Question

If A = BB’ + CC’, where B’ + C’ are respectively transpose of B and C respectively, $B=\left[\begin{array}{c}cos\theta \\ sin\theta \end{array}\right]$ and $C=\left[\begin{array}{c}sin\theta \\ -cos\theta \end{array}\right],\theta \in R$, then A =

• A. $\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]$
• B. $\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right]$
• C. $\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$
• D. $\left[\begin{array}{cc}-1& 0\\ 0& -1\end{array}\right]$

Answer 1

The correct option is C $\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$

$B=\left[\begin{array}{c}cos\theta \\ sin\theta \end{array}\right]C=\left[\begin{array}{c}sin\theta \\ -cos\theta \end{array}\right]\therefore A=B{B}^{\prime }+C{C}^{\prime }=\left[\begin{array}{c}cos\theta \\ sin\theta \end{array}\right]\left[cos\theta sin\theta \right]+\left[\begin{array}{c}sin\theta \\ -cos\theta \end{array}\right][sin\theta -cos\theta ]$
$=\left[\begin{array}{cc}co{s}^2\theta & cos\theta sin\theta \\ sin\theta cos\theta & si{n}^2\theta \end{array}\right]+=\left[\begin{array}{cc}si{n}^2\theta & -cos\theta sin\theta \\ -sin\theta cos\theta & co{s}^2\theta \end{array}\right]=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$