Question

18. Consider the system of equations Ax = b where | A= 1 -1 0 3 1 -2 -1 4 2 0 4 -1 –4 4 2 0 0 3 -2 2 2 and b = BENA 1 For each j, let a; denote the jth column of A.

Linear Algebra Question:

Answer 1

18. e). A is a 5 x 4 matrix. Therefore, a = 4 and b = 5.

The RREF of A is

1	0	2	0
0	1	-1	0
0	0	0	1
0	0	0	0
0	0	0	0

The kernel of T is the set of solutions to the equation T(X) = Ax = 0. In view of the RREF of A, if X = (x,y,z,w)^T, then this equation is equivalent to x+2z = 0 or, x = -2z, y-z = 0 or, y = z and w = 0. Then X = (-2z,z,z,0)^T = z(-2,1,1,0)^T = t( -2,1,1,0)^T where t = z is an arbitrary real number. This means that every solution to the equation T(X) = 0 is a scalar multiple of the vector ( -2,1,1,0)^T. Hence the set {( -2,1,1,0)^T } is a basis of the kernel of T.

The image of T is same as col(A). It may be observed from the RREF of A that the 1^st, 2^nd and 4^th columns of A are linearly independent and the 3^rd column is a linear combination of the first 2 columns.

Hence the set { ( 1,-1,0,3,1)^T,( -2,-1,4,2,0)^T, (0,0,3,-2,2)^T} is a basis of the image of T.

f). The RREF of [A|b] is

1	0	2	0	0
0	1	-1	0	-2
0	0	0	1	-1
0	0	0	0	0
0	0	0	0	0

This implies that b = 0a₁-2a₂-1a₃+0a₄. Hence b is in the span of the columns of A.

g). The vector ( 1,1,1,1,1)^T is not in the column space of A as it cannot be expressed as a linear combination of the columns of A.

h). A is a 4 x 5 matrix . In view of the RREF of A, the system AX = b is a system of 3 linearly independent equations in 4 variables so that there will always be 1 free variable if the equation AX = b is consistent. Hence there cannot be a unique solution for any vector v

for which AX = v.

i). The set {a₁,a₂, a₃, a₄ } is not linearly independent as a₃ = 2a₁-a₂.

j). It may be observed from the RREF of A that the set {a₁,a₂, a₄ } is linearly independent as none of these 3 vectors may be expressed either as a a linear combination of the other 2 vectors or a scalar multiple of another vector.

k ).Since the set {a₁,a₂, a₃, a₄ } is not linearly independent, hence the set {a₁,a₂, a₃, a₄ , b} cannot be linearly independent.