A = [1 4 2 2], [0 1 2 1], [-1 -7 -8 -5]
T is the linear transformation R^4->R^3 defined by T(x) = Ax
W is set of vectors x such that T(x) = 0
Looking for vectors u and v that span W
I came up with three equations and solved for x1 and x2:
Three equations:
x1+4x2+2x3=2x4
x2+2x3 = x4
-x1-7x4-8x3 = -5x4
I determined that x1 = 6x3-2x4 and x2 = -2x3 +x4
Thus giving me u and v: [6 -2 1 0] and [-2 1 0 1]
Obviously did more work than I detailed but this is the idea. However, I am wrong???? Help!!
T is the linear transformation R^4->R^3 defined by T(x) = Ax
W is set of vectors x such that T(x) = 0
Looking for vectors u and v that span W
I came up with three equations and solved for x1 and x2:
Three equations:
x1+4x2+2x3=2x4
x2+2x3 = x4
-x1-7x4-8x3 = -5x4
I determined that x1 = 6x3-2x4 and x2 = -2x3 +x4
Thus giving me u and v: [6 -2 1 0] and [-2 1 0 1]
Obviously did more work than I detailed but this is the idea. However, I am wrong???? Help!!
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Well, you introduced some sign error as soon as you set up the equations. You wrote:
x1+4x2+2x3=2x4
x2+2x3 = x4
-x1-7x4-8x3 = -5x4
The signs on the right are all incorrect. The first equation should be
x1 + 4x2 + 2x3 + 2x4 = 0.
If you move x4 to the right side, you would subtract giving
x1 + 4x2 + 2x3 = - 2x4.
All three of your equations have this same error. No doubt this is what messed up your results.
Since you are solving Ax = 0, I'd probably opt for just doing row reduction on A---the augmented matrix would have all 0s in the additional column, so you don't have to put it there.
The reduced row echelon form of A is
[1 0 -6 -2]
[0 1 2 1]
[0 0 0 0]
So if (x1, x2, x3, x4) is in the kernel, then
x1 - 6x3 - 2x4 = 0 ==> x1 = 6x3 + 2x4
x2 + 2x3 + x4 = 0 ==> x2 = -2x3 - x4.
(Your results are the same except for that pesky sign error!)
Letting x3 = 1 and x4 = 0 gives (6, -2, 1, 0). Letting x3 = 0 and x4 = 1 gives (2, -1, 0, 1).
x1+4x2+2x3=2x4
x2+2x3 = x4
-x1-7x4-8x3 = -5x4
The signs on the right are all incorrect. The first equation should be
x1 + 4x2 + 2x3 + 2x4 = 0.
If you move x4 to the right side, you would subtract giving
x1 + 4x2 + 2x3 = - 2x4.
All three of your equations have this same error. No doubt this is what messed up your results.
Since you are solving Ax = 0, I'd probably opt for just doing row reduction on A---the augmented matrix would have all 0s in the additional column, so you don't have to put it there.
The reduced row echelon form of A is
[1 0 -6 -2]
[0 1 2 1]
[0 0 0 0]
So if (x1, x2, x3, x4) is in the kernel, then
x1 - 6x3 - 2x4 = 0 ==> x1 = 6x3 + 2x4
x2 + 2x3 + x4 = 0 ==> x2 = -2x3 - x4.
(Your results are the same except for that pesky sign error!)
Letting x3 = 1 and x4 = 0 gives (6, -2, 1, 0). Letting x3 = 0 and x4 = 1 gives (2, -1, 0, 1).