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

• A. $a=-1$
• B. $a=1$
• C. $b=2$
• D. $b=-2$

Answer 1

Answer: (A), (D)

$a=-1$
$b=-2$

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

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

$A^2=\left(\begin{array}{cc}1& -1\\ 2& 1\end{array}\right)\left(\begin{array}{cc}1& -1\\ 2& 1\end{array}\right)=\left(\begin{array}{cc}-1& -2\\ 4& -1\end{array}\right)$

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

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

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

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

By equating the respective coefficients or elements.

we get $a=-1;b=-2$