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December 26th, 2014, 08:29 PM   #1
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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.
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December 28th, 2014, 01:54 AM   #2
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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
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January 9th, 2015, 05:31 AM   #3
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There is no matrix given. What matrix are you referring to?
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January 9th, 2015, 06:24 AM   #4
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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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