If $P = \left[ {\begin{array}{llllllllllllllllllll}x&0&0\\0&y&0\\0&0&z\end{array}} \right]$ and $Q = \left[ {\begin{array}{llllllllllllllllllll}a&0&0\\0&b&0\\0&0&c\end{array}} \right]$,

then prove that $PQ = \left[ {\begin{array}{cccccccccccccccccccc}{x{\rm{a}}}&0&0\\0&{{\rm{yb}}}&0\\0&0&{{\rm{zc}}}\end{array}} \right] = QP$

$PQ = \left[ {\begin{array}{llllllllllllllllllll}x&0&0\\0&y&0\\0&0&z\end{array}} \right]\left[ {\begin{array}{llllllllllllllllllll}a&0&0\\0&b&0\\0&0&c\end{array}} \right]$

$= \left[ {\begin{array}{cccccccccccccccccccc}{xa}&0&0\\0&{yb}&0\\0&0&{zc}\end{array}} \right]$

…..(i)
and $QP = \left[ {\begin{array}{llllllllllllllllllll}a&0&0\\0&b&0\\0&0&c\end{array}} \right]\left[ {\begin{array}{llllllllllllllllllll}x&0&0\\0&y&0\\0&0&z\end{array}} \right] = \left[ {\begin{array}{cccccccccccccccccccc}{ax}&0&0\\0&{by}&0\\0&0&{zc}\end{array}} \right]$

……(ii)
Thus, we see that, $PQ = QP$

[using Eqs. (i) and (ii)]