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  • Please Solve RD Sharma Class 12 Chapter 6 Adjoint and Inverese of Matrices Exercise 6.1 Question 16 Suquestion (iii) Maths Textbook Solution.

Please Solve RD Sharma Class 12 Chapter 6 Adjoint and Inverese of Matrices Exercise 6.1 Question 16 Suquestion (iii) Maths Textbook Solution.

Answers (1)

Answer:

Hence proved   [F(\alpha) G(\beta)]^{-1}=G(-\beta) F(-\alpha)

Hint:

Here, we use basic concept of determinant and inverse of matrix

A^{-1}=\frac{1}{|A|} \times \operatorname{Adj}(A)

Given:

F(\alpha)=\left[\begin{array}{ccc} \cos \alpha & -\sin \alpha & 0 \\ \sin \alpha & \cos \alpha & 0 \\ 0 & 0 & 1 \end{array}\right], G(\beta)=\left[\begin{array}{ccc} \cos \beta & 0 & \sin \beta \\ 0 & 1 & 0 \\ -\sin \beta & 0 & \cos \beta \end{array}\right] 

Solution:

We have to show

[F(\alpha) G(\beta)]^{-1}=G(-\beta) F(-\alpha)

We already know that,

{[G(\beta)]^{-1}=G(-\beta)}

{[F(\alpha)]^{-1}=F(-\alpha)}

L H S=[F(\alpha) G(\beta)]^{-1}

=[G(\beta)]^{-1}[F(\alpha)]^{-1}

=G(-\beta) F(-\alpha)

LHS=RHS  

Hence proved

 

Posted by

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