Matrix A =
|1 3 |
|-2 -6|
|-1 0 |
Vector u =
| 3|
|-1|
Determine if the vector u is in the kernel of A.
Also...
Vector v =
|1|
|2|
|0|
Determine if v is in the image of A
I have a test tomorrow on this and Im really behind. Can anyone explain how to do this, the answers and whats actually meant by the kernel and image??? Id be so greatful :)
|1 3 |
|-2 -6|
|-1 0 |
Vector u =
| 3|
|-1|
Determine if the vector u is in the kernel of A.
Also...
Vector v =
|1|
|2|
|0|
Determine if v is in the image of A
I have a test tomorrow on this and Im really behind. Can anyone explain how to do this, the answers and whats actually meant by the kernel and image??? Id be so greatful :)
-
A vector u is in the kernel of A if Au = 0. So all you have to do is take the product of A and u. If it's zero, the answer is yes, if it is not zero the answer is no.
Au = (0, 0, -3).
This is not zero, so u is not in the kernel.
A vector v is in the image of A if there exists a vector x such that Ax = v. This is a little tougher to determine because you have to figure out if such an x exists. One way to try is to answer is to set x = (x1, x2) and see if you can solve for the numbers x1 and x2.
(1) x1 + (3) x2 = 1
(-2) x1 + (-6)x2 = 2
(-1) x1 + (0) x2 = 0
The last equation gives x1 = 0. Plugging this into the first one gives x2 = 1/3, but into the second one gives x2 = -1/3. So, there is no solution to this system. That is, v is not in the image of A.
Au = (0, 0, -3).
This is not zero, so u is not in the kernel.
A vector v is in the image of A if there exists a vector x such that Ax = v. This is a little tougher to determine because you have to figure out if such an x exists. One way to try is to answer is to set x = (x1, x2) and see if you can solve for the numbers x1 and x2.
(1) x1 + (3) x2 = 1
(-2) x1 + (-6)x2 = 2
(-1) x1 + (0) x2 = 0
The last equation gives x1 = 0. Plugging this into the first one gives x2 = 1/3, but into the second one gives x2 = -1/3. So, there is no solution to this system. That is, v is not in the image of A.