Chapter 12, Problem 6MCQ

Solution

To find: To perform the operation $\left(AB\right)C$ .

  $\left(AB\right)C=\left[\begin{array}{cc}-47& 201\\ -105& 351\end{array}\right]$

Given information:

Given,matrices are:

  $A=\left[\begin{array}{cc}3& 1\\ 5& 7\end{array}\right],B=\left[\begin{array}{cc}4& 6\\ 1& 0\end{array}\right],C=\left[\begin{array}{cc}-5& 3\\ 1& 9\end{array}\right],D=\left[\begin{array}{cc}1.5& 2\\ 9& -6\end{array}\right]$

Calculation:

Performing matrix multiplication::

  $\left(AB\right)C=\left(\left[\begin{array}{cc}3& 1\\ 5& 7\end{array}\right]\left[\begin{array}{cc}4& 6\\ 1& 0\end{array}\right]\right)\left[\begin{array}{cc}-5& 3\\ 1& 9\end{array}\right]$

  $\left(AB\right)C=\left[\begin{array}{cc}\left(3\right)\left(4\right)+\left(1\right)\left(1\right)& \left(3\right)\left(6\right)+\left(1\right)\left(0\right)\\ \left(5\right)\left(4\right)+\left(7\right)\left(1\right)& \left(5\right)\left(6\right)+\left(7\right)\left(0\right)\end{array}\right]\left[\begin{array}{cc}-5& 3\\ 1& 9\end{array}\right]$

  $\left(AB\right)C=\left[\begin{array}{cc}12+1& 18+0\\ 20+7& 30+0\end{array}\right]\left[\begin{array}{cc}-5& 3\\ 1& 9\end{array}\right]$

  $\left(AB\right)C=\left[\begin{array}{cc}13& 18\\ 27& 30\end{array}\right]\left[\begin{array}{cc}-5& 3\\ 1& 9\end{array}\right]$

  $\left(AB\right)C=\left[\begin{array}{cc}\left(13\right)\left(-5\right)+\left(18\right)\left(1\right)& \left(13\right)\left(3\right)+\left(18\right)\left(9\right)\\ \left(27\right)\left(-5\right)+\left(30\right)\left(1\right)& \left(27\right)\left(3\right)+\left(30\right)\left(9\right)\end{array}\right]$

  $\begin{array}{l}\left(AB\right)C=\left[\begin{array}{cc}-65+18& 39+162\\ -135+30& 81+270\end{array}\right]\\ \left(AB\right)C=\left[\begin{array}{cc}-47& 201\\ -105& 351\end{array}\right]\end{array}$

Conclusion:

  $\left(AB\right)C=\left[\begin{array}{cc}-47& 201\\ -105& 351\end{array}\right]$