If A= [ ccc 1 & 2 & -3 5 & 0 & 2 1 & -1 & 1 ], B= [ ccc 3 & -1 & 2 4 & 2 & 5 2 & 0 & 3 ] and C= [ ccc 4 & 1 & 2 0 & 3 & 2 1 & -2 & 3 ], then compute (A+B) and (B- C). Also, verify that A+(B-C)=(A+B)-C.

A + B = [ * 20 c 1&2& - 3 5&0&2 1& - 1 &1 ] + [ * 20 c 3& - 1 &2 4&2&5 2&0&3 ] = [ * 20 c 1 + 3 & 2 - 1 & - 3 + 2 5 + 4 & 0 + 2 & 2 + 5 1 + 2 & - 1 + 0 & 1 + 3 ] = [ * 20 c 4&1& - 1 9&2&7 3& - 1 &4 ] B - C = [ * 20 c 3& - 1 &2 4&2&5 2&0&3 ] - [ * 20 c 4&1&2 0&3&2 1& - 2 &3 ] = [ * 20 c 3 - 4 & - 1 - 1 & 2 - 2 4 - 0 & 2 - 3 & 5 - 2 2 - 1 & 0 + 2 & 3 - 3 ] = [ * 20 c - 1 & - 2 &0 4& - 1 &3 1&2&0 ] Now, we show; A + (B – C) = (A + B) – C [ * 20 c 1&2& - 3 5&0&2 1& - 1 &1 ] + [ * 20 c - 1 & - 2 &0 4& - 1 &3 1&2&0 ] = [ * 20 c 4&1& - 1 9&2&7 3& - 1 &4 ] - [ * 20 c 4&1&2 0&3&2 1& - 2 &3 ] [ * 20 c 1 - 1 & 2 - 2 & - 3 + 0 5 + 4 & 0 - 1 & 2 + 3 1 + 1 & - 1 + 2 & 1 + 0 ] = [ * 20 c 4 - 4 & 1 - 1 & - 1 - 2 9 - 0 & 2 - 3 & 7 - 2 3 - 1 & - 1 + 2 & 4 - 3 ] [ * 20 c 0&0& - 3 9& - 1 &5 2&1&1 ] = [ * 20 c 0&0& - 3 9& - 1 &5 2&1&1 ] L.H.S. = R.H.S. Hence Proved.

If A= [ ccc 1 & 2 & -3 5 & 0 & 2 1 & -1 & 1 ], B= [ ccc 3 & -1 & …

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Question

If $A=\left[\begin{array}{ccc}1 & 2 & -3 \\ 5 & 0 & 2 \\ 1 & -1 & 1\end{array}\right], B=\left[\begin{array}{ccc}3 & -1 & 2 \\ 4 & 2 & 5 \\ 2 & 0 & 3\end{array}\right]$ and $C=\left[\begin{array}{ccc}4 & 1 & 2 \\ 0 & 3 & 2 \\ 1 & -2 & 3\end{array}\right]$,
then compute $(A+B)$ and $(B-$ C). Also, verify that $A+(B-C)=(A+B)-C$.

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Answer

$A + B = \left[ {\begin{array}{*{20}{c}} 1&2&{ - 3} \\ 5&0&2 \\ 1&{ - 1}&1 \end{array}} \right] + \left[ {\begin{array}{*{20}{c}} 3&{ - 1}&2 \\ 4&2&5 \\ 2&0&3 \end{array}} \right]$ $ = \left[ {\begin{array}{*{20}{c}} {1 + 3}&{2 - 1}&{ - 3 + 2} \\ {5 + 4}&{0 + 2}&{2 + 5} \\ {1 + 2}&{ - 1 + 0}&{1 + 3} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 4&1&{ - 1} \\ 9&2&7 \\ 3&{ - 1}&4 \end{array}} \right]$
$B - C = \left[ {\begin{array}{*{20}{c}} 3&{ - 1}&2 \\ 4&2&5 \\ 2&0&3 \end{array}} \right] - \left[ {\begin{array}{*{20}{c}} 4&1&2 \\ 0&3&2 \\ 1&{ - 2}&3 \end{array}} \right]$ $ = \left[ {\begin{array}{*{20}{c}} {3 - 4}&{ - 1 - 1}&{2 - 2} \\ {4 - 0}&{2 - 3}&{5 - 2} \\ {2 - 1}&{0 + 2}&{3 - 3} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} { - 1}&{ - 2}&0 \\ 4&{ - 1}&3 \\ 1&2&0 \end{array}} \right]$
Now, we show; A + (B – C) = (A + B) – C
$ \Rightarrow \left[ {\begin{array}{*{20}{c}} 1&2&{ - 3} \\ 5&0&2 \\ 1&{ - 1}&1 \end{array}} \right] + \left[ {\begin{array}{*{20}{c}} { - 1}&{ - 2}&0 \\ 4&{ - 1}&3 \\ 1&2&0 \end{array}} \right]$$ = \left[ {\begin{array}{*{20}{c}} 4&1&{ - 1} \\ 9&2&7 \\ 3&{ - 1}&4 \end{array}} \right] - \left[ {\begin{array}{*{20}{c}} 4&1&2 \\ 0&3&2 \\ 1&{ - 2}&3 \end{array}} \right]$
$ \Rightarrow \left[ {\begin{array}{*{20}{c}} {1 - 1}&{2 - 2}&{ - 3 + 0} \\ {5 + 4}&{0 - 1}&{2 + 3} \\ {1 + 1}&{ - 1 + 2}&{1 + 0} \end{array}} \right]$$ = \left[ {\begin{array}{*{20}{c}} {4 - 4}&{1 - 1}&{ - 1 - 2} \\ {9 - 0}&{2 - 3}&{7 - 2} \\ {3 - 1}&{ - 1 + 2}&{4 - 3} \end{array}} \right]$
$ \Rightarrow \left[ {\begin{array}{*{20}{c}} 0&0&{ - 3} \\ 9&{ - 1}&5 \\ 2&1&1 \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 0&0&{ - 3} \\ 9&{ - 1}&5 \\ 2&1&1 \end{array}} \right]$
$\Rightarrow$ L.H.S. = R.H.S. Hence Proved.

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