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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: 2,150 Thanks: 730 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!

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