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Home  >>  CBSE XII  >>  Math  >>  Model Papers
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Express \( \begin{bmatrix} 2 & -2 & -4 \\ -1 & 3 & 4 \\ 1 & -2 & -3 \end{bmatrix} \) as the sum of a symmetic and skew-symmetric matrix.

$\begin{array}{1 1} B=\begin{bmatrix} 2 & \frac{-3}{2} & \frac{-3}{2} \\ -\frac{3}{2} & 3 & 1 \\ -\frac{3}{2} & 1 & -3 \end{bmatrix} C = \begin{bmatrix} 0 & \frac{-1}{3} & \frac{-5}{2} \\ \frac{1}{3} & 0 & 3 \\ \frac{5}{3} & -3 & 0 \end{bmatrix} \\ B=\begin{bmatrix} 2 & \frac{-3}{2} & \frac{-3}{2} \\ -\frac{3}{2} & -3 & 1 \\ -\frac{-3}{2} & 1 & -3 \end{bmatrix} C = \begin{bmatrix} 0 & \frac{-1}{2} & \frac{-5}{2} \\ \frac{-1}{2} & 0 & 3 \\ \frac{-5}{2} & -3 & 0 \end{bmatrix} \\ B=\begin{bmatrix} 2 & \frac{-3}{2} & \frac{-3}{2} \\ -\frac{3}{2} & 3 & 1 \\ -\frac{3}{2} & 1 & -3 \end{bmatrix} C = \begin{bmatrix} 0 & \frac{-1}{2} & \frac{-5}{2} \\ \frac{1}{2} & 0 & 3 \\ \frac{5}{2} & -3 & 0 \end{bmatrix} \\ B=\begin{bmatrix} 2 & \frac{-3}{2} & \frac{-3}{2} \\ -\frac{3}{2} & 3 & 1 \\ -\frac{3}{2} & 1 & -3 \end{bmatrix} C = \begin{bmatrix} 1 & \frac{-1}{2} & \frac{-5}{2} \\ \frac{1}{2} & 1 & 3 \\ \frac{5}{2} & -3 & 1 \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:
Let A be a square matrix.
A= $\begin{bmatrix} 2 & -2 & -4 \\ -1 & 3 & 4 \\ 1 & -2 & -3 \end{bmatrix}$
Transpose can be obtained by changing the rows & column.
A'= $\begin{bmatrix} 2 & -1 & 1 \\ -2 & 3 & -2 \\ -4 & 4 & -3 \end{bmatrix}$
$\frac{1}{2}(A+A')$--B--Symmetric matrix.
A+A'=$\begin{bmatrix} 2 & -2 & -4 \\ -1 & 3 & 4 \\ 1 & -2 & -3 \end{bmatrix}+\begin{bmatrix} 2 & -1 & 1 \\ -2 & 3 & -2 \\ -4 & 4 & -3 \end{bmatrix}$
$\quad\;=\begin{bmatrix} 4 & -3 & -3 \\ -3 & 6 & 2 \\ -3 & 2 & -6 \end{bmatrix}$
$\frac{1}{2}(A+A')=\begin{bmatrix} 2 & \frac{-3}{2} & \frac{-3}{2} \\ -\frac{3}{2} & 3 & 1 \\ -\frac{3}{2} & 1 & -3 \end{bmatrix}$=B
B-Symmetric matrix.
Step2:
$\frac{1}{2}(A-A')$--C--Skew symmetric matrix.
A-A'=$\begin{bmatrix} 2 & -2 & -4 \\ -1 & 3 & 4 \\ 1 & -2 & -3 \end{bmatrix}+(-1)\begin{bmatrix} 2 & -1 & 1 \\ -2 & 3 & -2 \\ -4 & 4 & -3 \end{bmatrix}$
$\quad=\begin{bmatrix} 2 & -2 & -4 \\ -1 & 3 & 4 \\ 1 & -2 & -3 \end{bmatrix}+\begin{bmatrix} -2 & 1 & -1 \\ 2 & -3 & 2 \\ 4 & -4 & 3 \end{bmatrix}$
$\quad=\begin{bmatrix}0 & -1 & -5\\1 & 0 & 6\\5 &-6 & 0\end{bmatrix}$
$\frac{1}{2}(A-A')=\begin{bmatrix} 0 & \frac{-1}{2} & \frac{-5}{2} \\ \frac{1}{2} & 0 & 3 \\ \frac{5}{2} & -3 & 0 \end{bmatrix}$=C
C-Skew symmetric matrix.
Step3:
$B+C=\begin{bmatrix} 2 & \frac{-3}{2} & \frac{-3}{2} \\ -\frac{3}{2} & 3 & 1 \\ -\frac{3}{2} & 1 & -3 \end{bmatrix}+\begin{bmatrix} 0 & \frac{-1}{2} & \frac{-5}{2} \\ \frac{1}{2} & 0 & 3 \\ \frac{5}{2} & -3 & 0 \end{bmatrix}$
$\quad\;\;\;\;=\begin{bmatrix}2 & -2 & -4\\-1 & 3 & 4\\1 & -2 & -3\end{bmatrix}$=A.
answered Apr 10, 2013 by sharmaaparna1
 

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