Chapter 9, Problem 18CLT

To determine

To perform: the operation $BA-B^2$ if possible using the matrices below:

  $A=\left[\begin{array}{cc}-1& 3\\ 6& 2\end{array}\right]$ And $B=\left[\begin{array}{cc}-2& 5\\ 0& -1\end{array}\right]$

Answer to Problem 18CLT

  $\left[\begin{array}{cc}28& 19\\ -6& -3\end{array}\right]$

Explanation of Solution

The product of two matrices is defined only if the number of columns of the first matrix is equal the number of rows of the second matrix.

Here, dimension of both matrices A and B is $2\times 2$ .

So, both the products $BA$ and $B^2$ are defined.

Consider,

  $\begin{array}{c}BA=\left[\begin{array}{cc}-2& 5\\ 0& -1\end{array}\right]\left[\begin{array}{cc}-1& 3\\ 6& 2\end{array}\right]\\ =\left[\begin{array}{cc}-2\left(-1\right)+5\left(6\right)& -2\left(3\right)+5\left(2\right)\\ 0\left(-1\right)+\left(-1\right)\left(6\right)& 0\left(3\right)+\left(-1\right)\left(2\right)\end{array}\right]\\ =\left[\begin{array}{cc}2+30& -6+10\\ 0-6& 0-2\end{array}\right]\\ =\left[\begin{array}{cc}32& 4\\ -6& -2\end{array}\right]\end{array}$

  $\begin{array}{c}{B}^2=\left[\begin{array}{cc}-2& 5\\ 0& -1\end{array}\right]\left[\begin{array}{cc}-2& 5\\ 0& -1\end{array}\right]\\ =\left[\begin{array}{cc}-2\left(-2\right)+5\left(0\right)& -2\left(5\right)+5\left(-1\right)\\ 0\left(-2\right)+\left(-1\right)\left(0\right)& 0\left(5\right)+\left(-1\right)\left(-1\right)\end{array}\right]\\ =\left[\begin{array}{cc}4+0& -10-5\\ 0+0& 0+1\end{array}\right]\\ =\left[\begin{array}{cc}4& -15\\ 0& 1\end{array}\right]\end{array}$

So,

  $\begin{array}{c}BA-B^2=\left[\begin{array}{cc}32& 4\\ -6& -2\end{array}\right]-\left[\begin{array}{cc}4& -15\\ 0& 1\end{array}\right]\\ =\left[\begin{array}{cc}32-4& 4-\left(-15\right)\\ -6-0& -2-1\end{array}\right]\\ =\left[\begin{array}{cc}28& 19\\ -6& -3\end{array}\right]\end{array}$