Can RREF be used for any size matrix?

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Mr Davis 97
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I understand that the RREF algorithm can be used on matrices representing systems of equations to find the solution set of that system. However, can this algorithm be used for any matrix of any size? For example, what if we, what if we had a 1x1 mattix, or a 2x1? What is the minimum size of a matrix for which RREF makes sense?
 
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You can always apply the algorithm. The matrix might be in its final form already, then nothing changes.
 
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Mr Davis 97 said:
I understand that the RREF algorithm can be used on matrices representing systems of equations to find the solution set of that system. However, can this algorithm be used for any matrix of any size? For example, what if we, what if we had a 1x1 mattix, or a 2x1? What is the minimum size of a matrix for which RREF makes sense?
Think about the system of equations a matrix represents. A 1x1 matrix represents a single equation such as 3x = 0. For an equation like 3x = 6, you would need an augmented 1x2 matrix. You could solve this by row reduction, but it seems like massive overkill.

A 2x1 matrix would represent a system of two equations in one variable, such as
3x = 0
2x = 0
The matrix itself would consist of a single column whose entries are 3 and 2, respectively. Again, you could use row reduction, and find that (surprise!) x = 0. For such simple systems, you could use RREF, but it doesn't make much sense.
 
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Alright, it makes sense. Just one further question. How does a matrix with only one column represent a system of equations? I see that you use zero when you right it out in equation form, but where is that coming from?
 
Mr Davis 97 said:
Alright, it makes sense. Just one further question. How does a matrix with only one column represent a system of equations? I see that you use zero when you right it out in equation form, but where is that coming from?
It would represent a homogeneous system, a system of equations where the constant terms are all zero.

So the system of two equations in one unknown
3x = 0
2x = 0
could be represented by this matrix:
##\begin{bmatrix} 3 \\ 2 \end{bmatrix}##

A nonhomogeneous system such as
4x = 10
2x = 4
could be represented by the augmented matrix
##\begin{bmatrix} 4 & | & 10 \\ 2 & | & 4\end{bmatrix}##
I hope it's obvious that this system has no solution.
 
Wouldn't ##\begin{bmatrix} 3 & 0 \\ 2 & 0\end{bmatrix}## represent the homogeneous system that you are talking about, and not ##\begin{bmatrix} 3 \\ 2 \end{bmatrix}##?
 
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Mr Davis 97 said:
Wouldn't ##\begin{bmatrix} 3 & 0 \\ 2 & 0\end{bmatrix}## represent the homogeneous system that you are talking about, and not ##\begin{bmatrix} 3 \\ 2 \end{bmatrix}##?

That's what I think too.
 
Hey Mr Davis 97.

The algorithm can be applied to an arbitrary matrix since all operations are based on multiplication (scalar multiplication) and addition (and subtraction) which can be done for any set of values.
 
Mr Davis 97 said:
Wouldn't ##\begin{bmatrix} 3 & 0 \\ 2 & 0\end{bmatrix}## represent the homogeneous system that you are talking about, and not ##\begin{bmatrix} 3 \\ 2 \end{bmatrix}##?
Math_QED said:
That's what I think too.
Most of the linear algebra books I've seen use the matrix for the coefficients of the variables. If the system is nonhomogeneous (constants on the right sides of the equations, they use an augmented matrix.
If the system of equations is homogeneous, there's no point in dragging along a column of zeroes, none of which can change from any row operations.