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.
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 $z\neq x$.