1. Tardigrade
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  3. Mathematics
  4. The product of matrices A=[ cos 2 θ cos θ sin θ cos θ sin θ sin 2 θ] and sin B=[ cos 2 φ cos φ sin φ cos φ sin φ sin 2 φ] is a null matrix if θ-φ=

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Q. The product of matrices $A=\begin{bmatrix}\cos ^{2} \theta & \cos \theta \sin \theta \\ \cos \theta \sin \theta & \sin ^{2} \theta\end{bmatrix}$ and $\sin B=\begin{bmatrix}\cos ^{2} \phi & \cos \phi \sin \phi \\ \cos \phi \sin \phi & \sin ^{2} \phi\end{bmatrix}$ is a null matrix if $\theta-\phi=$

Matrices

Solution:

$A B=\begin{bmatrix}\cos ^{2} \theta & \cos \theta \sin \theta \\ \cos \theta \sin \theta & \sin ^{2} \theta\end{bmatrix}\begin{bmatrix}\cos ^{2} \phi & \cos \phi \sin \phi \\ \cos \phi \sin \phi & \sin ^{2} \phi\end{bmatrix}$
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$=\begin{bmatrix}\cos \theta \cos \phi(\cos (\theta-\phi)) & \cos \theta \sin \phi(\cos (\theta-\phi)) \\ \sin \theta \cos \phi(\cos (\theta-\phi)) & \sin \theta \sin \phi(\cos (\theta-\phi))\end{bmatrix}$
$=(\cos (\theta-\phi))\begin{bmatrix}\cos \theta \cos \phi & \cos \theta \sin \phi \\ \sin \theta \cos \phi & \sin \theta \sin \phi\end{bmatrix}$
Now $A B=O \Rightarrow \cos (\theta-\phi)=0$
$\Rightarrow \theta-\phi=(2 n+1) \frac{\pi}{2}, n \in Z$