bmatrix} 6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3 \end{bmatrix}$
$A'=\begin{bmatrix} 6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3 \end{bmatrix}$
$\frac{1}{2}(A+A')=\frac{1}{2}\left\{\begin{bmatrix} 6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3 \end{bmatrix}+\begin{bmatrix} 6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3 \end{bmatrix}\right\}$
$A+A'=\begin{bmatrix} 6+6 & -2-2 & 2+2 \\ -2-2 & 3+3 & -1-1 \\ 2+2 & -1-1 & 3+3 \end{bmatrix}$
$\;\;\;=\begin{bmatrix} 12 & -4 & 4 \\ -4 & 6 & -2 \\ 4 & -2 & 6 \end{bmatrix}$
$\frac{1}{2}(A+A')=\frac{1}{2}\begin{bmatrix} 12 & -4 & 4 \\ -4 & 6 & -2 \\ 4 & -2 & 6 \end{bmatrix}$
$\;\;\;=\begin{bmatrix} 6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3 \end{bmatrix}=B$ symmetric matrix2
Step2:
Let C=1/2(A-A') [Skew symmetric matrix]
$\frac{1}{2}(A-A')=\frac{1}{2}\left\{\begin{bmatrix} 6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3 \end{bmatrix}+(-1)\begin{bmatrix} 6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3 \end{bmatrix}\right\}$
$(A-A')=\begin{bmatrix} 6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3 \end{bmatrix}+\begin{bmatrix} -6 & 2 & -2 \\ 2 & -3 & 1 \\ -2 & 1 &- 3 \end{bmatrix}$
$\;\;\;=\begin{bmatrix} 6-6 & -2+2 & 2-2 \\ -2+2 & 3-3 & -1+1 \\ 2-2 & -1+1 & 3-3 \end{bmatrix}\Rightarrow \begin{bmatrix}0 & 0 &0\\0 & 0 & 0\\0 & 0 & 0\end{bmatrix}$
$\;\;\;=\begin{bmatrix} 6-6 & -2+2 & 2-2 \\ -2+2 & 3-3 & -1+1 \\ 2-2 & -1+1 & 3-3 \end{bmatrix}\Rightarrow \begin{bmatrix}0 & 0 &0\\0 & 0 & 0\\0 & 0 & 0\end{bmatrix}$
$\frac{1}{2}(A-A')=\frac{1}{2}\begin{bmatrix}0 & 0 &0\\0 & 0 & 0\\0 & 0 & 0\end{bmatrix}$
Step3:
$B+C=\begin{bmatrix} 6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3 \end{bmatrix} +\begin{bmatrix}0 & 0 &0\\0 & 0 & 0\\0 & 0 & 0\end{bmatrix}$
$\;\;\;=\begin{bmatrix} 6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3 \end{bmatrix} =A.$

edited Mar 21, 2013