Suppose that . In the following questions you may use the fact that the matrix B

Question

Suppose that

.

In the following questions you may use the fact that the matrix B is row-equivalent to A, where

.(a) Find:

(b) Find a basis for the nullspace of A. Enter each vector in the form [x1, x2, ...]; and enter your answer as a comma-separated list: for example if your answer is the set

then type "[1, 2, 3, 4, 5], [10, 20, 30, 40, 50]" WITHOUT the quote marks. { }

(c) Find a basis for the column space of A. As in part (b), enter each vector in the form [x1, x2, ...]; and enter your answer as a comma-separated list.   { }
(d) Let the columns of A be denoted by a1, a2, a3, a4, and a5. Which of the following sets is / are linearly independent? Select all that apply.
Note: three submissions allowed for this part.

a?{a1, a2, a3}

b?{a1, a2, a4}

c?{a1, a3, a5}

d:None of the above

A = -1 -2 3 4 21 -1 -1 1 2 10 -2 -2 2 2 14 -2 -2 2 3 17

.

Explanation / Answer

(a). Since matrix A is row-equivalent to B, hence rank(A) = rank(B) = no. of non-zero rows in B = 3. Further, as the dimension (rank-nullity) theorem, dim(null(A)) = nullity of A = no. of columns in A – rank(A) = 5-3 = 2.

(b). The null space of A is the set of solutions to the equation AX = 0. If X = (x,y,z,w,u)T, then in view of the fact that B is the RREF of A, this equation is equivalent to x +z+u = 0 or, x = -z-u, y-2z-5u = 0 or, y = 2z+5u, and w+3u = 0 or, w = -3u. Then, X = (-z-u, 2z+5u,z,-3u,u)T= z(-1,2,1,0,0)T+u(-1,5,0,-3,1)T. Thus, (-1,2,1,0,0)T,(-1,5,0,-3,1)T form a basis for null space of A.

( c). Since B is the RREF of A, it is apparent that only the 1st,2nd and 4th columns of A are linearly independent and the other columns of A are linear combinations of these 3 columns).Hence, (-1,-1,-2,-2)T, (-2,-1, -2,-2)T, (4,2,2,3)T form a basis for col(A).

(d). In view of part(c ) above, the option i.e. {a1,a2,a4} is the correct answer.

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