Question

If θ - ∅ = π/2, then show that $\left[\begin{array}{ccc}cos^{2}\theta&cos\thetasin\theta\\cos\thetasin\theta&sin^{2}\theta\end{array}\right] \left[\begin{array}{ccc}cos^{2}\phi&cos\phi sin\phi\\cos\phi sin\phi&sin^{2}\phi\end{array}\right] = O$

Answer 1

Matrix Multiplication

We are given a question involving Matrix Multiplication.

Matrix Product:

Suppose A is a matrix of order $\sf n\times m$ and B is a matrix of order $\sf m\times p$.

Let C be the product. That is, C = AB.

Then C is a matrix of order $\sf n\times p$ where each element is given by:

$\Large \displaystyle\sf c_{ij}=\sum\limits_{k=1}^{m} a_{ik}b_{kj}$

Now let us look at the question. Consider the LHS:

$\mathbb{LHS}\\\\\\ =\left[\begin{array}{ccc}\cos^{2}\theta&\cos\theta\sin\theta\\\cos\theta\sin\theta&\sin^{2}\theta\end{array}\right] \left[\begin{array}{ccc}\cos^{2}\phi&\cos\phi \sin\phi\\\cos\phi \sin\phi&\sin^{2}\phi\end{array}\right]$

$=\left[\begin{array}{ccc}\cos^2\theta\cos^2\phi+\cos\theta\sin\theta\cos\phi\sin\phi&\cos^2\theta\cos\phi\sin\phi+\cos\theta\sin\theta\sin^2\phi\\\cos\theta\sin\theta\cos^2\phi+\sin^2\theta\cos\phi\sin\phi&\cos\theta\sin\theta\cos\phi\sin\phi+\sin^2\theta\sin^2\phi\end{array}\right]$

$=\left[\begin{array}{ccc}\cos\theta\cos\phi(\cos\theta\cos\phi+\sin\theta\sin\phi)&\cos\theta\sin\phi(\cos\theta\cos\phi+\sin\theta\sin\phi)\\\sin\theta\cos\phi(\cos\theta\cos\phi+\sin\theta\sin\phi)&\sin\theta\sin\phi(\cos\theta\cos\phi+\sin\theta\sin\phi)\end{array}\right]$

$=(\cos\theta\cos\phi+\sin\theta\sin\phi)\left[\begin{array}{ccc}\cos\theta\cos\phi&\cos\theta\sin\phi\\\sin\theta\cos\phi&\sin\theta\sin\phi\end{array}\right]\\\\\\ \textsf{Use the identity: } \cos A\cos B+\sin A\sin B=\cos(A-B) \\\\\\ = \cos (\theta-\phi) \left[\begin{array}{ccc}\cos\theta\cos\phi&\cos\theta\sin\phi\\\sin\theta\cos\phi&\sin\theta\sin\phi\end{array}\right]\\\\\\ \textsf{We are given that } \theta-\phi=\dfrac{\pi}{2}$

$=\cos\dfrac{\pi}{2} \left[\begin{array}{ccc}\cos\theta\cos\phi&\cos\theta\sin\phi\\\sin\theta\cos\phi&\sin\theta\sin\phi\end{array}\right] \\\\\\ = 0 \times \left[\begin{array}{ccc}\cos\theta\cos\phi&\cos\theta\sin\phi\\\sin\theta\cos\phi&\sin\theta\sin\phi\end{array}\right] \\\\\\ = 0 \\\\\\ = \mathbb{RHS}$

$\mathcal{HENCE\ \ PROVED}$