Results 1 to 7 of 7
Like Tree3Thanks
  • 1 Post By HallsofIvy
  • 1 Post By Deveno
  • 1 Post By Deveno

Math Help - unique solution

  1. #1
    Junior Member
    Joined
    Aug 2012
    From
    us
    Posts
    66

    unique solution

    Does the matrix
    0 1 0 -15
    0 0 1 7
    has a unique solution?

    Thank you in advance
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,545
    Thanks
    1394

    Re: unique solution

    What do you mean by "solving" a matrix?? Is this the augmented matrix for two equations in three unknowns? Well, if so, two equations in three unknowns, no matter what the numbers are, cannot have a unique solution.
    Thanks from jojo7777777
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Aug 2012
    From
    us
    Posts
    66

    Re: unique solution

    "Is this the augmented matrix for two equations in three unknowns?" Yes, it is.
    So, does in this case the first (left) unknown is a parameter?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Mar 2011
    From
    Tejas
    Posts
    3,370
    Thanks
    739

    Re: unique solution

    if what you really mean is:

    "does

    \begin{bmatrix}0&1&0\\0&0&1 \end{bmatrix} \begin{bmatrix}x_1\\x_2\\x_3 \end{bmatrix} = \begin{bmatrix}-15\\7 \end{bmatrix}

    have a unique solution?"

    the answer is: no.

    (1,-15,7) is a solution, and so is (2,-15,7), and these are different. there are LOTS of different ways to look at this:

    1. we have 3 unknowns, and 2 equations. so IF the equations are consistent (and they are), the system is "under-determined" (more unknowns than equations).

    2. the rank of the matrix is 2, which is less than the dimension of the domain.

    3. as a rule: general solution = any specific solution + ALL homogeneous solutions. the homogenous case is found by replacing (-15,7) with (0,0) and solving THAT case.

    this is equivalent to finding the null space of the matrix. it is clear in this case, that the null space is all vectors of the form (t,0,0) (where t can be any real number, assuming that is the field you are working in).

    "free variables" correspond to the COLUMNS without a pivot (leading 1). in this case, that is column 1, so "x1 is free".

    let me explain number 2. above a bit more: when a matrix A "acts" on a vector space Rn, by taking x to Ax, it takes an n-vector to an m-vector (supposing that A is mxn). so A(Rn) (the image space of A) is some subspace of Rm. now if m < n (more columns than rows), A takes Rn to a "smaller" space.

    you can't take a "bigger" thing to a "smaller" thing without some "collapsing" occuring (two different vectors in Rn have to wind up "with the same image", since Rm doesn't have enough dimensions to reproduce all of Rn faithfully). this is what causes "non-uniqueness" (sorry for all the quotes, but i'm not speaking precisely, here).

    what rank(A) tells us, is "how big the image space is". in this case, it has dimension 2: and a basis for im(A) is: {(1,0),(0,1)} (in other words A is of FULL rank, the image of A is all of the target space...A is surjective, or "onto" R2. note that these are precisely the linearly independent COLUMNS of A: the image of A is also known as the COLUMN space).

    and THAT means that Ax = b is going to have a solution (in fact, many solutions) for any b, including b = (-15,7).

    going back to number 3: since a homogeneous solution is: (t,0,0) for any (real) number t, and (0,-15,7) is a PARTICULAR solution,

    the general solution is: (t,0,0) + (0,-15,7) = (t,-15,7), where we are "free" to pick t as we please.
    Last edited by Deveno; September 26th 2012 at 02:22 PM.
    Thanks from jojo7777777
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Aug 2012
    From
    us
    Posts
    66

    Re: unique solution

    I'm grateful to you for your profound explanation...!!!

    If you let me just ask one more question:
    according to
    in other words A is of FULL rank, the image of A is all of the target space...A is surjective, or "onto" R2.
    if A was
    010
    000
    and b was (-15,0)
    I should say "the image of A is just R"?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor

    Joined
    Mar 2011
    From
    Tejas
    Posts
    3,370
    Thanks
    739

    Re: unique solution

    no, because the image of A ISN'T R, it's a 1-dimensional subspace of R2.

    these are isomorphic (they "act alike") as VECTOR SPACES, but not as "other things".

    for example, in R, we have multiplication, and "fractions". what is:

    1/(a,0) ? what does that even MEAN (how do you "divide by a vector")?
    Thanks from jojo7777777
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Junior Member
    Joined
    Aug 2012
    From
    us
    Posts
    66

    Re: unique solution

    Thank you...You're a genius...
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 1
    Last Post: September 23rd 2011, 03:39 AM
  2. unique solution
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: January 2nd 2011, 09:34 AM
  3. unique solution
    Posted in the Algebra Forum
    Replies: 6
    Last Post: June 29th 2010, 01:10 PM
  4. Replies: 1
    Last Post: March 24th 2010, 12:14 AM
  5. Replies: 2
    Last Post: September 7th 2009, 02:01 PM

Search Tags


/mathhelpforum @mathhelpforum