Question

If $A=\left[\begin{array}{cc}1& 2\\ -1& -2\end{array}\right],B=\left[\begin{array}{cc}2& 9\\ -1& b\end{array}\right]$ and if $(A+B{)}^2=A^2+B^2$, find the value of $a$ and $b$.

Answer 1

Given that,

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

$B=\left[\begin{array}{cc}2& 9\\ -1& b\end{array}\right]$

$A+B=\left[\begin{array}{cc}1& 2\\ -1& -2\end{array}\right]$+ $\left[\begin{array}{cc}2& 9\\ -1& b\end{array}\right]$$=\left[\begin{array}{cc}3& 11\\ -2& -2+b\end{array}\right]$

$(A+B{)}^2=\left[\begin{array}{cc}3& 11\\ -2& -2+b\end{array}\right]$$\left[\begin{array}{cc}3& 11\\ -2& -2+b\end{array}\right]$

$=\left[\begin{array}{cc}9-22& 33-22+11b\\ -6+4-2b& -22+(b-2{)}^2\end{array}\right]$

$=\left[\begin{array}{cc}-13& 11+11b\\ -2-2b& -22+(b-2{)}^2\end{array}\right]$.......(i)

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

$B^2=\left[\begin{array}{cc}2& 9\\ -1& b\end{array}\right]$$\left[\begin{array}{cc}2& 9\\ -1& b\end{array}\right]$$=\left[\begin{array}{cc}4-9& 18+9b\\ -2-b& -9+b^2\end{array}\right]$$=\left[\begin{array}{cc}-5& 18+9b\\ -2-b& -9+b^2\end{array}\right]$

$A^2+B^2=\left[\begin{array}{cc}-1& -2\\ 1& 2\end{array}\right]$ + $\left[\begin{array}{cc}-5& 18+9b\\ -2-b& -9+b^2\end{array}\right]$$=\left[\begin{array}{cc}-6& 16+9b\\ -1-b& -7+b^2\end{array}\right]$.........(ii)

$\because (A+B{)}^2=A^2+B^2$

$⟹\left[\begin{array}{cc}-13& 11+11b\\ -2-2b& -22+(b-2{)}^2\end{array}\right]$$=\left[\begin{array}{cc}-6& 16+9b\\ -1-b& -7+b^2\end{array}\right]$

$⟹-2-2b=-1-b$

$⟹1-2=2b-b$

$\therefore b=-1$ (Ans)