Discussion
Given that α and β are complex numbers, which of the following sets of matrices forms a group under matrix multiplication ?
Answer: D
The set that forms a group is \(\begin{pmatrix} \alpha & \beta \\ -\beta^{*} & \alpha^{*} \end{pmatrix}\) with \(|\alpha|^{2} + |\beta|^{2} = 1\).
These are exactly the matrices of SU(2): unitary with determinant \(|\alpha|^{2} + |\beta|^{2} = 1\). The product of two such matrices has the same form, the identity is included (\(\alpha = 1, \beta = 0\)), and each has an inverse, its conjugate transpose, of the same form.
The others fail: the first has determinant 0, so no inverse. In the second, the product of two such matrices does not keep 1s on the diagonal, so the set is not closed. In the third, the determinant \(\alpha\beta^{*} - \alpha^{*}\beta\) is zero when \(\alpha\beta^{*}\) is real, so again there is no inverse.
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