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I´ve 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 I am still not sure how to proceed if we want to make sure we find all solutions. Also I would appreciate approach without using determinant.

Thank you.

user291002
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  • What definition of regular matrix are you using? regular = invertible ? or regular = http://math.stackexchange.com/questions/216015/understanding-regular-matrices ? – Nizar Nov 17 '15 at 16:10
  • Not that the definition really helps me here, at least not that I can see.. We can say that regular matrix is invertible matrix which is when 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 help me much, does it? – user291002 Nov 17 '15 at 16:40
  • Any hints or ideas? – user291002 Nov 17 '15 at 19:05

1 Answers1

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Trying to find the Row Echelon from of your matrix, we ma proceed as follows:

$R_2 \leftarrow R_2-xR_1 \text{ and } R_3 \leftarrow R_3-x^2R_1 $ implies $$ \begin{pmatrix} 1 & 1 & 1 \\ 0 & y-x & z-x \\ 0 & y^2-x^2 &z^2-x^2\\ \end{pmatrix} = \begin{pmatrix} 1 & 1 & 1 \\ 0 & y-x & z-x \\ 0 & (y-x)(y+x) &z^2-x^2\\ \end{pmatrix} $$ Now $R_3 \leftarrow R_3 -(x+y) R_2$ implies
\begin{pmatrix} 1 & 1 & 1 \\ 0 & y-x & z-x \\ 0 & 0 &(z^2-x^2)-(z-x)(x+y)\\ \end{pmatrix} Note that this last matrix and the original matrix $A$ have the same Rank, and $A$ is regular (invertible) if its rank is three. This last matrix has rank three if $ y-x\ neq 0 $ and $(z^2-x^2)-(z-x)(x+y) \neq 0 $ i.e. $x \neq y$ and $ (z-x)(z+x-x-y) \neq 0 $, i.e. $ x \neq y$ and $z \neq y$ and $z\neq x$.

Nizar
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