Q.Show that AB=BA where, $A=\begin{bmatrix}\cos\theta & \sin\theta\\ \sin\theta & \cos\theta\end{bmatrix}$, $B=\begin{bmatrix}\cos\phi & -\sin\phi\\ \sin\phi & \cos\phi\end{bmatrix}$

Given, exactly as printed, $A=\begin{bmatrix}\cos\theta&\sin\theta\\\sin\theta&\cos\theta\end{bmatrix}$, $B=\begin{bmatrix}\cos\phi&-\sin\phi\\\sin\phi&\cos\phi\end{bmatrix}$.

Compute AB (using $\cos\alpha\cos\beta\pm\sin\alpha\sin\beta=\cos(\alpha\mp\beta)$ and $\sin\alpha\cos\beta\pm\cos\alpha\sin\beta=\sin(\alpha\pm\beta)$):

$(AB)_{11}=\cos\theta\cos\phi+\sin\theta\sin\phi=\cos(\theta-\phi)$

$(AB)_{12}=\cos\theta(-\sin\phi)+\sin\theta\cos\phi=\sin(\theta-\phi)$

$(AB)_{21}=\sin\theta\cos\phi+\cos\theta\sin\phi=\sin(\theta+\phi)$

$(AB)_{22}=\sin\theta(-\sin\phi)+\cos\theta\cos\phi=\cos(\theta+\phi)$

$AB=\begin{bmatrix}\cos(\theta-\phi)&\sin(\theta-\phi)\\\sin(\theta+\phi)&\cos(\theta+\phi)\end{bmatrix}$

Compute BA:

$(BA)_{11}=\cos\phi\cos\theta+(-\sin\phi)\sin\theta=\cos(\theta+\phi)$

$(BA)_{12}=\cos\phi\sin\theta+(-\sin\phi)\cos\theta=\sin(\theta-\phi)$

$(BA)_{21}=\sin\phi\cos\theta+\cos\phi\sin\theta=\sin(\theta+\phi)$

$(BA)_{22}=\sin\phi\sin\theta+\cos\phi\cos\theta=\cos(\theta-\phi)$

$BA=\begin{bmatrix}\cos(\theta+\phi)&\sin(\theta-\phi)\\\sin(\theta+\phi)&\cos(\theta-\phi)\end{bmatrix}$

The off-diagonal entries of AB and BA match, but the two diagonal entries are swapped: $\cos(\theta-\phi)$ and $\cos(\theta+\phi)$ trade places. So AB=BA holds only when $\cos(\theta-\phi)=\cos(\theta+\phi)$, i.e. when $\sin\theta\sin\phi=0$ — not true in general. As literally printed, the exercise's claim 'Show that AB=BA' does not hold for arbitrary θ, φ.

This is because A as printed (with $+\sin\theta$ in the (1,2) slot) is a symmetric matrix, not the standard 2×2 rotation matrix. The standard rotation matrix is $R(\theta)=\begin{bmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{bmatrix}$ — B is already printed in exactly this form — and it is a standard fact that $R(\theta)R(\phi)=R(\phi)R(\theta)=R(\theta+\phi)$ for ALL θ, φ. This is almost certainly the identity intended (A is very likely missing a minus sign due to a printing/OCR slip). Re-deriving with A read as $R(\theta)$: …

Multiply the two 2×2 matrices both ways using compound-angle trig identities, then compare — and flag that A as printed l …

• Asserting AB=BA 'because it's a standard rotation-matrix identity' without actually multiplying out the matrices as printed.
• Missing that the printed A has $+\sin\theta$ in the top-right entry (not the usual $-\sin\theta$), which is what breaks the i …