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Is there a specific way to find out if a...
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Is there a specific way to find out if a matrix's rank depends on a variable inside said matrix?
In this example I've just been trying to replace beta (alpha being any value) with random values until I get the rank down to 2.
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Use GauB
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>>7767966
Even after Gauss it's hard to get to a conclusion, unless I'm doing it wrong.
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>>7767964
The rank of a matrix is equal to the size of the biggest non zero determinant you can calculate from it.

That matrix already has got at least two as rank, since it has a non zero 2x2 size determinant. To test if its rank is three, simply calculate the bigger determinant possible, which is the unique of size 3x3.
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>>7767974
Beta doesn't matter. The rank is 2 if alpha is 1 and 3 for all other values of alpha.
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>>7767988
I don't understand how can a matrix have different determinants, are you talking about cofactors, like the 2x2 determinants inside a 3x3 matrix?
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>>7767990
Why doesn't beta matter? how did you get to that conclusion?
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>>7767995
Call it as you like, but its accurate name is minor.

https://en.wikipedia.org/wiki/Minor_%28linear_algebra%29
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>>7767995
obviously he is. I didn't know those were called cofactors in english though, in my country we call them subdeterminants or minors
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>>7767990
No, the rank is 3 for α≠1. For α=1 it is 2.
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>>7767964
can the matrix have rank 1?
No because the first two columns are linearily independant whatever the value of beta is.

So the rank of your matrix is either 2 or 3.

Now just compute the determinant as a function of alpha and beta and find the values for which the det. is equal to 0 or different from 0. That will give you the cases where rank(A) is 2 and rank(A) is 3
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>>7768000
[eqn] \begin{bmatrix} -1 & 0 & 0 \\ 1 & 1 & \alpha \\ \beta & 1 & 1 \end{bmatrix} [/eqn]
Add the first row to the second row and add beta times the first row to the second row.
[eqn] \begin{bmatrix} -1 & 0 & 0 \\ 0 & 1 & \alpha \\ 0 & 1 & 1 \end{bmatrix} [/eqn]
Subtract the second row from the last row.
[eqn] \begin{bmatrix} -1 & 0 & 0 \\ 0 & 1 & \alpha \\ 0 & 0 & 1 - \alpha \end{bmatrix} [/eqn]

It's triangular iff alpha = 1.
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>>7767988
cool, I didn't know that was the case. thanks
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>>7768013
No, it is already triangular whatever alpha is.
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>>7768013
>It's triangular iff alpha = 1.
you mean singular