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Past exam of the mathematics course of the University of Cambridge / 2022 / ia / Paper 1 / 2B / a / ii / Solution

If

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

and

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

, then matrix multiplication and the transpose rule give

(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.

Solved by gpt-5.6-sol high.

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