# Linear Algebra solution to a system of equations

x + y+ z = 0
3x + 2y -2z = 0
4x + 3y -z = 0
6x + 5y + z = 0

## The Attempt at a Solution

I put the equations into a matrix and reduced to RREF. This is what I end up with:

x - 4z = 0
y + 5z = 0

The other two rows in the matrix are all zeroes.

I've never solved a system that had more equations than unknowns, so I'm confused on how many free variables I will need. Right now I have this as my solution:

x = 4t
y = -5t
z = t

## The Attempt at a Solution

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Dick
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x + y+ z = 0
3x + 2y -2z = 0
4x + 3y -z = 0
6x + 5y + z = 0

## The Attempt at a Solution

I put the equations into a matrix and reduced to RREF. This is what I end up with:

x - 4z = 0
y + 5z = 0

The other two rows in the matrix are all zeroes.

I've never solved a system that had more equations than unknowns, so I'm confused on how many free variables I will need. Right now I have this as my solution:

x = 4t
y = -5t
z = t

## The Attempt at a Solution

And that is just fine as a solution. There are really only two independent equations in there. The third one is the sum of the first two equations. The fourth is three times the first equation added to the second. They are redundant, as your RREF showed you.

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If you write the system of the equations in matrix form and you perform elementary row operations and put the matrix into row reduced echelon form then what is the rank of the matrix ? (Hint...which columns are independent and which are dependent ?) If the rank is r and the number of columns is n then n-r = the dimension of the nullspace for the coeficient matrix = the number of free variables. In this case the dimension for the nullspace is 1 since n - r = 3 - 2 = 1 and a basis for the nullspace is the vector c(4, -5, 1). So if c = 1 then x = 4, y = -5, z=1 is a solution. In other words you are correct.