3 and 4 .Determinants and Matrices
medium

If $A=\left[\begin{array}{lll}3 & \sqrt{3} & 2 \\ 4 & 2 & 0\end{array}\right]$ and $B=\left[\begin{array}{rrr}2 & -1 & 2 \\ 1 & 2 & 4\end{array}\right],$ verify that $(A+B)^{\prime}=A^{\prime}+B^{\prime}$.

Option A
Option B
Option C
Option D

Solution

We have

$A = \left[ {\begin{array}{*{20}{l}}
  3&{\sqrt 3 }&2 \\ 
  4&2&0 
\end{array}} \right],$ $B = \left[ {\begin{array}{*{20}{c}}
  2&{ – 1}&2 \\ 
  1&2&4 
\end{array}} \right]$ $ \Rightarrow A + B = \left[ {\begin{array}{*{20}{c}}
  5&{\sqrt 3  – 1}&4 \\ 
  5&4&4 
\end{array}} \right]$

Therefore      $(\mathrm{A}+\mathrm{B})^{\prime}=\left[\begin{array}{cc}5 & 5 \\ \sqrt{3}-1 & 4 \\ 4 & 4\end{array}\right]$

Now              $A^{\prime}=\left[\begin{array}{cc}3 & 4 \\ \sqrt{3} & 2 \\ 2 & 0\end{array}\right], B^{\prime}=\left[\begin{array}{cc}2 & 1 \\ -1 & 2 \\ 2 & 4\end{array}\right]$

So                $A^{\prime}+B^{\prime}=\left[\begin{array}{rr}5 & 5 \\ \sqrt{3} & -1 \\ 4 & 4\end{array}\right]$

Thus           $(A+B)^{\prime}=A^{\prime}+B^{\prime}$

Standard 12
Mathematics

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