Question

If $A=\left[\begin{array}{cc}1& -1\\ 2& -1\end{array}\right]$ and $B=\left[\begin{array}{cc}1& a\\ 4& b\end{array}\right]$ and ${(A+B)}^2=A^2+B^2$. Then, $a$ and $b$ are respectively

• A. $1,-1$
• B. $2,-3$
• C. $-1,1$
• D. $3,-2$

Answer 1

Answer: (A)

$1,-1$

Given, $A=\left[\begin{array}{cc}1& -1\\ 2& -1\end{array}\right]$ and $B=\left[\begin{array}{cc}1& a\\ 4& b\end{array}\right]$

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

$=\left[\begin{array}{cc}2& -1+a\\ 6& -1+b\end{array}\right]$

$\Rightarrow {(A+B)}^2=\left[\begin{array}{cc}2& -1+a\\ 6& -1+b\end{array}\right]\left[\begin{array}{cc}2& -1+a\\ 6& -1+b\end{array}\right]$

$=\left[\begin{array}{cc}4-6+6a& -2+2a+1-b-a+ab\\ 12-6+6b& -6+6a+1+b^2-2b\end{array}\right]$

$=\left[\begin{array}{cc}-2+6a& -1+a-b+ab\\ 6+6b& -5+6a-2b+b^2\end{array}\right]$

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

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

Also, $B^2=\left[\begin{array}{cc}1& a\\ 4& b\end{array}\right]\left[\begin{array}{cc}1& a\\ 4& b\end{array}\right]$

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

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

$\therefore \left[\begin{array}{cc}-2+6a& -1+a-b+ab\\ 6+6b& -5+6a-2b+b^2\end{array}\right]$

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

$\Rightarrow \left[\begin{array}{cc}-2+6a& -1+a-b+ab\\ 6+6b& -5+6a-2b+b^2\end{array}\right]$

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

On comparing both sides, we get

$-2+6a=4a$ and $6+6b=4+4b$

$\Rightarrow a=1$ and $b=-1$