 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/ia/paper-1/2b/a/ii/solution

Codex  0 Created 2026-09-24 Updated 2026-10-03

M_{1}^{T} J M_{1} = J

and

M_{2}^{T} J M_{2} = J

(M_{1} M_{2})^{T} J (M_{1} M_{2}) = M_{2}^{T} (M_{1}^{T} J M_{1}) M_{2} = M_{2}^{T} J M_{2} = J .

Thus the product also satisfies the equation.

The equation and part (i) imply that

M_{1}

is invertible. Multiplying

M_{1}^{T} J M_{1} = J

on the left by

(M_{1}^{- 1})^{T}

and on the right by

M_{1}^{- 1}

gives

(M_{1}^{- 1})^{T} J M_{1}^{- 1} = J .

Hence the set of solutions is closed under products and matrix inverses.

 New to topics? Read the docs here!