# Inconsistent vs consistent augmented matrices

1. Feb 5, 2017

### Arnoldjavs3

1. The problem statement, all variables and given/known data
Consider the following matrix where * indicates an arbitrary number and a ■ indicates a non zero number.

http://prntscr.com/e4xqkx

[■ * * * * | *]
[0 ■ * * 0 |* ]
[0 0 ■ * * | *]
[0 0 0 0 ■ | *]

(Sorry for poorly formatted matrix. The link above contains a screenshot if you can see it)

http://prntscr.com/e4xuoj
[■ * * | * ]
[0 ■ * | * ]
[0 0 ■ | * ]
2. Relevant equations

3. The attempt at a solution
Okay, so i know that a consistent system means that there has to be a unique solution or several of them. So in this isituation, i know that a non zero number(the last pivot) is equal to an arbritrary number. I believe this means that x5 is 'free'? This means that it is not constrained to any number and can be represented as x5 = t.
So how do I use this information to determine whether this augmented matrix is consistent or inconsistent? Are there any patterns to take note of here?

In the second matrix, it is similar in the sense that x3 is free and is not constrained. I'm having a hard time putting these matrices in perspective because they aren't actual numbers. Could something of this nature occur?:

x3 = t
x2 = s - t

2. Feb 5, 2017

### Staff: Mentor

No. Assuming you're talking about the first matrix above, x5 is specified, so you could solve for x5.

In the same matrix, x4 isn't specified, so it is free; i.e., x4 = t, where t can be any real value.
The system represented by this matrix is consistent, which means in this case there will be an infinite number of solutions.
Again, no. Since the reduced matrix (on the left) is diagonal, you could solve for unique solutions for each of the three variables. This system is also consistent, but there is a unique solution in this case.

From the last row, you could solve for x3, then back-substitute to solve for x2, and back-substitute these values to solve for x1.
Not for the second matrix.

For the first matrix, your solution would look something like $\vec{x} = \begin{bmatrix} c_1 \\ c_2 \\ c_3 \\ c_4 \\ c_5\end{bmatrix} + t\begin{bmatrix} 0 \\ 0 \\ 0 \\ 1 \\ 0\end{bmatrix}$, with t being a real parameter.

3. Feb 5, 2017

### Arnoldjavs3

So in the second matrix, since it is a diagonal with non-zero numbers and no variables being free, it is consistent with a unique solution as every variable can be pointed to a specific value(Hope that makes sense)?

And in the first matrix, there is a free variable. If there is a free variable(x4 like you said, i understand why now) then it is going to have infinite solutions?

4. Feb 5, 2017

### Staff: Mentor

Yes and yes. I wouldn't say "pointed to," though. Each variable has a unique value would be a better way to say this.

5. Feb 5, 2017

### Ray Vickson

Are all the * numbers the same, or are they possibly all different?