Question

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

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

Answer 1

The correct option is A

$1,-1$

Explanation for the correct option.

Step1. Given that,

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

Step2. Finding the value of a & b.

Now,

$$\begin{gathered} A=\left[\begin{array}{cc}1& -1\\ 2& -1\end{array}\right];B=\left[\begin{array}{cc}1& a\\ 4& b\end{array}\right] \\ \Rightarrow (A+B)=\left[\begin{array}{cc}2& a-1\\ 6& b-1\end{array}\right]\left[\begin{array}{cc}2& a-1\\ 6& b-1\end{array}\right] \\ \Rightarrow (A+B{)}^2=\left[\begin{array}{cc}-2+6a& a-1-b+ab\\ 6+6b& 5a-5-2b+b^2\end{array}\right] \\ {A}^2=\left[\begin{array}{cc}1& -1\\ 2& 1\end{array}\right]\left[\begin{array}{cc}1& -1\\ 2& 1\end{array}\right]=\left[\begin{array}{cc}-1& 0\\ 0& -1\end{array}\right] \\ {B}^2=\left[\begin{array}{cc}1& a\\ 4& b\end{array}\right]\left[\begin{array}{cc}1& a\\ 4& b\end{array}\right]=\left[\begin{array}{cc}1+4a& a+9b\\ 4+4b& 4a+b^2\end{array}\right] \\ As{(A+B)}^2=A^2+B^2 \\ \Rightarrow \left[\begin{array}{cc}-2+6a& a-1-b+ab\\ 6+6b& 5a-5-2b+b^2\end{array}\right]=\left[\begin{array}{cc}-1& 0\\ 0& -1\end{array}\right]+\left[\begin{array}{cc}1+4a& a+9b\\ 4+4b& 4a+b^2\end{array}\right] \\ \Rightarrow \left[\begin{array}{cc}-2+6a& a-1-b+ab\\ 6+6b& 5a-5-2b+b^2\end{array}\right]=\left[\begin{array}{cl}4a& a+9b\\ 4+4b& 4a+b^2-1\end{array}\right] \\ \therefore -2+6a=4a \\ \Rightarrow a=1 \\ \Rightarrow 6+6b=4+4b \\ \Rightarrow b=-1 \end{gathered}$$

Therefore the value of (a, b) is $(1,-1)$

Hence, correct option is (A)