3 and 4 .Determinants and Matrices
easy

Let $A=\left[\begin{array}{ll}2 & 4 \\ 3 & 2\end{array}\right], B=\left[\begin{array}{cc}1 & 3 \\ -2 & 5\end{array}\right], C=\left[\begin{array}{cc}-2 & 5 \\ 3 & 4\end{array}\right]$

Find $AB$

A

$\left[ {\begin{array}{*{20}{l}}
  { - 6}&{26} \\ 
  { - 1}&{19} 
\end{array}} \right]$

B

$\left[ {\begin{array}{*{20}{l}}
  { - 6}&{26} \\ 
  { - 1}&{19} 
\end{array}} \right]$

C

$\left[ {\begin{array}{*{20}{l}}
  { - 6}&{26} \\ 
  { - 1}&{19} 
\end{array}} \right]$

D

$\left[ {\begin{array}{*{20}{l}}
  { - 6}&{26} \\ 
  { - 1}&{19} 
\end{array}} \right]$

Solution

Matrix $A$ has $2$ columns. This number is equal to the number of rows in matrix $B$. Therefore, $AB $ is defined as :

$AB = \left[ {\begin{array}{*{20}{l}}
  2&4 \\ 
  3&2 
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
  1&3 \\ 
  { – 2}&5 
\end{array}} \right]$

$ = \left[ {\begin{array}{*{20}{l}}
  {2(1) + 4( – 2)}&{2(3) + 4(5)} \\ 
  {3(1) + 2( – 2)}&{3(3) + 2(5)} 
\end{array}} \right]$

$ = \left[ {\begin{array}{*{20}{l}}
  {2 – 8}&{6 + 20} \\ 
  {3 – 4}&{9 + 10} 
\end{array}} \right]$

$ = \left[ {\begin{array}{*{20}{l}}
  { – 6}&{26} \\ 
  { – 1}&{19} 
\end{array}} \right]$

Standard 12
Mathematics

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