Inconsistent vs consistent augmented matrices

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In summary, the first matrix represents a consistent system with an infinite number of solutions, with one free variable (x4). The second matrix represents a consistent system with a unique solution, as there are no free variables. In both cases, the system can be solved by using back-substitution to find the values of each variable.
  • #1
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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  • #2
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.
 
  • #3
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?
 
  • #4
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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  • #5
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?
 

FAQ: Inconsistent vs consistent augmented matrices

What is the difference between an inconsistent and consistent augmented matrix?

An inconsistent augmented matrix is one in which the system of equations represented by the matrix has no solution. This means that the equations are contradictory and cannot be satisfied simultaneously. On the other hand, a consistent augmented matrix has at least one solution, meaning that all the equations can be satisfied.

How can you tell if an augmented matrix is inconsistent or consistent?

An augmented matrix is inconsistent if it has a row of zeros that corresponds to a non-zero entry in the rightmost column. This indicates that the system of equations has no solution. Conversely, an augmented matrix is consistent if it does not have this pattern and has either a unique solution or infinitely many solutions.

Can an inconsistent augmented matrix have a solution?

No, an inconsistent augmented matrix cannot have a solution. This is because the rows of the matrix represent a set of equations that are contradictory and cannot be satisfied simultaneously. Therefore, there is no possible solution that would satisfy all the equations.

How can you solve a system of equations represented by an augmented matrix?

To solve a system of equations represented by an augmented matrix, you can use row operations to simplify the matrix and transform it into its row-echelon form. From there, you can easily determine if the matrix is consistent or inconsistent, and find the solutions to the system of equations if they exist. You can also use other methods such as substitution or elimination to solve the system.

Are there any real-life applications of inconsistent and consistent augmented matrices?

Yes, augmented matrices are commonly used in various fields such as engineering, physics, economics, and computer science to solve systems of equations. For example, in engineering, augmented matrices are used to model and solve systems of equations for designing structures and systems. In economics, they are used to analyze supply and demand equations. In computer science, augmented matrices are used in image processing and machine learning algorithms.

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