Where x, y, and z are unkown variables, find all values for a and b for which

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SUMMARY

The discussion focuses on solving a system of equations represented by an augmented matrix to determine the values of parameters a and b for which the system has no solution, infinitely many solutions, or a unique solution. The matrix provided is matrix=[2 -1 1 a; 1 1 2 1; b 3 3 a]. Participants suggest starting with row operations to achieve Row Echelon Form (REF) and emphasize the importance of calculating the determinant to identify conditions for solutions. The specific case of a=b=1 is also mentioned for finding a unique solution.

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  • Understanding of augmented matrices and systems of linear equations
  • Knowledge of Row Echelon Form (REF) and Gaussian elimination
  • Familiarity with determinants and their implications on the existence of solutions
  • Basic algebraic manipulation skills
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Homework Statement



Where x, y, and z are unknown variables and a and b are parameters, find all values a and b for which this system has (a) no solution (b) infinitely many solutions, and find the solutions (c) a unique solution, and find the solution when a=b=1.

matrix=[2 -1 1 a; 1 1 2 1; b 3 3 a] (where each ; denotes a new line in the matrix, so the matrix is 3 rows by 4 columns. And btw it is an augmented matrix).

Homework Equations





The Attempt at a Solution



I don't really know where to start! I know I have to somehow get the system into REF first, but it seems impossible because of the variables. I know that for (a) I have to somehow get a line that is something like [0 0 0 a] or [0 0 0 b] where a/b is not zero.
 
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It's not "impossible", just requires a bit of algebra.

[tex]\begin{bmatrix}2 & -1 & 1 & a \\ 1 & 1 & 2 & 1 \\ b & 3 & 3 & a \end{bmatrix}[/tex]
1a) swap the first two rows
1b) Subtract 2 times the new first row from the second row
1c) Subtract b times the new first row from the third row
That gives
[tex]\begin{bmatrix}1 & 1 & 2 & 1 \\ 0 & -3 & -3 & a- 2 \\ 0 & 3+ b & 1 & a- ab\end{bmatrix}[/tex]
Can you continue? The algebra gets complicated but not "impossible".

Alternatively, you could start by determining the values of a and b that make the determinant 0 or not 0.
 

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