Question

If $A=(\begin{array}{cc}1& -4\\ -2& 3\end{array})$ and $B=(\begin{array}{cc}-1& 6\\ 3& -2\end{array})$, then prove that $(A+B{)}^2\ne A^2+2AB+B^2$

Answer 1

$A=\left[\begin{array}{cc}1& -4\\ -2& 3\end{array}\right],B=\left[\begin{array}{cc}-1& 6\\ 3& -2\end{array}\right]$

To prove : ${(A+B)}^2\ne A^2+2AB+B^2$

L.H.S $\Rightarrow {(A+B)}^2=(A+B)(A+B)$

$\Rightarrow A+B=\left[\begin{array}{cc}1& -4\\ -2& 3\end{array}\right]\left[\begin{array}{cc}-1& 6\\ 3& -2\end{array}\right]$

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

$\therefore (A+B)(A+B)=\left[\begin{array}{cc}0& 2\\ 1& 1\end{array}\right]\left[\begin{array}{cc}0& 2\\ 1& 1\end{array}\right]$

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

$\Rightarrow {(A+B)}^2=\left[\begin{array}{cc}2& 2\\ 1& 3\end{array}\right]$

Now, R.H.S $A^2+2AB+B^2$

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

$=\left[\begin{array}{cc}1+8& -4-12\\ -2-6& 8+9\end{array}\right]$

$=\left[\begin{array}{cc}9& -16\\ -8& 17\end{array}\right]$

$\Rightarrow 2AB=2\left[\begin{array}{cc}1& -4\\ -2& 3\end{array}\right]\left[\begin{array}{cc}-1& 6\\ 3& -2\end{array}\right]$

$=2\left[\begin{array}{cc}-1-12& 6+8\\ 2+9& -12-6\end{array}\right]=2\left[\begin{array}{cc}-13& 14\\ 11& -18\end{array}\right]$

$=\left[\begin{array}{cc}-26& 28\\ 22& -36\end{array}\right]$

$\Rightarrow B^2=B.B=\left[\begin{array}{cc}-1& 6\\ 3& -2\end{array}\right]\left[\begin{array}{cc}-1& 6\\ 3& -2\end{array}\right]=\left[\begin{array}{cc}1+18& -6-12\\ -3-6& 18+4\end{array}\right]$

$\Rightarrow B^2=\left[\begin{array}{cc}19& -18\\ -9& 22\end{array}\right]$

$\therefore A^2+2AB+B^2=\left[\begin{array}{cc}9& -16\\ -8& 17\end{array}\right]+\left[\begin{array}{cc}-26& 28\\ 22& -36\end{array}\right]+\left[\begin{array}{cc}19& -18\\ -9& 22\end{array}\right]$

$=\left[\begin{array}{cc}9-26+19& -16+28-18\\ -8+22-9& 17-36+22\end{array}\right]$

$=\left[\begin{array}{cc}2& -6\\ 5& 3\end{array}\right]$

Hence, proved ${(A+B)}^2\ne A^2+2AB+B^2$