Tamil Nadu Board of Secondary EducationHSC Arts Class 12

# If FF(α)=[cosα0sinα010-sinα0cosα], show that FF[F(α)]-1=F(-α) - Mathematics

Sum

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

#### Solution

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

("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)

Concept: Inverse of a Non-singular Square Matrix
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Chapter 1: Applications of Matrices and Determinants - Exercise 1.1 [Page 15]

#### APPEARS IN

Tamil Nadu Board Samacheer Kalvi Class 12th Mathematics Volume 1 and 2 Answers Guide
Chapter 1 Applications of Matrices and Determinants
Exercise 1.1 | Q 3 | Page 15
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