Answer
Take any $\mathrm{b}$ in $\mathrm{R}^{m}$.
$AD\mathrm{b}=I_{m}\mathrm{b}=\mathrm{b}$.
$A(D\mathrm{b})=\mathrm{b}$.
Thus, the vector $\mathrm{x}=D\mathrm{b}$ satisfies $A\mathrm{x}=\mathrm{b}$.
This proves that the equation $A\mathrm{x}=\mathrm{b}$ has a solution for each $\mathrm{b} \in \mathrm{R}^{m}$.
By Theorem 4 in Section 1.4, since $A\mathrm{x}=\mathrm{b}$ has a solution for all $\mathrm{b} \in \mathrm{R}^{m}$ (Th4.a.),
then $A$ has a pivot position in each row (Th4.d).
Since each pivot is in a different column, $A$ must have at least as many columns as rows.
Work Step by Step
The answer contains the proof of the statement.
