MATH-205 Midterm: Bates MATH 205 032202johnson205examsolns
Document Summary
The fourth column has no pivot, so x4 is a free variable. The corresponding system is x1 + 3x4 = 10, which we solve for the pivot variables: x2 = 2, x3 x4 = 2. Therefore x1 = 10 3x4 x2 = 2 x3 = 2 +x4 x4 = x4 x1 x2 x3 x4. A basis for the row space is the pivot rows of r, or of a. A basis for the column space is the pivot columns of a (but not of r). A basis for the nullspace can be found as in problem 1 or by taking the negative of the upper right corner. 5 of r, putting a 3 3 identity matrix below it, and taking the three columns of that. So the only basis that requires more work is the left nullspace. To get it we transpose the pivot columns of a and eliminate: 1 3 2 1 2 1.
