Math 240 fall 2011 final exam solutions. You may assume given matrix equations are well de ned (i. e. the matrix sizes are compatible): (24 points) following the earlier command. >> y = [1 2 3 4 5; the matlab command. 0 0 0 0 0 (a) [8 pts] write down a basis for the row space of y . The rst three rows of rref(y) are such a basis: (cid:0)1 2 3 0 0(cid:1) ,(cid:0)0 0 0 1 0(cid:1) ,(cid:0)0 0 0 0 1(cid:1) (b) [8 pts] write down a basis for the column space of y . The pivot columns of y (not rref(y)) are such a basis: 69 (c) [8 pts] write down a basis for the null space of y . Here is a basis of two vectors obtained in the standard way by setting one free variable equal to one and the others zero:
