Thread: Which of the following vectors belong to the null space of A

Hey everyone,

I'm stuck on this question I'll show you what I got so far, but first I'll write out the question...

Let A = |2 4 -5|.Which of the following vectors belongs to the null space of A?
|1 3 -3|

u1 = (1, 2, 3), u2 = (3, 1, 2), u3 = (0, 0, 0), u4 = (0, 0), u5 = (6, 2, 4)

|2 4 -5 | R1 --> 1/2 * R1
|1 3 -3 |

|1 2 -5/2|
|1 3 -3 | R2 --> -1 * R1 + R2

|1 2 -5/2| R1 ---> -2 * R2 + R1
|0 1 -1/2|

|1 0 -3/2|
|0 1 -1/2|

this corresponds to the system

1x1 +(-3/2)x3 = 0
1x2 +(-1/2)x3 = 0

the system has infinitely many solutions:

x1 = +(3/2) x3
x2 = +(1/2) x3

the solution can be written in vector form:

(3/2, 1/2, 1) therefore the null space has a basis formed by this set.

This is what I got, but I do not know how to use the U1, U2, U3, U4, U5 from the question above. Did I make a mistake or what?

Thanks for you help!

Originally Posted by Theaisback

Hey everyone,

I'm stuck on this question I'll show you what I got so far, but first I'll write out the question...

Let A = |2 4 -5|.Which of the following vectors belongs to the null space of A?
|1 3 -3|

u1 = (1, 2, 3), u2 = (3, 1, 2), u3 = (0, 0, 0), u4 = (0, 0), u5 = (6, 2, 4)

|2 4 -5 | R1 --> 1/2 * R1
|1 3 -3 |

|1 2 -5/2|
|1 3 -3 | R2 --> -1 * R1 + R2

|1 2 -5/2| R1 ---> -2 * R2 + R1
|0 1 -1/2|

|1 0 -3/2|
|0 1 -1/2|

this corresponds to the system

1x1 +(-3/2)x3 = 0
1x2 +(-1/2)x3 = 0

the system has infinitely many solutions:

x1 = +(3/2) x3
x2 = +(1/2) x3

the solution can be written in vector form:

(3/2, 1/2, 1) therefore the null space has a basis formed by this set.

This is what I got, but I do not know how to use the U1, U2, U3, U4, U5 from the question above. Did I make a mistake or what?

Thanks for you help!

I did not check, but if the null space of that matrix is spanned by then

only the scalar multiples of this vector belongs to the null space...

Tonio

The question does not ask for a basis or other general information about the null space of A but only which of these given vectors is in the null space- so it might be simplest just to multiply A by each of them.


(We could have stopped after seeing that 2(1)+ 4(2)- 5(3) is not 0.)



etc.

(0, 0, 0) and (0, 0) are obvious!