Chapter 16.3, Problem 23WE

To determine

To find: the product $D^2$ .

Answer to Problem 23WE

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

Explanation of Solution

Given information:

Let $A=\left[\begin{array}{cc}2& -3\\ 2& 1\end{array}\right],B=\left[\begin{array}{cc}\frac{1}{8}& \frac{3}{8}\\ -\frac{1}{4}& \frac{1}{4}\end{array}\right]$ and $D=\left[\begin{array}{ccc}1& 0& -1\\ 0& 2& -1\\ 1& 0& 0\end{array}\right]$

Calculation:

Substitute that,

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

Now,

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

Now, the product is,

  $\begin{array}{c}{D}^2=\left[\begin{array}{ccc}1& 0& -1\\ 0& 2& -1\\ 1& 0& 0\end{array}\right]\left[\begin{array}{ccc}1& 0& -1\\ 0& 2& -1\\ 1& 0& 0\end{array}\right]\\ =\left[\begin{array}{ccc}1\left(1\right)+0\left(0\right)+\left(-1\right)\left(1\right)& 1\left(0\right)+0\left(2\right)+\left(-1\right)0& 1\left(-1\right)+0\left(-1\right)+\left(-1\right)0\\ 0\left(1\right)+2\left(0\right)+\left(-1\right)\left(1\right)& 0\left(0\right)+2\left(2\right)+\left(-1\right)0& 0\left(-1\right)+2\left(-1\right)+\left(-1\right)\left(0\right)\\ 1\left(1\right)+0\left(0\right)+0\left(1\right)& 1\left(0\right)+0\left(2\right)+0\left(0\right)& 1\left(-1\right)+0\left(-1\right)+0\left(0\right)\end{array}\right]\\ =\left[\begin{array}{ccc}0& 0& -1\\ -1& 4& -2\\ 1& 0& -1\end{array}\right]\end{array}$

Therefore, $D^2=\left[\begin{array}{ccc}0& 0& -1\\ -1& 4& -2\\ 1& 0& -1\end{array}\right]$ .