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Home  >>  CBSE XII  >>  Math  >>  Matrices
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If $ A = \begin{bmatrix} 1 & 2 & -3 \\ 5 & 0 & 2 \\ 1 & -1 & 1 \end{bmatrix} , B = \begin{bmatrix} 3 & -1 & 2 \\ 4 & 2 & 5 \\ 2 & 0 & 3 \end{bmatrix} \text{ and } C = \begin{bmatrix} 4 & 1 & 2 \\ 0 & 3 & 2 \\ 1 & -3 & 2 \end{bmatrix} $, then compute $( A + B )$ and $( B - C )$. Also verify that $A + ( B - C ) = ( A + B ) - C. $

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Toolbox:
  • The sum $A+B$ 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.
  • The 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.
$A+B=\begin{bmatrix}1 & 2 & -3\\5 & 0 & 2\\1 & -1 & 1\end{bmatrix}+\begin{bmatrix}3 & -1 & 2\\4 & 2 & 5\\2 & 0 & 3\end{bmatrix}$
$A+B = \begin{bmatrix}1+3 & 2-1 & -3+2\\5+4 & 0+2 & 2+5\\1+2 & -1+0 &1+3\end{bmatrix}$
$A+B = \begin{bmatrix}4 & 1 & -1\\9 & 2 & 7\\3 & -1 &4\end{bmatrix}$($R_1$)
$B-C=\begin{bmatrix}3 & -1 & 2\\4 & 2 & 5\\2 & 0 &3\end{bmatrix}-\begin{bmatrix}4 & 1 & 2\\0 & 3 & 2\\1 & -2 &3\end{bmatrix}$
$B-C= \begin{bmatrix}3+(-4) & -1+(-1) & 2+(-2)\\4+0 & 2+(-3) & 5+(-2)\\2+(-1) & 0+2 &3+(-3)\end{bmatrix}$
$B-C = \begin{bmatrix}-1 & -2 & 0\\4 & -1 & 3\\1 & 2 &0\end{bmatrix}$ ($R_2$)
Verify that $LHS: A+(B-C)$ = $RHS: (A+B) - C$
Given $A=\begin{bmatrix}1 & 2 & -3\\5 & 0 & 2\\1 & 1 &1\end{bmatrix}$ and from $R_2$ above, we have $B-C=\begin{bmatrix}-1 & -2 & 0\\4 & -1 & 3\\1 & 2 &0\end{bmatrix}:$
$LHS: A+(B-C)=\begin{bmatrix}1 & 2 & -3\\5 & 0 & 2\\1 & 1 &1\end{bmatrix}+\begin{bmatrix}-1 & -2 & 0\\4 & -1 & 3\\1 & 2 &0\end{bmatrix}$
$LHS: A+(B-C) = \begin{bmatrix}1-1 & 2-2 &-3+ 0\\5+4 & 0-1 & 2+3\\1+1 &1+ 2 &1+0\end{bmatrix}$
$LHS: A+(B-C) = \begin{bmatrix}0 & 0 & -3\\9 & -1 & 5\\2 & 3 &1\end{bmatrix}$
Given that $C = \begin{bmatrix}-4 & -1 & -2\\0 & -3 & -2\\-1 & 2 &-3\end{bmatrix}$ and from $R_1$ above, $A+B= \begin{bmatrix}4 & 1 & -1\\9 & 2 & 7\\3 & 1 &4\end{bmatrix}: $
$RHS: (A+B) - C = \begin{bmatrix}4 & 1 & -1\\9 & 2 & 7\\3 & 1 &4\end{bmatrix}+\begin{bmatrix}-4 & -1 & -2\\0 & -3 & -2\\-1 & 2 &-3\end{bmatrix}$
$RHS: (A+B)-C = \begin{bmatrix}4-4 & 1-1 & -1-2\\9+0 & 2-3 & 7-2\\3-1 & 1+2 &4-3\end{bmatrix}$
$RHS: (A+B) - C = \begin{bmatrix}0 & 0 & -3\\9 & -1 & 5\\2 & 3 &1\end{bmatrix}$
Therefore, we can see that $LHS: A+(B-C)$ = $RHS: (A+B) - C$
answered Feb 27, 2013 by balaji.thirumalai
 

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