1. (14 points) Consider the system of equations represented by Ax -b where A and
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1. (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 4-3 0 0 5 10 0 15 2 6 0 8 4 18 1304 2 0 001 2 0 0 00 00 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? Explain. 2, (6 points) Let T : R8 R9 be a linear transformation. Let K be the set of all vectors u in R8 such that T(u)-0 (written mathematically in short-hand as: K = {u E RnIT(u) = 0)) Show that K is a subspace of R3 by showing that this set satisfies the three conditions needed for a subset to be a subspace. 3. (4 points) What can be said about the Nul(A) if A is a 4 x 7 matrix? Could it be the zero 4. (4 points) Suppose H is a four-dimensional subspace of Rn and {U1,U2, U3:span H. Is 5. (4 points) Could a 3 x 6 matrix have dim Col(A)5 and dim Nul(A)1 Justify your subspace? Why or why not? (v1, U2, U3,U4} a basis of H? Explain without stating Theorern 1.5 answerExplanation / Answer
(a). A scrutiny of the RREF of the matrix[A|b] reveals that rank(A) = no. of non-zero rows in the RREF of A = 2.
(b). As per the rank-nullity theorem, dim(nullA) = no. of columns in A – rank(A) = 5-2 = 3.
(c ). As per the RREF of A, only the 1stand the 3rd columns of A are linearly independent and the other 3 columns are linear combinations or scalar multiples of these 2 columns. Hence { ( 1,2,02)T,(-2,-5,5,0)T} is a basis for col(A).
(d). If X = (x,y,z,w,u)T satisfies the equation AX = 0, then as per the RREF of the matrix[A|b] , we have 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,-2,1,0)T+u(-2,0,0,0,1)T. Thus {(-3,1,0,0,0)T,(-4,0,-2,1,0)T,(-2,0,0,0,1)T } is a basis for Null(A).
( e).We know that dim( R4) = 4. Since dim(col(A)) = 2,the basis for col(A) cannot span R4.
(f). We know that dim( R6) = 6. Since dim(Null(A)) = 3,the basis for Null(A) cannot span R6.
(g). Since col(A) does not span R4, the system Ax = b will not always be consistent. Even for the given b, this system is inconsistent as we cannot have 0 = 1 ( please see the 3rd row of the RREF of the matrix[A|b]).
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