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

1. ## 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!

2. 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 $\begin{pmatrix}3/2\\1/2\\1\end{pmatrix}$ then

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

Tonio

3. 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.

$\begin{bmatrix}2 & 4 & - 5\\ 1 & 3 & -3\end{bmatrix}\begin{bmatrix}1 \\ 2 \\ 3\end{bmatrix}= \begin{bmatrix}-5 \\ -2\end{bmatrix}$
(We could have stopped after seeing that 2(1)+ 4(2)- 5(3) is not 0.)

$\begin{bmatrix}2 & 4 & - 5\\ 1 & 3 & -3\end{bmatrix}\begin{bmatrix}3 \\ 1 \\ 2\end{bmatrix}= \begin{bmatrix}0 \\ 0}\end{bmatrix}$

etc.

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