Chapter 16.3, Problem 15WE

To determine

To find: the value of $\left(AB\right)C\text{ and }A\left(BC\right)$ using given matrices.

Answer to Problem 15WE

Matrices $\left(AB\right)C=A\left(BC\right)=\left[\begin{array}{cc}-4& 1\\ 2& -1\end{array}\right]$ .

Explanation of Solution

Given:

Matrices $A$ :

  $\left[\begin{array}{cc}1& 2\\ 0& -1\end{array}\right]$ .

Matrices $B$ :

  $\left[\begin{array}{cc}0& -1\\ 1& 1\end{array}\right]$ .

Matrices $C$ :

  $\left[\begin{array}{cc}-2& 0\\ 0& 1\end{array}\right]$ .

Concept used:

Two multiply the two matrices column of the first matrices must equals to the raw of the second matrices.

  $\left[\begin{array}{cc}{a}_{11}& a_{12}\\ a_{21}& a_{22}\end{array}\right]\times \left[\begin{array}{cc}{b}_{11}& b_{12}\\ b_{21}& b_{22}\end{array}\right]=\left[\begin{array}{cc}{a}_{11}{b}_{11}+a_{12}{b}_{21}& a_{11}{b}_{12}+a_{12}{b}_{22}\\ a_{21}{b}_{11}+a_{22}{b}_{21}& a_{21}{b}_{12}+a_{22}{b}_{22}\end{array}\right]$ .

Calculation:

Matrices $\left(AB\right)C$:

  $\begin{array}{l}\left[\begin{array}{cc}2& 1\\ -1& -1\end{array}\right]\times \left[\begin{array}{cc}-2& 0\\ 0& 1\end{array}\right]=\left[\begin{array}{cc}-4+0& 0+1\\ 2+0& 0-1\end{array}\right].\\ \Rightarrow \left[\begin{array}{cc}-4& 1\\ 2& -1\end{array}\right].\end{array}$

Matrices $A\left(BC\right)$ :

  $\begin{array}{l}\left[\begin{array}{cc}1& 2\\ 0& -1\end{array}\right]\times \left[\begin{array}{cc}0& -1\\ -2& 1\end{array}\right]=\left[\begin{array}{cc}0-4& -1+2\\ 0+2& 0-1\end{array}\right].\\ \Rightarrow \left[\begin{array}{cc}-4& 1\\ 2& -1\end{array}\right].\end{array}$

Hence, Matrices $\left(AB\right)C=A\left(BC\right)=\left[\begin{array}{cc}-4& 1\\ 2& -1\end{array}\right]$ .