I. (14 points) Consider the system of equations represented by Ax b where A and
ID: 3137070 • Letter: I
Question
I. (14 points) Consider the system of equations represented by Ax b where A and its row- reduced echelon form are given by 1 3 -2 0 2 0 2 6 -5 -2 43 0 0 5 10 0 15 2 6 0 84 18 1 3 0 4 2 0 0 01 2 0 0 0 0 0 1 (a) (2 points) What is the rank(A)? (b) (2 points) What is the dim Null(A)? (c) (2 points) What is a basis for Col(A)? (d) (2 points) What is a basis for Null(A)? (e) (2 points) Does your basis for Col(A) span R4? Explain (f) (2 points) Does your basis for Null(A) span R6? Explain (g) (2 points) Will the system of equations always have at least one solution for any given b in R4? ExplainExplanation / Answer
1.(a).The matrix A has 2 non-zero rows so that rank (AT) = rank(A) = 2.
(b). dim(null A) = No. of columns in A –rank(A) = 5-2 = 3.
(c ).It is apparent from the RREF of A that only its 12st and the 3rd columns are linearly independent. Hence { (1,2,0,2)T,(-2,-5,5,0)T}is a basis for col(A).
(d). Null(A) is the set of solutions to the equation AX = 0. If X = (x,y,z,w,u)T, then this equation is equivalent to x +3y +4w +2u = 0 or, x = -3y-4w-2u and z+2w = 0 or, z = -2w. Then X = (-3y-4w-2u,y,-2w,w,u)T = y(-3,1,0,0,0)T+w(-4,0,-2,0)T+u(-2,0,0,0,1)T. Hence, {(-3,1,0,0,0)T,(-4,0,-2,0)T,(-2,0,0,0,1)T } is a basis for Null(A).
(e ).Since dim( R4 ) = 4 and dim(col(A)) = 2, hence the basis for col(A) does not span R4.
(f). Since dim( R6 ) = 6 and dim(Null(A)) = 3, hence the basis for Null(A) does not span R6.
(g). Since col(A) does not span R4, hence the system AX = b will not have solution for all b. The given system itself is inconsistent as we cannot have 0 = 1.