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Home  >>  CBSE XII  >>  Math  >>  Matrices
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Show that $$ \begin{array}{l} (i) \qquad \begin{bmatrix} 5 & -1 \\ 6 & 7 \end{bmatrix} \begin{bmatrix} 2 & 1 \\ 3 & 4 \end{bmatrix} \: \neq \: \begin{bmatrix} 2 & 1 \\ 3 & 4 \end{bmatrix} \begin{bmatrix} 5 & -1 \\ 6 & 7 \end{bmatrix} \\[0.5em] (ii) \qquad \begin{bmatrix} 1 & 2 & 3 \\ 0 & 1 & 0 \\ 1 & 1 & 0 \end{bmatrix} \begin{bmatrix} -1 & 1 & 0 \\ 0 & -1 & 1 \\ 2 & 3 & 4 \end{bmatrix} \: \neq \: \begin{bmatrix} -1 & 1 & 0 \\ 0 & -1 & 1 \\ 2 & 3 & 4 \end{bmatrix} \begin{bmatrix} 1 & 2 & 3 \\ 0 & 1 & 0 \\ 1 & 1 & 0 \end{bmatrix} \end{array} $$

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(i)Consider LHS
 
$\begin{bmatrix}5 &-1\\6 & 7\end{bmatrix}\begin{bmatrix}2 & 1\\3 & 4\end{bmatrix}.$
 
$\Rightarrow \begin{bmatrix}5\times 2+(-1)\times 3 & 5\times 1+(-1)\times 4\\6\times 2+7\times 3 & 6\times 1+7\times 4\end{bmatrix}$
 
$\Rightarrow \begin{bmatrix}10-3 & 5-4\\12+21 & 6+28\end{bmatrix}\Rightarrow \begin{bmatrix}7 & 1\\33 & 34\end{bmatrix}$
 
Consider RHS:
 
$\begin{bmatrix}2 & 1\\ 3&4 \end{bmatrix}\begin{bmatrix}5 & -1\\ 6&7 \end{bmatrix}$
 
$\begin{bmatrix}2\times 5+1\times 6 &2\times(- 1)+1\times 7\\ 3\times 5+4\times 6&3(-1)+4\times 7 \end{bmatrix}$
 
$\Rightarrow \begin{bmatrix}10+6 & -2+7\\ 15+24&-3+28 \end{bmatrix}$
 
$\Rightarrow \begin{bmatrix}16 & 5\\ 39&25\end{bmatrix}$
 
LHS$\neq$ RHS
 
Hence proved.
 
(ii)LHS:
 
$(ii) \qquad \begin{bmatrix} 1 & 2 & 3 \\ 0 & 1 & 0 \\ 1 & 1 & 0 \end{bmatrix} \begin{bmatrix} -1 & 1 & 0 \\ 0 & -1 & 1 \\ 2 & 3 & 4 \end{bmatrix}$
 
$ \Rightarrow \begin{bmatrix} 1\times(-1)+2\times 0+3\times 2 & 1\times 1+2(-1)+3\times 3 & 1\times 0+2\times 1+3\times 4 \\ 0\times(-1)+1\times 0+0\times 2 & 0\times 1+1\times (-1)+0\times 3 & 0\times 0+1\times 1+0\times 4 \\ 1\times (-1)+1\times 0+0\times 2 & 1\times 1+1\times (-1)+0\times 4 & 1\times 0+1\times 1+0\times 4 \end{bmatrix} $
 
$\Rightarrow \begin{bmatrix}-1+0+6 & 1-2+9 & 0+2+12\\0+0+0 & 0-1+0 & 0+1+0\\-1+0+0 & 1-1+0 & 0+1+0\end{bmatrix}$
 
$\Rightarrow \begin{bmatrix}5 & 8 & 14\\0 & -1 & 1\\-1& 0 &1 \end{bmatrix}$
 
RHS:
 
$ \begin{bmatrix} -1 & 1 & 0 \\ 0 & -1 & 1 \\ 2 & 3 & 4 \end{bmatrix} \begin{bmatrix} 1 & 2 & 3 \\ 0 & 1 & 0 \\ 1 & 1 & 0 \end{bmatrix} $
 
$ \begin{bmatrix} -1\times1+1\times 0+0\times 1 & -1\times 2+1\times 1+0\times 1 & -1\times 3+1\times 0+0\times 0 \\ 0\times(1)+-1\times 0+1\times 1 & 0\times 2+-1\times (1)+1\times 1 & 0\times 3+(-1)\times 0+1\times 0 \\ 2\times (1)+2\times 0+2\times 1 & 3\times 2+3\times (1)+4\times 1 & 2\times 3+3\times 0+4\times 0 \end{bmatrix} $
 
$\Rightarrow \begin{bmatrix}-1+0+0 & -2+1+0 & -3+0+0\\0-1+1 & 0-1+1 & 0+0+0\\2+0+2 & 6+3+4 & 6+0+0\end{bmatrix}$
 
$\Rightarrow \begin{bmatrix}-1 & -1 & -3\\0 & 0 & 0\\4& 13 &6 \end{bmatrix}$
 
LHS$\neq$ RHS
 
Hence proved.

 

answered Feb 13, 2013 by sreemathi.v
 

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