Discuss a system of equations by Gauss

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The discussion focuses on analyzing a system of equations using Gauss's method, specifically through row reduction to reduced row echelon form. Participants emphasize the importance of determining the values of m that lead to no solutions, a unique solution, or infinitely many solutions. It is noted that for m not equal to 2, an inverse can be calculated, and checking the determinant is crucial for confirming the nature of the solutions. Special cases for m, such as m = 2 and m = -1, are examined to illustrate conditions leading to inconsistent or compatible systems. The conversation highlights the analytical approach needed to understand the implications of different values of m on the system's solvability.
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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
 
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inverse said:
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 <number of unknowns Undetermined System Compatible but m = -1 is a row which becomes zero, then range is 2 <number of unknowns should be compatible system Undetermined, but as I see a non-zero constant term and the others are zeros, I deduce that it's an imcompatible system, but as you can known analytically?
 
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