# Let M and N be matrices , M = [(a, b),(c,d)] and N =[(e, f),(g, h)], and v a vector v = [(x), (y)]. Show that M(Nv) = (MN)v?

Aug 4, 2016

This is called an associative law of multiplication.
See the proof below.

#### Explanation:

(1) $N v = \left[\begin{matrix}e & f \\ g & h\end{matrix}\right] \cdot \left[\begin{matrix}x \\ y\end{matrix}\right] = \left[\begin{matrix}e x + f y \\ g x + h y\end{matrix}\right]$

(2) $M \left(N v\right) = \left[\begin{matrix}a & b \\ c & d\end{matrix}\right] \cdot \left[\begin{matrix}e x + f y \\ g x + h y\end{matrix}\right] = \left[\begin{matrix}a e x + a f y + b g x + b h y \\ c e x + c f y + \mathrm{dg} x + \mathrm{dh} y\end{matrix}\right]$

(3) $M N = \left[\begin{matrix}a & b \\ c & d\end{matrix}\right] \cdot \left[\begin{matrix}e & f \\ g & h\end{matrix}\right] = \left[\begin{matrix}a e + b g & a f + b h \\ c e + \mathrm{dg} & c f + \mathrm{dh}\end{matrix}\right]$

(4) $\left(M N\right) v = \left[\begin{matrix}a e + b g & a f + b h \\ c e + \mathrm{dg} & c f + \mathrm{dh}\end{matrix}\right] \cdot \left[\begin{matrix}x \\ y\end{matrix}\right] = \left[\begin{matrix}a e x + b g x + a f y + b h y \\ c e x + \mathrm{dg} x + c f y + \mathrm{dh} y\end{matrix}\right]$

Notice that the final expression for vector in (2) is the same as the final expression for vector in (4), just the order of summation is changed.

End of proof.