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#?

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.

To show that $M(Nv) = (MN)v$, we first need to compute each of these expressions.

Starting with $M(Nv)$:

$M(Nv) = M\left(N\begin{pmatrix}x\\y\end{pmatrix}\right)$ $= M\left(\begin{pmatrix}e & f\\g & h\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}\right)$ $= M\left(\begin{pmatrix}ex + fy\\gx + hy\end{pmatrix}\right)$ $= \begin{pmatrix}a & b\\c & d\end{pmatrix}\begin{pmatrix}ex + fy\\gx + hy\end{pmatrix}$ $= \begin{pmatrix}a(ex + fy) + b(gx + hy)\\c(ex + fy) + d(gx + hy)\end{pmatrix}$ $= \begin{pmatrix}aex + afy + bgx + bhy\\cex + cfy + dgx + dhy\end{pmatrix}$

Now, moving on to $(MN)v$:

$(MN)v = \left(\begin{pmatrix}a & b\\c & d\end{pmatrix}\begin{pmatrix}e & f\\g & h\end{pmatrix}\right)\begin{pmatrix}x\\y\end{pmatrix}$ $= \begin{pmatrix}(ae + bg) & (af + bh)\$ce + dg) & (cf + dh)\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}$ $= \begin{pmatrix}(ae + bg)x + (af + bh)y\$ce + dg)x + (cf + dh)y\end{pmatrix}$

Comparing the two results, we see that $M(Nv) = (MN)v$. Hence, the equality holds.