 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2023/paper-2/8/d/solution

Write a matrix in the set as

M (a, b)

M (0, 0)

has no multiplicative inverse matrix, so

S

is not a group. Its determinant is

det M (a, b) = ∣ a ∣^{2} + ∣ b ∣^{2},

which vanishes only at that zero matrix. It is therefore the only element that needs to be removed.

For closure, direct matrix multiplication gives

M (a, b) M (c, d) = M (a c - b d^{*}, a d + b c^{*}) .

The product of nonzero determinants is nonzero, so two nonzero elements never multiply to zero. The identity matrix is

M (1, 0)

M (a, b)^{- 1} = \frac{1}{∣ a ∣ ^{2} + ∣ b ∣ ^{2}} (a^{*} b^{*} - b a) = M (\frac{a ^{*}}{∣ a ∣ ^{2} + ∣ b ∣ ^{2}}, \frac{- b}{∣ a ∣ ^{2} + ∣ b ∣ ^{2}}) .

Thus

S ∖ {0_{2 \times 2}} is a group

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