Linear Equations - Cramer's Rule

[ZedCar]

Homework Statement

Does the following set of linear equations have a unique solution?

(excluding any trivial solutions when x=y=z=0)

Do not attempt to formally solve the equations.

x+2y-4z=8
4x-6y+12z=19
-6x+3y-6z=-20

Homework Equations

The Attempt at a Solution

My immediate thought would be to solve the equations using Cramers Rule. However, it states not to formally solve the equations. So I assume I'm not allowed to do this.

So what does the question mean me to do?

 

[Mark44]

Homework Statement

Does the following set of linear equations have a unique solution?

(excluding any trivial solutions when x=y=z=0)

Do not attempt to formally solve the equations.

x+2y-4z=8
4x-6y+12z=19
-6x+3y-6z=-20

Homework Equations

The Attempt at a Solution

My immediate thought would be to solve the equations using Cramers Rule. However, it states not to formally solve the equations. So I assume I'm not allowed to do this.

So what does the question mean me to do?

Rewrite the system of three equations as a matrix equation in the form Ax = b. Here x represents the vector <x, y, z>^T, and b represents the constant vector <8, 19, -20>^T. What conditions on matrix A guarantee a unique solution to the equation?

 

[ZedCar]

By ^T do you mean transpose?

Just wanted to check before attempting to try and solve this.

 

[ZedCar]

If I take the matrix;

1, 2, -4
4, -6, 12
-6, 3, -6

The determinant = 0

Therefore it is not invertible.

Therefore this implies it has no unique solution.

Is this correct?

 

[Mark44]

By ^T do you mean transpose?

Just wanted to check before attempting to try and solve this.

Yes.

If I take the matrix;

1, 2, -4
4, -6, 12
-6, 3, -6

The determinant = 0

Therefore it is not invertible.

Therefore this implies it has no unique solution.

Is this correct?

A better way is to say that it has no solution. A system of equations can have
1) a unique solution
2) multiple solutions (an infinite number of them)
3) no solutions

If you say "no unique solution" this might be interpreted as multiple solutions.

 

[ZedCar]

Okay. Thank you.

 

[HallsofIvy]

But isn't "no unique solution" what he wants to say? The question, after all, was "Does the following set of linear equations have a unique solution?" The fact that the determinant of this 3 by 3 set of equations is 0 means it maps R³ into a proper subspace of R³. If the right hand side doesn't happen to be in that subspace, there is no solution but if it does, an entire subspace will be mapped to it.

 

[Mark44]

But isn't "no unique solution" what he wants to say? The question, after all, was "Does the following set of linear equations have a unique solution?" The fact that the determinant of this 3 by 3 set of equations is 0 means it maps R³ into a proper subspace of R³. If the right hand side doesn't happen to be in that subspace, there is no solution but if it does, an entire subspace will be mapped to it.

For this particular problem, there is no solution. I am distinguishing between "no unique solution" and "no solution" as the former term might be interpreted by some to mean that there are multiple solutions (i.e., not a unique solution). That was the distinction I was trying to make.