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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? 8x48 x+24 x+24 x+24 x+24 x+24 8x48 x+24 x+24 x+24 x+24 x+24 8x48 x+24 x+24 x+24 x+24 x+24 8x48 x+24 x+24 x+24 x+24 x+24 8x48 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 $\displaystyle A = \left[\begin{array}{ccccc} 8x48 & x+24 & x+24 & x+24 & x+24 \\ x+24 & 8x48 & x+24 & x+24 & x+24 \\ x+24 & x+24 & 8x48 & x+24 & x+24 \\ x+24 & x+24 & x+24 & 8x48 & x+24 \\ x+24 & x+24 & x+24 & x+24 & 8x48 \\ \end{array}\right]$ 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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