1. Let A = [v1, v2, v3] with the columns v1 = (0, 1, 2)T , v2 = (3, 1, -1)T , v3

Question

1. Let A = [v1, v2, v3] with the columns v1 = (0, 1, 2)T , v2 =
(3, 1, -1)T , v3 = (3, 2, 1)T , x = (x1, x2, x3)T and h = (a, b, c)T .
a) Is the system Ax = h is consistent for all h? If it is consistent, can it have a
unique solution?
b) Are the vectors u = (-1, 2, 5)T and v(-3, 1, 3)T in the span of the columns
of A?

c) Do the columns of A span R3?

2. Let A = [v1v2v3] with the columns v1 = (1, 2, 3)T , v2 = (1, 3, 5)T
and v3 = (1, 1, 1)T be the standard matrix of the linear map T : R3 -> R3.
a) Is T onto?
b) Is T one-to-one?
c) Is v = (-1, 1, 2)T in the range of T?

Explanation / Answer

1)

a) The determinant is zero since it's obvious that the third column is the sum of the first two. So the system is not consistent and cannot have an unique solution.

b) You need to solve :

i) a*(0,1,2)+b*(3,1,-1)+c*(3,2,1) = (-1,2,5) that is 3b+3c=-1, a+b+2c=2, 2a-b+c=5 , set a=0 then b=-8/3 and c=7/3

So (-1,2,5) = -8/3*(3,1,-1) + 7/3(3,2,1) so u is in the span of the columns of A.

ii) a*(0,1,2)+b*(3,1,-1)+c*(3,2,1) = (1,3,5) that is 3b+3c=-3, a+b+2c=1, 2a-b+c=3

So 2*(a+b+2c)-(2a-b+c) = 3b+3c = -1 which is inconsistent with 3b+3c=-3, so there is no solution.

So v is not the span of the columns of A.

c) No, the matrix is not full rank since not invertible (and thus not onto) so it cannot be the case.

2)

The matrix is again clearly not invertible, since we see that column(1) = 1/2 * (Column(2) + Column(3)), so determinant is zero, so it cannot be one-to-one nor onto.

a) It's not onto

b) It's not one-to-one

c) You need to solve A(x,y,z)^T = (-1,1,2) which gives the system x+y+z=-1,2x+3y+z=1,3x+5y+z=2

So (2x+3y+z) - 2(x+y+z) =y-z = 3 and (3x+5y+z)-3(x+y+z)=2(y-z)=5 which is incosistent with y-z=3, so no solution exist.

So v is not in the range of T

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