Question

If $A=\left[\begin{array}{cc}1& -1\\ 2& -1\end{array}\right],B=\left[\begin{array}{cc}a& 1\\ b& -1\end{array}\right]$ and $(A+B{)}^2=A^2+B^2$ ,find $a$ and $b$.

Answer 1

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

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

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

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

and $B=\left[\begin{array}{cc}a& 1\\ b& -1\end{array}\right]$

$B^2=B.B=\left[\begin{array}{cc}a& 1\\ b& -1\end{array}\right]\left[\begin{array}{cc}a& 1\\ b& -1\end{array}\right]$

$\therefore B^2=\left[\begin{array}{cc}{a}^2+b& a-1\\ ba-b& b+1\end{array}\right]$

$A+B=\left[\begin{array}{cc}1& -1\\ 2& -1\end{array}\right]+\left[\begin{array}{cc}a& 1\\ b& -1\end{array}\right]$

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

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

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

$=\left[\begin{array}{cc}1+a^2+2a+0& 0\\ 2+2a-b+ab-4& 0+4\end{array}\right]$

$\therefore {(A+B)}^2=\left[\begin{array}{cc}1+a^2+2a& 0\\ 2a-b+ab-2& 4\end{array}\right]$

$A^2+B^2=\left[\begin{array}{cc}-1& 0\\ 0& -1\end{array}\right]+\left[\begin{array}{cc}{a}^2+b& a-1\\ ba-b& b+1\end{array}\right]$

$=\left[\begin{array}{cc}-1+a^2+b& a-1\\ ba-b& -1+b+1\end{array}\right]$

$=\left[\begin{array}{cc}{a}^2+b-1& a-1\\ ba-b& b\end{array}\right]$

Given:${(A+B)}^2=A^2+B^2$

$\Rightarrow \left[\begin{array}{cc}1+a^2+2a& 0\\ 2a-b+ab-2& 4\end{array}\right]=\left[\begin{array}{cc}{a}^2+b-1& a-1\\ ba-b& b\end{array}\right]$

$\Rightarrow a-1=0$ or $a=1$

$\Rightarrow b=4$

$\therefore a=1,b=4$