The Total solution for NCERT class 6-12
Answer - 11 : -
Given
Answer - 12 : -
Answer - 13 : -
Answer - 14 : -
Answer - 15 : -
Here, (AB) – 1 = B – 1 A –1
|A| = – 5 + 4 = – 1
Cofactors of A are
C11 = – 1
C21 = 8
C31 = – 12
C12 = 0
C22 = 1
C32 = – 2
C13 = 1
C23 = – 10
C33 = 15
(i) [F (α)]-1 =F (-α)
(ii) [G(β)]-1 = G (-β)
(iii) [F(α) G (β)]-1 = G (-β) F (-α)
Answer - 16 : -
(i) Given
F (α) =
|F (α)| = cos2 α + sin2 α= 1≠ 0
C11 = cos α
C21 = sin α
C31 = 0
C12 = – sin α
C22 = cos α
C32 = 0
C13 = 0
C23 = 0
C33 = 1
(ii) We have
|G (β)| = cos2 β + sin2 β =1
C11 = cos β
C21 = 0
C31 = -sin β
C13 = sin β
C33 = cos β
(iii) Now we have to show that
[F (α) G(β)] – 1 = G (– β) F (– α)
We have already know that
[G(β)] – 1 = G (– β)
[F(α)] – 1 = F (– α)
And LHS = [F (α) G (β)] – 1
= [G (β)] – 1 [F (α)] –1
= G (– β) F (– α)
Hence = RHS
Answer - 17 : -
Consider,
Answer - 18 : -
Answer - 19 : - Given
Answer - 20 : -