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Home  >>  CBSE XII  >>  Math  >>  Matrices
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Find $ \frac{1}{2}(A + A')$ and $\frac{1}{2}(A - A')$ , when $ A = \begin{bmatrix} 0 & a & b \\ -a & 0 & c \\ -b & -c & 0 \end{bmatrix} $

$\begin{array}{1 1} I and \begin{bmatrix}0 & a & b \\ -a & 0 & c \\ 0 &-c& 0 \end{bmatrix} \\ \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} and \begin{bmatrix}0 & a & b \\ -a & 0 & c \\ 0 &-c& 0 \end{bmatrix} \\ \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 1 & 1 & 1 \end{bmatrix} and \begin{bmatrix}0 & a & b \\ -a & 0 & c \\ 0 &-c& 0 \end{bmatrix} \\ \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} and \begin{bmatrix}0 & a & a \\ -a & 0 & c \\ 0 &-c& 0 \end{bmatrix}\end{array} $

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Toolbox:
  • 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 $1 \leq i \leq m$ and $1 \leq j \leq n.$
Given $A = \begin{bmatrix} 0 & a & b \\ -a & 0 & c \\ -b & -c & 0 \end{bmatrix}, A '= \begin{bmatrix} 0 & -a & -b \\ a & 0 & -c \\ b & c & 0 \end{bmatrix} $
$A+A'=\begin{bmatrix} 0 & a & b \\ -a & 0 & c \\ -b & -c & 0 \end{bmatrix}+\begin{bmatrix} 0 & -a & -b \\ a & 0 & -c \\ b & c & 0 \end{bmatrix} $
$\;\;\;=\begin{bmatrix} 0+0 & a-a & b-b \\ -a+a & 0+0 & c-c \\ -b+b & -c+c & 0+0 \end{bmatrix} $
$\;\;\;=\begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} $
$\Large \frac{1}{2}$$(A+A')=\Large \frac{1}{2}$$\begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} =\begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} $
Given $A = \begin{bmatrix} 0 & a & b \\ -a & 0 & c \\ -b & -c & 0 \end{bmatrix}, A '= \begin{bmatrix} 0 & -a & -b \\ a & 0 & -c \\ b & c & 0 \end{bmatrix} $
$A - A' = (A+(-1)A')=\begin{bmatrix} 0 & a & b \\ -a & 0 & c \\ -b & -c & 0 \end{bmatrix}+(-1)\begin{bmatrix} 0 & -a & -b \\ a & 0 & -c \\ b & c & 0 \end{bmatrix} $
$\;\;\;=\begin{bmatrix} 0 & a & b \\ -a & 0 & c \\ -b & -c & 0 \end{bmatrix}+\begin{bmatrix} 0 & a & b \\ -a & 0 & c \\ -b & -c & 0 \end{bmatrix} $
$\;\;\;= \begin{bmatrix} 0+0 & a+a & b+b \\ -a-a & 0+0 & c+c \\ b-b &- c -c& 0+0 \end{bmatrix} $
$\;\;\;= \begin{bmatrix}0 & 2a & 2b \\ -2a & 0 & 2c \\ 0 &-2c& 0 \end{bmatrix} $
$\Large \frac{1}{2}$$(A-A')=\Large \frac{1}{2}$$ \begin{bmatrix}0 & 2a & 2b \\ -2a & 0 & 2c \\ 0 &-2c& 0 \end{bmatrix} $
$\;\;\;= \begin{bmatrix}0 & a & b \\ -a & 0 & c \\ 0 &-c& 0 \end{bmatrix} $
answered Mar 3, 2013 by sharmaaparna1
edited Mar 13, 2013 by balaji.thirumalai
 

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