Inconsistent vs consistent augmented matrices

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Homework Help Overview

The discussion revolves around the consistency of augmented matrices in linear algebra, specifically examining two matrices with arbitrary and non-zero elements. Participants are exploring the implications of free variables and the conditions for a system to be consistent or inconsistent.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the nature of free variables in the context of the first matrix, questioning whether certain variables can be considered free and how this affects the consistency of the system. They also explore the implications of having unique versus infinite solutions based on the structure of the matrices.

Discussion Status

The conversation is ongoing, with participants providing insights into the nature of the matrices and the solutions they imply. Some guidance has been offered regarding the identification of free variables and the conditions for consistency, but there is no explicit consensus on the interpretations of the matrices.

Contextual Notes

Participants are considering the implications of arbitrary numbers represented by * and non-zero numbers represented by ■ in the matrices. There is uncertainty about whether all * numbers are the same or different, which may affect the analysis of the systems.

Arnoldjavs3
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Homework Statement


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 ■ | * ]

Homework Equations

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
 
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Arnoldjavs3 said:

Homework Statement


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 ■ | * ]

Homework Equations

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'?
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.
Arnoldjavs3 said:
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?
The system represented by this matrix is consistent, which means in this case there will be an infinite number of solutions.
Arnoldjavs3 said:
In the second matrix, it is similar in the sense that x3 is free and is not constrained.
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.
Arnoldjavs3 said:
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
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.
 
Mark44 said:
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.

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?
 
Arnoldjavs3 said:
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?
Yes and yes. I wouldn't say "pointed to," though. Each variable has a unique value would be a better way to say this.
 
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Arnoldjavs3 said:

Homework Statement


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 ■ | * ]

Homework Equations

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

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

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