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Home  >>  CBSE XII  >>  Math  >>  Matrices
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Express the following matrices as the sum of a symmetric and a skew symmetric matrix: $(iii) \begin{bmatrix} 3 & 3 & -1 \\ -2 & -2 & 1 \\ -4 & -5 & 2 \end{bmatrix} $

This question has multiple parts. Therefore each part has been answered as a separate question on Clay6.com
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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:
(iii)Let B=1/2(A+A') [Symmetric matrix]
$A=\begin{bmatrix} 3 & 3 & -1 \\ -2 & -2 & 1 \\ -4 & -5 & 2\end{bmatrix}$
$A'=\begin{bmatrix} 3 & -2 & -4 \\ 3& -2 & -5 \\ -1 & 1 & 2\end{bmatrix}$
$\frac{1}{2}(A+A')=\frac{1}{2}\left\{\begin{bmatrix} 3 & 3 & -1 \\ -2 & -2 & 1 \\ -4 & -5 & 2 \end{bmatrix}+\begin{bmatrix} 3 & -2 & -4 \\ 3 & -2 & -5 \\ -1 & 1 & 2 \end{bmatrix}\right\}$
$(A+A')=\begin{bmatrix} 3 & 3 & -1 \\ -2 & -2 & 1 \\ -4 & -5 & 2 \end{bmatrix}+\begin{bmatrix} 3 & -2 & -4 \\ 3 & -2 & -5 \\ -1 & 1 & 2 \end{bmatrix}$
$\;\;\;=\begin{bmatrix} 3+3 & 3-2 &-1 -4 \\ -2+3 & -2-2 & 1-5 \\ -4-1 &-5 +1 & 2+2 \end{bmatrix}$
$\;\;\;=\begin{bmatrix} 6 & 1 & -5 \\ 1 & -4 & -4 \\ -5 & -4 & 4 \end{bmatrix}$
$\;\;\;=1/2\begin{bmatrix} 6 & 1 & -5 \\ 1 & -4 & -4 \\ -5 & -4 & 4 \end{bmatrix}$
$\;\;\;=\begin{bmatrix} 3 & \frac{1}{2} & \frac{-5}{2} \\ \frac{1}{2} & -2 & -2 \\ \frac{-5}{2} & -2 & 2 \end{bmatrix}=B$ symmetric matrix.
Step2:
Let C=1/2(A-A') [Skew symmetric matrix]
$(A-A')=\begin{bmatrix} 3 & 3 & -1 \\ -2 & -2 & 1 \\ -4 & -5 & 2 \end{bmatrix}+(-1)\begin{bmatrix} 3 & -2 & -4 \\ 3 & -2 & -5 \\ -1 & 1 & 2 \end{bmatrix}$
$\;\;\;=\begin{bmatrix} 3 & 3 & -1 \\ -2 & -2 & 1 \\ -4 & -5 & 2 \end{bmatrix}+\begin{bmatrix} -3 & 2 & 4 \\ -3 & 2 & 5 \\ 1 & -1 & -2 \end{bmatrix}$
$\;\;\;=\begin{bmatrix} 3-3 & 3+2 & -1+4 \\ -2-3 & -2+2 & 1+5 \\ -4+1 & -5-1 & 2-2 \end{bmatrix}$
$\;\;\;=\begin{bmatrix} 0 & 5 & 3 \\ -5 & 0 & 6 \\ -3 & -6 & 0 \end{bmatrix}$
$\frac{1}{2}(A-A')=\frac{1}{2}\begin{bmatrix} 0 & 5 & 3 \\ -5 & 0 & 6 \\ -3 & -6 & 0 \end{bmatrix}$
$\;\;\;=\begin{bmatrix}0 & \frac{5}{2} & \frac{3}{2}\\\frac{-5}{2} & 0 & 3\\\frac{-3}{2} & -3 & 0\end{bmatrix}\Rightarrow C\rightarrow $skew symmetric.
Step3:
$B+C=\begin{bmatrix} 3 & \frac{1}{2} & \frac{-5}{2} \\ \frac{1}{2} & -2 & -2 \\ \frac{-5}{2} & -2 & 2 \end{bmatrix}+\begin{bmatrix}0 & \frac{5}{2} & \frac{3}{2}\\\frac{-5}{2} & 0 & 3\\\frac{-3}{2} & -3 & 0\end{bmatrix}$
$\;\;\;=\begin{bmatrix}3 & \frac{6}{2} & \frac{2}{2}\\\frac{-4}{2} & -2& 1\\\frac{-8}{2} & -5 & 2\end{bmatrix}=\begin{bmatrix}3 &3&1\\-2&-2&1\\-4&-5&2\end{bmatrix}=A.$

 

answered Mar 14, 2013 by sharmaaparna1
edited Mar 21, 2013 by sharmaaparna1
 

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