Results 1 to 2 of 2

Math Help - Challenging MATRIX QUESTION

  1. #1
    Newbie
    Joined
    Nov 2011
    Posts
    1

    Challenging MATRIX QUESTION

    Show working to sole for x in the equation

    (x-3 1 -1)
    (1 x-5 1 ) =0
    ( -1 1 x-3)

    THSI IS A 3x3 MATRIX
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,386
    Thanks
    1322

    Re: Challenging MATRIX QUESTION

    Quote Originally Posted by prittyboy View Post
    Show working to sole for x in the equation

    (x-3 1 -1)
    (1 x-5 1 ) =0
    ( -1 1 x-3)

    THSI IS A 3x3 MATRIX
    In which case this is not "challenging", it is simply non-sense. A matrix cannot be equal to a number. I suspect this is not a matrix question at all but is a deteriminant equation:
    \left|\begin{array}{ccc}x- 3 & 1 & -1 \\ 1 & x- 5 & 1 \\ -1 & 1 & x- 3\end{array}\right|= 0

    Expanding on the top row,
    \left|\begin{array}{ccc}x- 3 & 1 & -1 \\ 1 & x- 5 & 1 \\ -1 & 1 & x- 3\end{array}\right|= (x- 3)\left|\begin{array}{cc}x- 5 & 1 \\ 1 & x- 3\end{array}\right|- \left|\begin{array}{cc}1 & 1 \\ -1 & x- 3\end{array}\right| - \left|\begin{array}{cc}1 & x- 5 \\ -1 & 1\end{array}\right|
    = (x- 3)((x-5)(x-3)- 1)- (1(x-3)+ 1)- (1+ 1(x- 5))= (x- 3)(x^2- 8x+ 14)- x+ 2- x+ 4
    = (x^3- 8x^2+14x)- (3x^2+ 24x- 42)-2x+ 6= x^3- 11x^2- 12x+ 48= 0
    a cubic equation. There is a general formula for solving cubic equations but it is a bit complicated. You might try checking for rational roots. The "rational root theorem" says that any rational number root of the polynomial equation a_nx^n+ a_{n-1}x^{n-1}+\cdot\cdot\cdot+ a_1x+ a_0= 0 must be of the form \frac{m}{n} where the numerator, m, divides the constant term, a_0 and the denominator, n, divides the leading coefficent, a_n.

    Since, here, the leading coefficient is 1, any rational number solution to the equation must be an integer that evenly divides 48. Try those to see if there is a rational root.
    Last edited by HallsofIvy; November 8th 2011 at 03:36 PM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. A Challenging PDE question
    Posted in the Differential Equations Forum
    Replies: 5
    Last Post: August 30th 2011, 05:18 AM
  2. (challenging) question about cards
    Posted in the Advanced Statistics Forum
    Replies: 0
    Last Post: June 16th 2010, 01:14 PM
  3. A Challenging Question
    Posted in the Advanced Statistics Forum
    Replies: 10
    Last Post: August 15th 2009, 06:03 PM
  4. Challenging question?
    Posted in the Algebra Forum
    Replies: 1
    Last Post: July 1st 2008, 02:34 AM
  5. Please help... challenging Integral Question
    Posted in the Calculus Forum
    Replies: 10
    Last Post: November 20th 2007, 06:06 PM

Search Tags


/mathhelpforum @mathhelpforum