Chapter 10, Problem 53RE

To determine

To carry out: the indicated operation

Answer to Problem 53RE

The product of $BC$ is $\left[\begin{array}{cc}-\frac{7}{2}& 2\\ 1& -\frac{9}{2}\end{array}\right]$

Explanation of Solution

Given:

  $BC$

  $A=\left[\begin{array}{ccc}2& 0& -1\end{array}\right]$

  $B=\left[\begin{array}{ccc}1& 2& 4\\ -2& 1& 0\end{array}\right]$

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

  $D=\left[\begin{array}{cc}1& 4\\ 0& -1\\ 2& 0\end{array}\right]$

  $E=\left[\begin{array}{cc}2& -1\\ -\frac{1}{2}& 1\end{array}\right]$

  $F=\left[\begin{array}{ccc}4& 0& 2\\ -1& 1& 0\\ 7& 5& 0\end{array}\right]$

  $G=\left[5\right]$

Consider the matrices,

  $B=\left[\begin{array}{ccc}1& 2& 4\\ -2& 1& 0\end{array}\right]$ and $C=\left[\begin{array}{cc}\frac{1}{2}& 3\\ 2& \frac{3}{2}\\ -2& 1\end{array}\right]$

Find theproduct of $BC$ .

Since $B$ has dimension $2\times 3$ and $C$ has dimension $3\times 2$ , the product $GA$ is defined and has dimension $2\times 2$ .

Therefore we can write

  $\begin{array}{l}BC=\left[\begin{array}{ccc}1& 2& 4\\ -2& 1& 0\end{array}\right]\left[\begin{array}{cc}\frac{1}{2}& 3\\ 2& \frac{3}{2}\\ -2& 1\end{array}\right]\\ =A\\ =\left[\begin{array}{cc}{a}_{11}& a_{12}\\ a_{21}& a_{22}\end{array}\right]\end{array}$

The entry $a_{11}$ is the inner product of the first row of $B$ and the first column of $C$ .

  $\begin{array}{l}{a}_{11}=\left[\begin{array}{ccc}1& 2& 4\\ -2& 1& 0\end{array}\right]\left[\begin{array}{cc}\frac{1}{2}& 3\\ 2& \frac{3}{2}\\ -2& 1\end{array}\right]\\ =1\left(\frac{1}{2}\right)+2\left(2\right)+4\left(-2\right)\\ =\frac{1}{2}+4-8\\ =\frac{1}{2}-4\\ =\frac{1-8}{2}\\ =\frac{-7}{2}\end{array}$

Similarly, we can calculate the remaining entries of the product as follows

Find $a_{12}$ as follows:

  $\begin{array}{l}{a}_{12}=\left[\begin{array}{ccc}1& 2& 4\\ -2& 1& 0\end{array}\right]\left[\begin{array}{cc}\frac{1}{2}& 3\\ 2& \frac{3}{2}\\ -2& 1\end{array}\right]\\ =1\left(3\right)+2\left(\frac{3}{2}\right)+4\left(1\right)\\ =3+3-4\\ =2\end{array}$

Find $a_{21}$ as follows:

  $\begin{array}{l}{a}_{21}=\left[\begin{array}{ccc}1& 2& 4\\ -2& 1& 0\end{array}\right]\left[\begin{array}{cc}\frac{1}{2}& 3\\ 2& \frac{3}{2}\\ -2& 1\end{array}\right]\\ =-2\left(\frac{1}{2}\right)+1\left(2\right)+0\left(1\right)\\ =-1+2\\ =1\end{array}$

Similarly find $a_{22}$ as follows:

  $\begin{array}{l}{a}_{22}=\left[\begin{array}{ccc}1& 2& 4\\ -2& 1& 0\end{array}\right]\left[\begin{array}{cc}\frac{1}{2}& 3\\ 2& \frac{3}{2}\\ -2& 1\end{array}\right]\\ =-2\left(3\right)+1\left(\frac{3}{2}\right)+0\left(1\right)\\ =-6+\frac{3}{2}+0\\ =\frac{-9}{2}\end{array}$

The product of $BC$ is $\left[\begin{array}{cc}-\frac{7}{2}& 2\\ 1& -\frac{9}{2}\end{array}\right]$

Conclusion:

Therefore, the product of $BC$ is $\left[\begin{array}{cc}-\frac{7}{2}& 2\\ 1& -\frac{9}{2}\end{array}\right]$