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 December 26th, 2014, 08:29 PM #1 Newbie   Joined: Dec 2014 From: Romania Posts: 5 Thanks: 0 Math Focus: linear algebra what a way to solve the determinant of this kind of a matrix? i tried raw operations. but not that confident about it. someone can show me some guidness please? -8x-48 x+24 x+24 x+24 x+24 x+24 -8x-48 x+24 x+24 x+24 x+24 x+24 -8x-48 x+24 x+24 x+24 x+24 x+24 -8x-48 x+24 x+24 x+24 x+24 x+24 -8x-48 Last edited by greg1313; December 27th, 2014 at 06:54 AM. December 28th, 2014, 01:54 AM #2 Newbie   Joined: Dec 2014 From: Romania Posts: 5 Thanks: 0 Math Focus: linear algebra i already solved it . but no one can give a second idea ? i just added the columns and took -4x+48 as a common facor January 9th, 2015, 05:31 AM #3 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 There is no matrix given. What matrix are you referring to? January 9th, 2015, 06:24 AM #4 Senior Member   Joined: Apr 2014 From: Glasgow Posts: 2,157 Thanks: 731 Math Focus: Physics, mathematical modelling, numerical and computational solutions I presume he means $\displaystyle A = \left[\begin{array}{ccccc} -8x-48 & x+24 & x+24 & x+24 & x+24 \\ x+24 & -8x-48 & x+24 & x+24 & x+24 \\ x+24 & x+24 & -8x-48 & x+24 & x+24 \\ x+24 & x+24 & x+24 & -8x-48 & x+24 \\ x+24 & x+24 & x+24 & x+24 & -8x-48 \\ \end{array}\right]$ You can simplify it a little bit using a scalar operation before attempting to find the determinant: $\displaystyle A = (x+24)\left[\begin{array}{ccccc} -8\left(\frac{x+6}{x+24}\right) & 1 & 1 & 1 & 1 \\ 1 & -8\left(\frac{x+6}{x+24}\right) & 1 & 1 & 1 \\ 1 & 1 & -8\left(\frac{x+6}{x+24}\right) & 1 & 1 \\ 1 & 1 & 1 & -8\left(\frac{x+6}{x+24}\right) & 1 \\ 1 & 1 & 1 & 1 & -8\left(\frac{x+6}{x+24}\right) \\ \end{array}\right]$ Because $\displaystyle det(kA) = k^ndet(A)$, where k is a scalar multiple and n is the rank of the square matrix, $\displaystyle det(A) = (x+24)^5\left|\begin{array}{ccccc} -8\left(\frac{x+6}{x+24}\right) & 1 & 1 & 1 & 1 \\ 1 & -8\left(\frac{x+6}{x+24}\right) & 1 & 1 & 1 \\ 1 & 1 & -8\left(\frac{x+6}{x+24}\right) & 1 & 1 \\ 1 & 1 & 1 & -8\left(\frac{x+6}{x+24}\right) & 1 \\ 1 & 1 & 1 & 1 & -8\left(\frac{x+6}{x+24}\right) \\ \end{array}\right|$ Maybe this one will be easier to manage! Tags determinant, kind, matrix, solve, yhe Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post jones123 Linear Algebra 3 April 24th, 2013 03:16 AM helloprajna Linear Algebra 2 December 8th, 2012 06:31 AM bull-roarer Algebra 2 May 23rd, 2009 02:44 PM battery Linear Algebra 1 November 21st, 2008 11:45 AM jones123 Calculus 1 December 31st, 1969 04:00 PM

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