Question

If an augmented matrix [A b] has a pivot position in every row, then the equation Ax=b is consistent. (T/F).

The answer is F because in an augmented matrix having a pivot point in every row could be inconsistent OR consistent. But how could it be inconsistent?

Answer 1

\[\left[\begin{matrix}a_1 & a & b\\ 0 & a_2 & b\\0& 0& b_3\end{matrix}\right]\]
A solution exists if b3 is zero, otherwise the system is inconsistent.
I'm thinking the pivots are a1, a2, and b3. b3 is a pivot, so it can't be zero.
So the system is inconsistent.

Answer 2

\[\left[\begin{matrix}a_1 & a & b\end{matrix}\right]\]
In this case, the system is consistent with 1 free variable.