## discuss a system of equations by Gauss

Hello,

How would you discuss this system of equations by Gauss's method?

$\begin{bmatrix}{x}&{y}&{(m-1)z}&{1}\\{x}&{(m-1)y}&{z}&{m-1}\\{(m-1)x}&{y}&{z}&{m+2}\end{bmatrix}$

NOTE: the last column are the independent terms

Thank you very much

 Quote by inverse Hello, How would you discuss this system of equations by Gauss's method? $\begin{bmatrix}{x}&{y}&{(m-1)z}&{1}\\{x}&{(m-1)y}&{z}&{m-1}\\{(m-1)x}&{y}&{z}&{m+2}\end{bmatrix}$ NOTE: the last column are the independent terms Thank you very much
Hey inverse and welcome to the forums.

If this is an augmented system [M | v], then you can reduce this whole thing to reduced row echelon form and then consider what values of m actually make sense in the context of there being no solutions, a unique solution, or infinitely many solutions if any of those categories exist.

The easiest way to check that you have done the reduction carefully, is to multiply your calculated inverse by your original matrix and you should get the identity if you end up getting a properly row-reduced system. It looks like you should get an inverse as long as m <> 2 by visual inspection, but you would have to check algebraically.

If you post your final row-reduced system and thus your calculated inverse for a general m, then as long as the determinant is non-zero for valid m, we can double check your inverse by multiplying that by the original matrix to get an identity.

This is really the hardest part since checking for inconsistent solutions is basically looking for 0 = 1 type arguments and infinite-solutions happens when you have 0 determinant and not a 0 = 1 situation.

 Thank you chiro Otherwise, to stagger the matrix, one can argue for m = 2 and m = -1, for m = 2 is a row which becomes zero, therefore range