My Math Forum Regular matrix - LU decomposition

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 November 17th, 2015, 10:26 AM #1 Newbie   Joined: Oct 2015 From: CR Posts: 10 Thanks: 0 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.
 November 18th, 2015, 06:36 AM #2 Banned Camp   Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 126 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.
 November 18th, 2015, 10:41 AM #3 Banned Camp   Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 126 $\displaystyle \begin{vmatrix} 1 &1 &1 \\ x&y &z \\ x^{2}&y^{2} &z^{2} \end{vmatrix}=\begin{vmatrix} 1 & 1 & 1\\ x& y &z \\ 0 &y^{2}-xy & z^{2}-xz \end{vmatrix}=\begin{vmatrix} 1 &1 & 1\\ 0 &y-x &z-x \\ 0 & (y-x)y & (z-x)z \end{vmatrix}=\\(y-x)(z-x)\begin{vmatrix} 1 &1 \\ y & z \end{vmatrix}=(z-x)(z-y)(y-x)$ 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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