Question

If `"F"(alpha) = [(cosalpha, 0, sinalpha),(0, 1, 0),(-sinalpha, 0, cosalpha)]`, show that `["F"(alpha)]^-1 = "F"(- alpha)`

Answer 1

Let A = F(α)

So `["F"(alpha)]^-1 = "A"^-1`

Now A = `[(cosalpha, 0, sinalpha),(0, 1, 0),(-sinalpha, 0, cosalpha)]`

|A| = `[(cosalpha, 0, sinalpha),(0, 1, 0),(-sinalpha, 0, cosalpha)]`

Expanding the determinant - along R₂ we get

`- (0) + 1[cos^2alpha + sin^-2alpha] - (0)` = 1 ≠ 0

So `"A"^-1` exists

Now `"A"^-1 = 1/|"A"| ("adj A") = 1/1 ("adj A")`= adj A

To FInd adj A: adj A = (A_ij)^T

`("A"_"ij")^"T" = [(+|(1, 0),(0, cos alpha)|, -|(0, 0),(-sinalpha, cosalpha)|, +|(0, 1),(-sinalpha, 0)|),(-|(0, sinalpha)|, +|(cosalpha, sinalpha), (- sinalpha, cosalpha)|, -|(cosalpha, 0),(- sinalpha, 0)|),(+|(0, sinalpha),(1, 0)|, -|(cosalpha, sinalpha),(0, 0)|, +|(cosalpha, 0),(0, 1)|)]`

= `[(+(cosalpha), -(0), +(sinalpha)),(-(0), +(1), -(0)),(+(-sinalpha), -(0), +(cosalpha))]`

= `[(cosalpha, 0, sinalpha),(0, 1, 0),(-sinalpha, 0, cosalpha)]`

∴ adj A = `("A"^-1)`

= `("A"_"ij")^"T"`

= `[(cosalpha, 0, -sinalpha),(0, 1, 0),(sinalpha, 0, cosalpha)]`

(i.e) `"A"^-1 = ["F"(alpha)]^-1 = [(cosalpha, 0, -sinalpha),(0, 1, 0),(sinalpha, 0, cosalpha)]`

Given `"F"(alpha) = [(cosalpha, 0, sinalpha),(0, 1, 0),(-sinalpha, 0, cosalpha)]`

So `"F"(- alpha) = [(cos(- alpha), 0, sin(- alpha)),(0, 1, 0),(-sin(- alpha), 0, cos(- alpha))]`

= `[(cosalpha, 0, sinalpha),(0, 1, 0),(sinalpha, 0, cosalpha)]`

∴  `cos(- theta) = cos theta` and `sin(- theta) = - sin theta)`

Here (1) = (2)

⇒ `["F"(alpha)]^-1 = "F"(- alpha)`