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Home  >>  CBSE XII  >>  Math  >>  Matrices
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If $ A' = \begin{bmatrix} 3 & 4 \\ -1 & 2 \\ 0 & 1 \end{bmatrix} \text{ and } B = \begin{bmatrix} -1 & 2 &1 \\ 1 & 2 & 3 \end{bmatrix} \text{ , then verify that } $$ \text{ (i) } (A + B )' = A' + B' \qquad \qquad $

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Toolbox:
  • 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.
  • 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.
(i) Given $A'=\begin{bmatrix} 3 & 4 \\ -1 & 2 \\ 0 & 1 \end{bmatrix}$
$A=\begin{bmatrix} 3 & -1 &0 \\ 4 & 2 & 1 \end{bmatrix}$
$B=\begin{bmatrix} -1 & 2 &1 \\ 1 & 2 & 3 \end{bmatrix}$
$B'=\begin{bmatrix} -1 & 1 \\ 2 & 2\\1 & 3 \end{bmatrix}$
Consider LHS:
$(A+B)=\begin{bmatrix} 3 & -1 &0 \\ 4 & 2 & 1 \end{bmatrix}+\begin{bmatrix} -1 & 2 &1 \\ 1 & 2 & 3 \end{bmatrix}$
$\;\;\;=\begin{bmatrix} 3-1 & -1+2 &0+1 \\ 4+1 &2+ 2 & 1+3 \end{bmatrix}$
$\;\;\;=\begin{bmatrix} 2 & 1 &1 \\ 5 & 4 & 4 \end{bmatrix}$
$(A+B)'=\begin{bmatrix} 2 & 5 \\ 1 & 4\\1 & 4 \end{bmatrix}$
RHS:
$A'=\begin{bmatrix} 3 & 4 \\ -1 & 2 \\ 0 & 1 \end{bmatrix}$
$B'=\begin{bmatrix} -1 & 1 \\ 2 & 2\\1 & 3 \end{bmatrix}$
$A'+B'=\begin{bmatrix} 3 & 4 \\ -1 & 2 \\ 0 & 1 \end{bmatrix}+\begin{bmatrix} -1 & 1 \\ 2 & 2\\1 & 3 \end{bmatrix}$
$\;\;\;=\begin{bmatrix}3 -1 & 4+1 \\-1+ 2 &2+ 2\\0+1 & 1+3 \end{bmatrix}\Rightarrow \begin{bmatrix} 2 & 5 \\ 1 & 4\\1 & 4 \end{bmatrix}$
Hence LHS=RHS.
answered Mar 14, 2013 by sharmaaparna1
 

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