Matrices — Class 12 Maths Solution

ncert exercise SA NCERT,Ex.3.3,Q.No.3,Page.88
Question

If $A' = \left[ {\begin{array}{cccccccccccccccccccc}3&4\\{ - 1}&2\\0&1\end{array}} \right]$ and $B = \left[ {\begin{array}{cccccccccccccccccccc}{ - 1}&2&1\\1&2&3\end{array}} \right]$, then verify that

(i) $(A + B)' = A' + B'$

(ii) $(A - B)' = A' - B'$

Step-by-step Solution

.:

$A' = \left[ {\begin{array}{cccccccccccccccccccc}3&4\\{ - 1}&2\\0&1\end{array}} \right] \Rightarrow A = \left[ {\begin{array}{cccccccccccccccccccc}3&{ - 1}&0\\4&2&1\end{array}} \right]$

and $B = \left[ {\begin{array}{cccccccccccccccccccc}{ - 1}&2&1\\1&2&3\end{array}} \right] \Rightarrow B' = \left[ {\begin{array}{cccccccccccccccccccc}{ - 1}&1\\2&2\\1&3\end{array}} \right]$

(i) Now, $A + B = \left[ {\begin{array}{cccccccccccccccccccc}3&{ - 1}&0\\4&2&1\end{array}} \right] + \left[ {\begin{array}{cccccccccccccccccccc}{ - 1}&2&1\\1&2&3\end{array}} \right]$
$= \left[ {\begin{array}{cccccccccccccccccccc}{3 - 1}&{ - 1 + 2}&{0 + 1}\\{4 + 1}&{2 + 2}&{1 + 3}\end{array}} \right] = \left[ {\begin{array}{cccccccccccccccccccc}2&1&1\\5&4&4\end{array}} \right]$

$\Rightarrow$ $(A + B)' = \left[ {\begin{array}{cccccccccccccccccccc}2&5\\1&4\\1&4\end{array}} \right]$

$A' + B' = \left[ {\begin{array}{cccccccccccccccccccc}3&4\\{ - 1}&2\\0&1\end{array}} \right] + \left[ {\begin{array}{cccccccccccccccccccc}{ - 1}&1\\2&2\\1&3\end{array}} \right] = \left[ {\begin{array}{cccccccccccccccccccc}{3 - 1}&{4 + 1}\\{ - 1 + 2}&{2 + 2}\\{0 + 1}&{1 + 3}\end{array}} \right] = \left[ {\begin{array}{cccccccccccccccccccc}2&5\\1&4\\1&4\end{array}} \right]$

$\therefore$ $(A + B)' = A' + B'.$

(ii) $A - B = \left[ {\begin{array}{cccccccccccccccccccc}3&{ - 1}&0\\4&2&1\end{array}} \right] - \left[ {\begin{array}{cccccccccccccccccccc}{ - 1}&2&1\\1&2&3\end{array}} \right]$

$= \left[ {\begin{array}{cccccccccccccccccccc}{3 - ( - 1)}&{ - 1 - 2}&{0 - 1}\\{4 - 1}&{2 - 2}&{1 - 3}\end{array}} \right]$

$= \left[ {\begin{array}{cccccccccccccccccccc}4&{ - 3}&{ - 1}\\3&0&{ - 2}\end{array}} \right]$

$\Rightarrow$ $(A - B)' = \left[ {\begin{array}{cccccccccccccccccccc}4&3\\{ - 3}&0\\{ - 1}&{ - 2}\end{array}} \right],A' - B'$

$= \left[ {\begin{array}{cccccccccccccccccccc}3&4\\{ - 1}&2\\0&1\end{array}} \right] - \left[ {\begin{array}{cccccccccccccccccccc}{ - 1}&1\\2&2\\1&3\end{array}} \right]$

$= \left[ {\begin{array}{cccccccccccccccccccc}{3 + 1}&{4 - 1}\\{ - 1 - 1}&{2 - 2}\\{0 - 1}&{1 - 3}\end{array}} \right]$

$= \left[ {\begin{array}{cccccccccccccccccccc}4&3\\{ - 3}&0\\{ - 1}&{ - 2}\end{array}} \right]$
$\therefore$ $(A - B)' = A' = B'.$

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