MATH 2360 Midterm: MATH 2360 TTU Exam 2 Solutions
31 Jan 2019
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0: find a basis for the nullspace of a. Note that a and r have the same nullspace. Looking at r we see that the variables x1, x2 and x4 are leading variables and x3, x5 and x6 are free variables, say x3 = , x5 = and x6 = . Thus, the nullspace of a is parametrized by x1 x2 x3 x4 x5 x6. Hence, a basis of the nullspace of a is given by the vectors. 1: find a basis for the rowspace of a. A basis of the rowspace of a is given by the nonzero rows in the rref of. A, i. e. , r. thus, a basis of the rowspace of a is (cid:2) 1 0 1 0 1 3 (cid:3) , (cid:2) 0. 2 1 (cid:3) : find a basis for the columnspace of a. The columns in r that contain the leading entries are 1, 2 and 4.
