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Question : 113 of 160
Marks:
+1,
-0
Solution:
Given,
A(α,β)=[| cos‌α | sin‌α | 0 |
| −sin‌α | cos‌α | 0 |
| 0 | 0 | eβ |
] Now,
|A(α,β)|=eβ(cos2α+sin2α)=eβ C11=eβ‌cos‌α,C12=eβ‌sin‌α,C13=0 C21=−eβ‌sin‌α,C22=eβ‌cos‌α,C23=0 C31=0,C32=0,C33=cos2α+sin2α=1 ∴λ‌adj(A(α,β)) =[| eβ‌cos‌α | −eβ‌sin‌α | 0 |
| eβ‌sin‌α | −eβ‌cos‌α | 0 |
| 0 | 0 | 1 |
] ∴[A(α,β)]−1=×eβ [| cos‌α | −sin‌α | 0 |
| sin‌α | cos‌α | 0 |
| 0 | 0 | e−β |
] =[| cos‌α | −sin‌α | 0 |
| sin‌α | cos‌α | 0 |
| 0 | 0 | e−β |
] =A(−α,−β)
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