Past exam of the mathematics course of the University of Cambridge / 2019 / ia / Paper 1 / 8A / c

Table of contents
- i c
  - Solution i
- ii c
  - Solution ii

i

Solution

A

fixes the first coordinate and restricts to one half of the rotation generator on the last two coordinates. Hence

e^{x A} = 100 0 cos (x /2) sin (x /2) 0 - sin (x /2) cos (x /2) .

ii

Solution

B

fixes the second coordinate and satisfies the hyperbolic-generator relation on the first and third coordinates. Therefore

e^{x B} = cosh x 0 sinh x 010 sinh x 0 cosh x .

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