Question

If `cos theta [(cos theta, sin theta),(-sin theta, cos theta)] + sin theta[(x, -cos theta),(cos theta, x)]` = I₂, find x.

Answer 1

`[(cos theta, sin theta),(-sin theta,  cos theta)] xx [(cos theta, - x),(x, cos theta)]` = I₂ 

`[(cos^2theta + xsintheta, -xsintheta + sinthetacostheta),(-sinthetacostheta + xcostheta, xsintheta + cos^2theta)] = [(1, 0),(0, 1)]`

`[(cos^2theta + xsintheta, 0),(0, xsintheta + cos^2theta)] = [(1, 0),(0, 1)]`

cos²θ + x sinθ = 1

x sinθ = 1 – cos²θ

x sinθ = sin²θ

x = `(sin^2theta)/(sin theta)`

x = sinθ

– x cosθ + sinθ cosθ = 0

sinθ cosθ = x cosθ

⇒ sinθ = x ...(÷ cosθ) 

∴ x = sinθ

The value of x = sinθ