Linear Algebra - checking a system for consistence.

[MJay82]

Homework Statement

Determine if the system is consistent. Do not completely solve the system.
I don't know if there's an easy way to input a matrix. These are the numbers from an augmented matrix:
2 / 0 / 0 / -4 / -10
0 / 3 / 3 / 0 / 0
0 / 0 / 1 / 4 / -1
-3 / 2 / 3 / 1 / 5

Homework Equations

Row operations used: Replacement and scaling.

The Attempt at a Solution

So I think I did this right, but the problem assigned didn't have a solution in the back of the book.

My answer is that the system is not consistent. The operations that I performed, in order, are:
1) Scale (1/2)R1 and (1/3)R2
2) Replace R4 with SUM (R4 + (3)R1)
3) Replace R4 with SUM (R4 + (-2)R2)
4) Replace R4 with SUM (R4 + (-1)R3)
5) Replace R4 with SUM (R4 + R3)

This yielded an R4 of:
0 0 0 0 -9, which cannot be true.

 

[MJay82]

Apparently the board reduces more than 1 space to 1 space. Maybe I'll go back and put in slashes.

 

[Mark44]

Homework Statement

Determine if the system is consistent. Do not completely solve the system.
I don't know if there's an easy way to input a matrix. These are the numbers from an augmented matrix:
2 / 0 / 0 / -4 / -10
0 / 3 / 3 / 0 / 0
0 / 0 / 1 / 4 / -1
-3 / 2 / 3 / 1 / 5

Homework Equations

Row operations used: Replacement and scaling.

The Attempt at a Solution

So I think I did this right, but the problem assigned didn't have a solution in the back of the book.

My answer is that the system is not consistent. The operations that I performed, in order, are:
1) Scale (1/2)R1 and (1/3)R2
2) Replace R4 with SUM (R4 + (3)R1)
3) Replace R4 with SUM (R4 + (-2)R2)
4) Replace R4 with SUM (R4 + (-1)R3)
5) Replace R4 with SUM (R4 + R3)

This yielded an R4 of:
0 0 0 0 -9, which cannot be true.

I'm pretty sure you made a mistake. I get a solution for the system, which means that it is consistent. In any case, you are not asked for a solution - just determine whether the system is consistent. One way to do that is to take the determinant of the 4 x 4 matrix that is the left part of your augmented matrix. If the determinant is nonzero, the system is consistent; otherwise the system is inconsistent.

 

[MJay82]

Thanks for the quick response, Mark. I'm going to rework it. Am I wrong to think that if a row (x1 - x4 in this case) is all 0's, and the other side (the constant - I guess) is non-zero, that the system is inconsistent?
In other words, I thought that by rule
0(x1) + 0(x2) + 0(x3) + 0(x4) = C
meant that the system is inconsistent.

 

[JonF]

Yes you are correct, that would mean that it is inconsistent.

Taking the determinant of a 4x4 matrix is a pain. I suggest thinking about what this row tells you for a sec: 0 / 3 / 3 / 0 / 0. It should give you the information needed to make this 4x4 a 3x3.

 

[MJay82]

Jon, I haven't learned how to perform an operation like that yet. I obviously notice that x2 = -x3, but don't know what that means yet.

I reworked the problem and very quickly found where I made a careless mistake, and amend my answer to "The system is consistent."
Mark, you said you took it to a solution, right? I didn't finish it, but I found x4 = 1. Is this correct?
Thanks so much for all the help!

 

[JonF]

How did you get the system is consistent without solving it? Have you learned about determinants? If so that is probably the method they want you to use

All I suggested was realizing x3 = -x2, so you could clean up the system a bit to make it easier on yourself before you throw it in a matrix. Just perform the substitution of x3 = -x2.

So that means you could deal with this instead:
2 / 0 / -4 / -10
0 / -1 / 4 / -1
-3 / -1 / 1 / 5

 

[Ray Vickson]

I'm pretty sure you made a mistake. I get a solution for the system, which means that it is consistent. In any case, you are not asked for a solution - just determine whether the system is consistent. One way to do that is to take the determinant of the 4 x 4 matrix that is the left part of your augmented matrix. If the determinant is nonzero, the system is consistent; otherwise the system is inconsistent.

This last statement is incorrect: whether or not a system with zero coefficient determinant is inconsistent depends on the right-had-side. For some right-hand-sides the system will have a (non-unique) solution, so will be consistent, but for others there will be no solution. Simple example: x+y=1 and 2x=2y=2 are consistent, while x=y=1 and 2x+2y=3 are inconsistent.

RGV

 

[MJay82]

How did you get the system is consistent without solving it? Have you learned about determinants? If so that is probably the method they want you to use

All I suggested was realizing x3 = -x2, so you could clean up the system a bit to make it easier on yourself before you throw it in a matrix. Just perform the substitution of x3 = -x2.

So that means you could deal with this instead:
2 / 0 / -4 / -10
0 / -1 / 4 / -1
-3 / -1 / 1 / 5

I should have said that I didn't solve it COMPLETELY. I took it to reduced echelon form and confirmed that there was no row with a pivot in the augmented column, or in other words, of the form [0 0 0 b] where b is nonzero.

Thanks for the tip about what seems like some sort of column simplification step. Using your roadmap, I'll work out how you simplified the matrix.

 

[Mark44]

This last statement is incorrect: whether or not a system with zero coefficient determinant is inconsistent depends on the right-had-side. For some right-hand-sides the system will have a (non-unique) solution, so will be consistent, but for others there will be no solution. Simple example: x+y=1 and 2x=2y=2 are consistent, while x=y=1 and 2x+2y=3 are inconsistent.

What I said about consistency and the determinant can be applied only to systems of n equations in n unknowns, so the the related matrices are square.

In your first example, the system could be written as
x + y = 1
x + 0y= 1
0x + y = 1

The system for the second example could be written as
x + y = 3/2
x + 0y= 1
0x + y = 1

Both systems have three equations in two unknowns, so determinant methods aren't applicable.

If you have a system of n equations in n unknowns (as in the OP's problem, with n = 4), a nonzero determinant means that the system is consistent and that there is a unique solution.

 

[Ray Vickson]

What I said about consistency and the determinant can be applied only to systems of n equations in n unknowns, so the the related matrices are square.

In your first example, the system could be written as
x + y = 1
x + 0y= 1
0x + y = 1

The system for the second example could be written as
x + y = 3/2
x + 0y= 1
0x + y = 1

Both systems have three equations in two unknowns, so determinant methods aren't applicable.

If you have a system of n equations in n unknowns (as in the OP's problem, with n = 4), a nonzero determinant means that the system is consistent and that there is a unique solution.

Sorry: those were typos; I am using a new, unfamiliar keyboard. I meant: first system is x+y=1 and 2x+2y=2; the second system is x+y=1 and 2x+2y=3. Anyway, there is a theorem in linear algebra that says for an nxn system, the equations Ax=b have a unique solution if det(A) is nonzero, and have either no solution or infinitely many solutions if det(A) = 0. Whether or not there are any solutions in the latter case depends on b.

RGV

 

[Mark44]

Sorry: those were typos; I am using a new, unfamiliar keyboard. I meant: first system is x+y=1 and 2x+2y=2; the second system is x+y=1 and 2x+2y=3. Anyway, there is a theorem in linear algebra that says for an nxn system, the equations Ax=b have a unique solution if det(A) is nonzero, and have either no solution or infinitely many solutions if det(A) = 0. Whether or not there are any solutions in the latter case depends on b.

I thought the first one might be a typo, but the second system had what turns out to be the same typo, so I thought you meant x = y = 1. Anyway, now I get what happened.

Regarding the LA theorem, that's what I was using when I was talking about determinants.