Chapter 8.2, Problem 84E

To determine

Toshow: ${\left(A-B\right)}^2\ne A^2-2AB+B^2$ by taking $A=\left[\begin{array}{cc}2& -1\\ 1& 3\end{array}\right]\text{and}B=\left[\begin{array}{cc}-1& 1\\ 0& -2\end{array}\right]$ .

Explanation of Solution

Given information: $A=\left[\begin{array}{cc}2& -1\\ 1& 3\end{array}\right]\text{and}B=\left[\begin{array}{cc}-1& 1\\ 0& -2\end{array}\right]$

Proof:

As we have $A=\left[\begin{array}{cc}2& -1\\ 1& 3\end{array}\right]\text{and}B=\left[\begin{array}{cc}-1& 1\\ 0& -2\end{array}\right]$

To prove ${\left(A-B\right)}^2\ne A^2-2AB+B^2$

Consider left hand side part:

  $\begin{array}{l}{\left(A-B\right)}^2={\left(\left[\begin{array}{cc}2& -1\\ 1& 3\end{array}\right]-\left[\begin{array}{cc}-1& 1\\ 0& -2\end{array}\right]\right)}^2\\ ={\left[\begin{array}{cc}3& -2\\ 1& 5\end{array}\right]}^2\\ =\left[\begin{array}{cc}3& -2\\ 1& 5\end{array}\right]\times \left[\begin{array}{cc}3& -2\\ 1& 5\end{array}\right]\\ =\left[\begin{array}{cc}9-2& -6-10\\ 3+5& -2+25\end{array}\right]\\ {\left(A-B\right)}^2=\left[\begin{array}{cc}7& 16\\ 8& 23\end{array}\right]\mathrm{........}(1)\end{array}$

Now consider right hand side part :

  $\begin{array}{l}{A}^2-2AB+B^2\\ \text{To}\text{finding}{A}^2={\left[\begin{array}{cc}2& -1\\ 1& 3\end{array}\right]}^2\text{and}{B}^2={\left[\begin{array}{cc}-1& 1\\ 0& -2\end{array}\right]}^2\\ A^2=\left[\begin{array}{cc}2& -1\\ 1& 3\end{array}\right]\times \left[\begin{array}{cc}2& -1\\ 1& 3\end{array}\right]and{B}^2=\left[\begin{array}{cc}-1& 1\\ 0& -2\end{array}\right]\times \left[\begin{array}{cc}-1& 1\\ 0& -2\end{array}\right]\end{array}$

  $\begin{array}{l}{A}^2=\left[\begin{array}{cc}4-1& -2-3\\ 2+3& -1+9\end{array}\right]and{B}^2=\left[\begin{array}{cc}1+0& -1-2\\ 0+0& 0+4\end{array}\right]\\ A^2=\left[\begin{array}{cc}3& -5\\ 5& 8\end{array}\right]and{B}^2=\left[\begin{array}{cc}1& -3\\ 0& 4\end{array}\right]\\ AB=\left[\begin{array}{cc}2& -1\\ 1& 3\end{array}\right]\left[\begin{array}{cc}-1& 1\\ 0& -2\end{array}\right]\\ =\left[\begin{array}{cc}-2-0& 2+2\\ -1+0& 1-6\end{array}\right]\\ AB=\left[\begin{array}{cc}-2& 4\\ -1& -5\end{array}\right]\end{array}$

  $\begin{array}{l}\text{Thus},A^2-2AB+B^2=\left[\begin{array}{cc}3& -5\\ 5& 8\end{array}\right]-2\left[\begin{array}{cc}3& 4\\ -1& -5\end{array}\right]+\left[\begin{array}{cc}1& -3\\ 0& 4\end{array}\right]\\ =\left[\begin{array}{cc}4& -8\\ 5& 12\end{array}\right]-\left[\begin{array}{cc}6& 8\\ -2& -10\end{array}\right]\\ \therefore A^2-2AB+B^2=\left[\begin{array}{cc}10& 0\\ 3& 2\end{array}\right]\mathrm{........}(2)\end{array}$

Therefore from equations $(1)$ and $(2)$ , ${\left(A-B\right)}^2\ne A^2-2AB+B^2$

Hence, theorem proved.