Why Are 3x3 Systems with Arithmetically Increasing Constants Always Infinite?

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The discussion centers on the observation that systems of linear equations with arithmetically increasing constants yield infinite solutions. The example provided is a system of three equations: 3x + 5y + 7z = 9, 7x + 3y - z = -5, and 12x + 13y + 14z = 15. The participant demonstrates the use of column operations on the augmented matrix to simplify the system, ultimately revealing that the last two columns become zero, allowing for z to take any value. This leads to the conclusion that such systems inherently possess infinite solutions due to their structural properties.

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  • Understanding of linear algebra concepts, specifically systems of equations.
  • Familiarity with augmented matrices and their manipulation.
  • Knowledge of column operations in matrix algebra.
  • Basic skills in elimination methods for solving equations.
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AlphaNoodle
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I encountered a system:
3x+5y+7z = 9
7x+3y-z=-5
12x+13y+14z=15
And the solution was infinite solutions.

However, when looking at each equation, the constants (including coefficients) increase/decrease by a constant amount.
3x+5y+7z = 9 (+2)
7x+3y-z=-5 (-4)
12x+13y+14z=15 (+1)

And I made other systems using that same format, where all the equations' constants increased/decreased arithmetically (by a constant)
And the solution was always infinity.

I am curious to why this is and is there any proof behind this?
I got as far as:

Ax+(A+m)y+(A+2m)z=(A+3m)
Bx+(B-n)y+(B-2n)z=(B-3n)
Cx+(C+k)y+(C+2k)z=(C+3k)

But I am clueless to why there is always infinite solutions to these types of systems. I always get to the point where a number = a number (using elimination) to solve the system, and form there the solutions are always infinite. I don't even know how to proceed with a proof, and I was hoping for some help. I am sure I am interpreting something wrong or missing something, but anything would be appreciated. Thanks in advanced.
 
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One way to solve a system of equations is to "column" reduce the augmented matrix.

Here, the augmented matrix is
[tex]\begin{bmatrix}A & A+ n & A+ 2n & A+ 3n \\ B & B- m & B- 2m & B- 3m \\ C & C+ k & C+ 2k & C+ 3k\end{bmatrix}[/tex]

If we add -1 times the first column to each of the other columns we get
[tex]\begin{bmatrix}A & n & 2n & 3n \\ B & -m & -2m & -3m \\ C & k & 2k & 3k\end{bmatrix}[/tex]

Now add -2 times the second column to the third column and -3 times the second column to the fourth column to get
[tex]\begin{bmatrix}A & n & 0 & 0 \\ B & -m & 0 & 0 \\ C & k & 0 & 0\end{bmatrix}[/tex]
and now we can see that, since the last two columns are all "0"s, the last unknown value, z, can be anything at all and still satisfy these equations.
 
I did not know there were column operations, still in high school haha. But a question I have is how do you represent column operations? in other words row operations are simply L1 -> L1+3 or something like that, but how are column operations represented?
 

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