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November 17th, 2015, 10:26 AM   #1
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Regular matrix - LU decomposition

I have the following math problem:

Determime for which ${x,y,z}$ is the matrix $A$ regular and then find the LU decomposition.

$A=\begin{pmatrix}
1 & 1 & 1 \\
x & y & z \\
x^2 & y^2 &z^2\\
\end{pmatrix}$

Determining when it is regular... I have tried using the definition of regular matrix. For instance that its rows and columns must be linearly independent but that didn´t take me far. We know the rank is 3 but we canť really use that anywhere..
From the definition we know that when a matrix is regular it is also invertible.
That implies there exists some matrix $X$ (inverse of $A$) such that $AX=XA=I_n$.. In this particular case $I_3$
Knowing that, if I now use the definition of matrix multiplication we get 9 equations with 12 variables? That doesn´t do anything for me either, does it?

I am still not sure how to proceed to find the general solution (all triplets of x, y, z).
I would also appreciate an approach without using determinant.

Thank you.
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November 18th, 2015, 06:36 AM   #2
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Det A=(x-y)(x-z)(y-z) which is non-singular if x, y, and z are unequal.

A is called a Van der Monde determinant.

For the LU decomposition (a row reduction procedure) google"LU decomposition" and watch one of the utubes.
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November 18th, 2015, 10:41 AM   #3
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$\displaystyle \begin{vmatrix}
1 &1 &1 \\
x&y &z \\
x^{2}&y^{2} &z^{2}
\end{vmatrix}=\begin{vmatrix}
1 & 1 & 1\\
x& y &z \\
0 &y^{2}-xy & z^{2}-xz
\end{vmatrix}=\begin{vmatrix}
1 &1 & 1\\
0 &y-x &z-x \\
0 & (y-x)y & (z-x)z
\end{vmatrix}=\\(y-x)(z-x)\begin{vmatrix}
1 &1 \\
y & z
\end{vmatrix}=(z-x)(z-y)(y-x)$

You didn't see that? I had no problem once I saw someone else do it. By the way, that's basically how you prove the general case by induction.
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