If  A=sinθ−cosθcosθsinθ and  B=cosθsinθ−sinθcosθ, then  (sinθ)A+(cosθ)B

# If  $A=\left[\begin{array}{cc}\mathrm{sin}\theta & -\mathrm{cos}\theta \\ \mathrm{cos}\theta & \mathrm{sin}\theta \end{array}\right]$ and  $B=\left[\begin{array}{cc}\mathrm{cos}\theta & \mathrm{sin}\theta \\ -\mathrm{sin}\theta & \mathrm{cos}\theta \end{array}\right]$, then  $\left(\mathrm{sin}\theta \right)A+\left(\mathrm{cos}\theta \right)B$

1. A

$\left[\begin{array}{cc}1& 0\\ 1& 0\end{array}\right]$

2. B

$\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$

3. C

$\left[\begin{array}{cc}1& 1\\ 1& 1\end{array}\right]$

4. D

$\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]$

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### Solution:

$\begin{array}{l}\left(\mathrm{sin}\theta \right)\left[\begin{array}{cc}\mathrm{sin}\theta & -\mathrm{cos}\theta \\ \mathrm{cos}\theta & \mathrm{sin}\theta \end{array}\right]+\left(\mathrm{cos}\theta \right)\left[\begin{array}{cc}\mathrm{cos}\theta & \mathrm{sin}\theta \\ -\mathrm{sin}\theta & \mathrm{cos}\theta \end{array}\right]\\ =\left[\begin{array}{cc}{\mathrm{sin}}^{2}\theta +{\mathrm{cos}}^{2}\theta & -\mathrm{sin}\theta \mathrm{cos}\theta +\mathrm{cos}\theta \mathrm{sin}\theta \\ \mathrm{sin}\theta \mathrm{cos}\theta -\mathrm{cos}\theta \mathrm{sin}\theta & {\mathrm{sin}}^{2}\theta +{\mathrm{cos}}^{2}\theta \end{array}\right]=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]\end{array}$

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