Here's an illustration of our principle that, if \(A \vec{x} = \vec{b} \) is consistent, then the solutions of \(A \vec{x} = \vec{b} \) are one solution + \(\ker A\).

Quiz 3 will be given in problem sessions Monday 9/26 through Sunday 10/2.

9/30 - We've posted some recommended practice problems on Chapter 3 (corresponding to the material on Problem Sets 8 - 11).