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Math Help - rank and nullity

  1. #1
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    rank and nullity

    Compute the rank and nullity of the matrix A and state what this implies about the existence and/or uniqueness of solutions x of Ax=b.

    A = [1 1 3]
    1 2 3
    3 2 1

    Sol: Reduced row echelon form = [1 0 -1]
    0 1 2
    0 0 0

    Therefore, Rank(A)=2, and Nullity (A) = 1. But I don't get the second part though. Does it mean there's only one solution?
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  2. #2
    MHF Contributor harish21's Avatar
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    Firstly, your rank is not correct..

    The row reduced echleon form of the given matrix should be a I_3

    try doing it again..
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  3. #3
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    my bad. The first row should be 1 1 1. Not 1 1 3.
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  4. #4
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    Quote Originally Posted by Taurus3 View Post
    Compute the rank and nullity of the matrix A and state what this implies about the existence and/or uniqueness of solutions x of Ax=b.

    A = [1 1 3]
    1 2 3
    3 2 1

    Sol: Reduced row echelon form = [1 0 -1]
    0 1 2
    0 0 0

    Therefore, Rank(A)=2, and Nullity (A) = 1. But I don't get the second part though. Does it mean there's only one solution?
    The zero vector is also a solution and a scalar times your nullspace vector are solutions.
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  5. #5
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    Can I also give the same reason if
    A= [0 1 1 0, 2 -1 -1 2, 1 2 2 4] where the reduced row echelon form is [1 0 0 0, 0 1 1 0, 0 0 0 1] as the rank would be 4 and nullity would be 0.
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  6. #6
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    Quote Originally Posted by Taurus3 View Post
    Can I also give the same reason if
    A= [0 1 1 0] where the reduced row echelon form is [1 0 0 0] as the rank would be 4 and nullity would be 0.
    2 -1 -1 2 0 1 1 0
    1 2 2 4 0 0 0 1
    You need to learn latex. You have a row vector and then some random numbers mashed together.
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  7. #7
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    I've edited my previous post. Is it better now?
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