Discussion
The matrices \(\bar{A} = \begin{bmatrix} a & b \\ -b & a \end{bmatrix}\) and \(B = \begin{bmatrix} c & d \\ -d & c \end{bmatrix}\), for some values of a, b, c, d :
Answer: B
The two matrices commute for all values of a, b, c, d.
Multiplying in both orders:
\[ \begin{aligned} AB &= \begin{bmatrix} ac - bd & ad + bc \\ -(bc + ad) & ac - bd \end{bmatrix} \\ BA &= \begin{bmatrix} ca - db & cb + da \\ -(da + cb) & ca - db \end{bmatrix} \end{aligned} \]
The entries are equal because ordinary numbers commute, whether real or complex. Both matrices have the form \(aI + bJ\) with the same \(J = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}\), and any two polynomials in the same matrix commute; such matrices represent complex numbers.
So the restriction to real numbers is unnecessary, and they do not anticommute, since \(AB + BA = 2AB\) is not zero in general.
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