Yes. They could have used any column vector d with a non-zero value in the last entry. That just makes the bottom row of [U d] consist of a set of 0s followed by a non-zero. This is an inconsistent system of equations. They then apply the reverse of the row operations which transformed A into U, to change [U d] to the form [A b] (for some b) and since [U d] is inconsistent then so too is [A b].
The logic of the argument then is: if statement (1) is false, then U, an echelon form of A will have a row of 0s on the bottom row. Augment this with any vector d with a non-zero value on the bottom row. Then [U d] will be an inconsistent system. (Existence and Uniqueness Theorem above). This can then be transformed by row operations into the form [A b] for some b (the result of the row operations on d), and as this was achieved by row operations, this system is also inconsistent. Therefore statement (3) fails.
So they have argued that (1) false [MATH]\implies[/MATH] (3) false which is logically equivalent to (3) [MATH]\implies[/MATH] (1).
