3 and 4 .Determinants and Matrices
medium

If $A^{\prime}=\left[\begin{array}{cc}-2 & 3 \\ 1 & 2\end{array}\right]$ and $B=\left[\begin{array}{cc}-1 & 0 \\ 1 & 2\end{array}\right],$ then find $(\mathrm{A}+2 \mathrm{B})^{\prime}$

A

$\left[\begin{array}{cc}-4 & 5 \\ 1 & 6\end{array}\right]$

B

$\left[\begin{array}{cc}-4 & 5 \\ 1 & 6\end{array}\right]$

C

$\left[\begin{array}{cc}-4 & 5 \\ 1 & 6\end{array}\right]$

D

$\left[\begin{array}{cc}-4 & 5 \\ 1 & 6\end{array}\right]$

Solution

We know that $A=\left(A^{\prime}\right)^{\prime}$

$\therefore A=\left[\begin{array}{cc}-2 & 1 \\ 3 & 2\end{array}\right]$

$\therefore \,\,A + 2B = \left[ {\begin{array}{*{20}{c}}
  { – 2}&1 \\ 
  3&2 
\end{array}} \right]$ $ + 2\left[ {\begin{array}{*{20}{c}}
  { – 1}&0 \\ 
  1&2 
\end{array}} \right] = $ $\left[ {\begin{array}{*{20}{c}}
  { – 2}&1 \\ 
  3&2 
\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}
  { – 2}&0 \\ 
  2&4 
\end{array}} \right]$ $ = \left[ {\begin{array}{*{20}{c}}
  { – 4}&1 \\ 
  5&6 
\end{array}} \right]$

$\therefore(A+2 B)^{\prime}=\left[\begin{array}{cc}-4 & 5 \\ 1 & 6\end{array}\right]$

Standard 12
Mathematics

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