Past exam of the mathematics course of the University of Cambridge / 2022 / ia / Paper 1 / 2B / b

Table of contents
- Solution b

Solution

Write

M = (p r q s) .

The first column has unit length for the indefinite quadratic form:

p^{2} - r^{2} = 1.

Because

p = M_{11} > 0

, there is a unique

u \in R

such that

p = cosh u, r = sinh u .

The two columns are orthogonal for the same form and the second has squared length

- 1

:

pq - rs = 0, q^{2} - s^{2} = - 1.

The vectors with these properties are

(q, s) = \pm (sinh u, cosh u) .

Using the hyperbolic cosine and hyperbolic sine, the two possible families are therefore

(cosh u sinh u sinh u cosh u), (cosh u sinh u - sinh u - cosh u), u \in R .

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