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Math Help - Which of the following vectors belong to the null space of A

  1. #1
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    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!
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  2. #2
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    Quote Originally Posted by Theaisback View Post
    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
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  3. #3
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    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!
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