Chapter 16.4, Problem 7ST

To determine

To find: $A^2$

Answer to Problem 7ST

  $A^2=\left[\begin{array}{cc}1& 16\\ 0& 9\end{array}\right]$

Explanation of Solution

Given information:

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

Calculation:

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

  $A^2=\left[\begin{array}{cc}1& 4\\ 0& 3\end{array}\right]\left[\begin{array}{cc}-1& 2\\ 1& 0\end{array}\right]$

Now, simplify

  $\begin{array}{c}{A}^2=\left[\begin{array}{cc}1& 4\\ 0& 3\end{array}\right]\left[\begin{array}{cc}-1& 2\\ 1& 0\end{array}\right]\\ =\left[\begin{array}{cc}1\left(1\right)+4\left(0\right)& 1\left(4\right)+4\left(3\right)\\ 0\left(1\right)+3\left(0\right)& 0\left(4\right)+3\left(3\right)\end{array}\right]\\ =\left[\begin{array}{cc}1& 16\\ 0& 9\end{array}\right]\end{array}$

Therefore, $A^2=\left[\begin{array}{cc}1& 16\\ 0& 9\end{array}\right]$ .