Results 1 to 9 of 9

Math Help - Augmented Matrix with constants

  1. #1
    Newbie
    Joined
    Sep 2011
    Posts
    4

    Augmented Matrix with constants

    How would you solve a matrix with constants inside of it (such as h,k). An example I was given was
    -7x + 3y + 4z = 3
    -4x - 4y - 4z = -5
    29x -y + h = k
    and was told to find "This system has a unique solution whenever "

    What I did was bring the matrix to a form as close as I could to row echelon form, but I have no idea what the conditions are to find what k isnt equal to for a unique solution to be found.

    The near row echelon form I found was

    x + 11y + 12z = 13
    0x + 1y + 11/10z = 47/40
    0x + 0y + (9/5 + h)z = k + 67/20

    and from there I'm at a complete loss as to how to solve the problem

    Edit: It'd also be appreciated if someone could let me know the conditions for a similar problem, but with infinite solutions and with no solutions

    Edit 2: The question was actually when , not k, sorry for the inconvenience!

    Thanks!
    Last edited by virus1019; September 3rd 2011 at 09:01 AM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,407
    Thanks
    1294

    Re: Augmented Matrix with constants

    Quote Originally Posted by virus1019 View Post
    How would you solve a matrix with constants inside of it (such as h,k). An example I was given was
    -7x + 3y + 4z = 3
    -4x - 4y - 4z = -5
    29x -y + h = k
    and was told to find "This system has a unique solution whenever "

    What I did was bring the matrix to a form as close as I could to row echelon form, but I have no idea what the conditions are to find what k isnt equal to for a unique solution to be found.

    The near row echelon form I found was

    x + 11y + 12z = 13
    0x + 1y + 11/10z = 47/40
    0x + 0y + (9/5 + h)z = k + 67/20

    and from there I'm at a complete loss as to how to solve the problem
    Whenever \displaystyle k \neq what?

    Have you tried evaluating the determinant of this matrix and setting it equal to 0? That will tell you what h can't be at least...
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Sep 2011
    Posts
    4

    Re: Augmented Matrix with constants

    oh sorry, i posted the wrong thing :S it was supposed to be h, not k

    I don't think I've learned determinants in class yet though, so I'm not quite sure what you mean
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,407
    Thanks
    1294

    Re: Augmented Matrix with constants

    Have you learnt about the identity matrix or the inverse matrix?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Sep 2011
    Posts
    4

    Re: Augmented Matrix with constants

    yes, ive learnt the identity matrix and the inverse matrix
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,407
    Thanks
    1294

    Re: Augmented Matrix with constants

    Quote Originally Posted by virus1019 View Post
    yes, ive learnt the identity matrix and the inverse matrix
    OK, well if you think about a 2x2 matrix \displaystyle \mathbf{A} = \left[\begin{matrix}a & b \\ c & d\end{matrix}\right], the inverse matrix is \displaystyle \mathbf{A}^{-1} = \frac{1}{ad - bc}\left[\begin{matrix}\phantom{-}d & -b \\ -c & \phantom{-}a\end{matrix}\right]. Clearly, this inverse matrix only exists if \displaystyle ad-bc \neq 0, so it determines if there is a solution to the matrix equation \displaystyle \mathbf{A}\mathbf{x} = \mathbf{b}. This is why that quantity is called the DETERMINANT, and is given the notation \displaystyle |\mathbf{A}|, or \displaystyle \left|\begin{matrix}a&b\\c&d\end{matrix}\right|.

    Square matrices of higher dimension can also have inverses, but again, only as long as its determinant is nonzero. Higher determinants can be evaluated by reducing it to smaller determinants, e.g. for a 3x3 matrix

    \displaystyle \left|\begin{matrix}a&b&c\\d&e&f\\g&h&j\end{matrix  }\right| = a\left|\begin{matrix}e&f\\h&j\end{matrix}\right| - b\left|\begin{matrix}d&f\\g&j\end{matrix}\right| + c\left|\begin{matrix}d&e\\g&h\end{matrix}\right|

    The easy way to reduce a large determinant to the smaller determinants is to choose a row (like the top row), start with the first element, visualise the row and column that it lies in being erased, and the remaining elements becoming the smaller determinant. Move to the next element in the top row, mentally erase the row and column it lies in, and the remaining elements become the smaller determinant, etc. As you move along the row, the signs alternate (+,-,+,-,...)

    In summary, a matrix equation only has solution when the inverse matrix exists, and the inverse matrix exists only when its determinant is nonzero.

    So to answer your original question, if you evaluate the determinant and set it equal to 0, you'll find the value that h can not take for the matrix equation to have solution.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Newbie
    Joined
    Sep 2011
    Posts
    4

    Re: Augmented Matrix with constants

    But then what happens to the answers (to the individual equations) in the matrices? Do they just disappear while finding the determinant? Or do they play a role in finding the determinant as well?
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,407
    Thanks
    1294

    Re: Augmented Matrix with constants

    Quote Originally Posted by virus1019 View Post
    But then what happens to the answers (to the individual equations) in the matrices? Do they just disappear while finding the determinant? Or do they play a role in finding the determinant as well?
    The determinant tells you IF there will be a solution. The determinant is also a part of the inverse matrix that you can premultiply both sides by to solve the matrix equation.

    I suggest you Google for some information about determinants.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,405
    Thanks
    1328

    Re: Augmented Matrix with constants

    You reduced to
    x+ 11y+ 12z= 13
    0x+ y+ \frac{11}{10}z= \frac{47}{40}
    0x+ 0y+ \left(\frac{9}{5}+ h\right)z= k+ \frac{67}{20}

    Look at that last equation. Obviously you solve for z by dividing both sides by \frac{9}{5}+ h.

    For what value of h are you not allowed to do that?
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Augmented matrix
    Posted in the Advanced Algebra Forum
    Replies: 11
    Last Post: July 13th 2010, 07:14 PM
  2. Augmented matrix
    Posted in the Algebra Forum
    Replies: 1
    Last Post: March 3rd 2009, 10:01 AM
  3. Augmented matrix
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: September 9th 2008, 08:17 PM
  4. Augmented matrix
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: September 19th 2006, 11:10 AM
  5. Augmented matrix
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: July 20th 2006, 12:44 AM

Search Tags


/mathhelpforum @mathhelpforum