Question

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

• A. $4$
• B. $5$
• C. $6$
• D. $7$

Answer 1

The correct option is B $5$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

$\Rightarrow a-1=0$

$\Rightarrow a=1$

$\Rightarrow b=4$

$\therefore a=1,b=4$

Hence $a+b=1+4=5$