 Solution

The inner product of columns one and three isThe leading terms cancel, establishing compatibility with unitarity at cubic order. Direct multiplication of the printed truncation leaves $-A{\lambda }^5(\rho -i\eta )/2$, so it must not be treated as exactly unitary beyond its stated accuracy.

For the exact matrix, conjugate the closure relation and defineDivide by $-b$ and let $z=-a/b=\bar{\rho }+i\bar{\eta }$. Then $a/(-b)=z$ and $c/(-b)=1-z$, giving a CKM unitarity triangle with verticesAt leading Wolfenstein order these are $(0,0),(1,0),(\rho ,\eta )$. The barred coordinates define the exact triangle; they differ from the unbarred parameters beyond leading order.

For positive $\bar{\eta }$, define $\gamma$ at the origin, $\beta$ at $(1,0)$, and $\alpha$ at the apex. Comparing the directions of the incident complex sides gives the rephasing-invariant relationsTo leading order $\gamma =\arg (\rho +i\eta )$ and $\beta =atan2(\eta ,1-\rho )$; the remaining angle follows by subtraction. Each ratio cancels the unphysical common phase of the corresponding column-orthogonality terms.