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Home  >>  CBSE XII  >>  Math  >>  Model Papers
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Let $ A = \begin{bmatrix} 3 & 2 & 5 \\[0.3em] 4 & 1 & 3 \\[0.3em] 0 & 6 & 7 \end{bmatrix}$ Express $A$ as sum of two matrices such that one is symmetric and the other is skew symmetric.

$\begin{array}{1 1}B=\begin{bmatrix} 3 & 3 & \frac{5}{2} \\ 3 & 1 & \frac{9}{2} \\ \frac{5 }{2} & \frac{9}{2} & 7 \end{bmatrix} C = \begin{bmatrix} 0 & -1 & \frac{5}{2} \\ 1 & 0 & \frac{-3}{2} \\ \frac{-5 }{2} & \frac{3}{2} & 0 \end{bmatrix} \\ B=\begin{bmatrix} 3 & 3 & \frac{5}{2} \\ 3 & 2 & \frac{9}{3} \\ \frac{5 }{2} & \frac{9}{2} & 7 \end{bmatrix} C = \begin{bmatrix} 0 & 1 & \frac{5}{2} \\ -1 & 0 & \frac{-3}{2} \\ \frac{-5 }{2} & \frac{3}{2} & 0 \end{bmatrix} \\ B=\begin{bmatrix} 3 & 3 & \frac{5}{2} \\ 3 & 1 & \frac{9}{2} \\ \frac{5 }{2} & \frac{9}{2} & 7 \end{bmatrix} C = \begin{bmatrix} 1 & -1 & \frac{5}{2} \\ 1 & 1 & \frac{-3}{2} \\ \frac{-5 }{2} & \frac{3}{2} & 0 \end{bmatrix} \\ B=\begin{bmatrix} 1 & 1 & \frac{5}{2} \\ 2 & 1 & \frac{9}{2} \\ \frac{5 }{2} & \frac{9}{2} & 7 \end{bmatrix} C = \begin{bmatrix} 0 & -1 & \frac{5}{2} \\ 1 & 0 & \frac{-3}{2} \\ \frac{-5 }{2} & \frac{3}{2} & 0 \end{bmatrix}\end{array} $

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Toolbox:
  • Any square matrix can be expressed as the sum of a symmetric and a skew symmetric matrix by A=1/2(A+A') +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.
  • The sum / difference $A(+/-)B$ of two $m$-by-$n$ matrices $A$ and $B$ is calculated entrywise: $(A (+/-) B)_{i,j} = A_{i,j} +/- B_{i,j}$ where 1 ≤ i ≤ m and 1 ≤ j ≤ n.
Step1:
Given
A=$ \begin{bmatrix} 3 & 2 & 5 \\ 4 & 1 & 3 \\ 0 & 6 & 7 \end{bmatrix}$
Let A be a square matrix.
We can write A=$\frac{1}{2}(A+A')+\frac{1}{2}(A-A')$
$\frac{1}{2}(A+A')$-Symmetric matrix.
Let $\frac{1}{2}(A+A')$=B.
A'=$ \begin{bmatrix} 3 & 4 & 0 \\ 2 & 1 & 6 \\ 5 & 3 & 7 \end{bmatrix}$
$A+A'= \begin{bmatrix} 3 & 2 & 5 \\ 4 & 1 & 3 \\ 0 & 6 & 7 \end{bmatrix}+ \begin{bmatrix} 3 & 4 & 0 \\ 2 & 1 & 6 \\ 5 & 3 & 7 \end{bmatrix}$
$\qquad\;\;= \begin{bmatrix} 6 & 6 & 5 \\ 6 & 2 & 9 \\ 5 & 9 & 14 \end{bmatrix}$
$\frac{1}{2}(A+A')=\frac{1}{2}\begin{bmatrix} 6 & 6 & 5 \\ 6 & 2 & 9 \\ 5 & 9 & 14 \end{bmatrix}$
$\qquad\qquad\;\;=\begin{bmatrix} 3 & 3 & \frac{5}{2} \\ 3 & 1 & \frac{9}{2} \\ \frac{5 }{2} & \frac{9}{2} & 7 \end{bmatrix}$
Let $\frac{1}{2}(A+A')$=B=Symmetric matrix.
Step2:
$\frac{1}{2}(A-A')$-Skew symmetric matrix.
Let $\frac{1}{2}(A-A')$ = C
$A-A'= \begin{bmatrix} 3 & 2 & 5 \\ 4 & 1 & 3 \\ 0& 6 & 7 \end{bmatrix}+(-1) \begin{bmatrix} 3 & 4 & 0 \\ 2 & 1 & 6 \\ 5 & 3 & 7 \end{bmatrix}$
$\qquad\;\;= \begin{bmatrix} 3 & 2 & 5 \\ 4 & 1 & 3 \\ 0& 6 & 7 \end{bmatrix}+ \begin{bmatrix} -3 & -4 & 0 \\ -2 & -1 & -6 \\ -5 & -3 &- 7 \end{bmatrix}$
$\qquad\;\;= \begin{bmatrix} 0 & -2 & 5 \\ 2 & 0 & -3 \\ - 5 & 3 & 0 \end{bmatrix}$
$\frac{1}{2}(A-A')=\frac{1}{2}\begin{bmatrix} 0 & -2 & 5 \\ 2 & 0 & -3 \\ -5 & 3 & 0 \end{bmatrix}$
$\qquad\qquad\;\;=\begin{bmatrix} 0 & -1 & \frac{5}{2} \\ 1 & 0 & \frac{-3}{2} \\ \frac{-5 }{2} & \frac{3}{2} & 0 \end{bmatrix}$
$\frac{1}{2}(A-A')$=C=Skew symmetric matrix.
Step3:
$B+C=\begin{bmatrix} 3 & 3 & \frac{5}{2} \\ 3 & 1 & \frac{9}{2} \\ \frac{5 }{2} & \frac{9}{2} & 7 \end{bmatrix}+\begin{bmatrix} 0 & -1 & \frac{5}{2} \\ 1 & 0 & \frac{-3}{2} \\ \frac{-5 }{2} & \frac{3}{2} & 0 \end{bmatrix}$
$\Rightarrow \begin{bmatrix}3 & 2& 5\\4 & 1 & 3\\0 & 6& 7\end{bmatrix}=A.$
answered Apr 6, 2013 by sharmaaparna1
 

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