Matrix Determinants: Find x for Invertibility

AI Thread Summary
A matrix is invertible when its determinant is non-zero, and the discussion revolves around finding the values of x that satisfy this condition. The user encountered confusion regarding the factoring of (1 - x) from a minor matrix, questioning whether it should be factored out from both rows. Clarification was provided that only one instance of (1 - x) can be factored out, as it is common to all members of the matrix. The conversation also referenced the impact of scalar multiplication on the determinant, confirming that scaling a row by a factor m scales the determinant by the same factor. Understanding these concepts is crucial for determining the invertibility of the matrix.
SherlockOhms
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Homework Statement


For which values of x is the matrix (see attachment) invertible?


Homework Equations


Row ops. Cofactors etc..


The Attempt at a Solution


Well, a matrix is only invertible when it's determinant is non zero. I've begun doing some row ops and have just hit a little snag. If you look at the attachment you'll see that I can facto (1 - x) out from the minor matrix. I remember hearing in a lecture that you have to factor out (1 - x) from both the top and bottom row of the matrix (i.e. you'll have (1 -x)^2 factored out instead of just (1 - x). Could somebody explain why you don't just factor out (1 - x) one like you would with factoring a scalar out of a matrix as normal?x
 
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In the reduced 2x2 matrix, the factor (1-x) is common to all of the members of the matrix. You can only factor it out once. Whatever you heard about factoring the matrix was incorrect.

See this article: http://en.wikipedia.org/wiki/Matrix_(mathematics)

specific topic: scalar multiplication ofa matrix
 
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Thanks for clearing that up.
 
Even with a scalar, wikipedia confirms that to scale a row by m scales the determinant by m, which is clear if you think of the formula for the determinant.
 
Yeah. That's actually what got me thinking about it in the first place.
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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