Pre Calc Help D:

In rref form the left side of the matrix should be an identity matrix.

Actually it only needs to be in "Reduced Row Echelon" form. The Identity matrix is part of that group of matrices =P

• All nonzero rows are above any rows of all zeroes.
• The leading coefficient (also called pivot) of a row is always strictly to the right of the leading coefficient of the row above it.
• Every leading coefficient is 1 and is the only nonzero entry in its column.
  

In the matrix:

| 1 0 -3 -10|
| 0 1 3 34|
| 0 0 0 0 |

the threes in column three just signify that it's a free variable. You can define the set of equations being represented by the matrix as a 1-parameter family of solutions, or a line in 3-space dimension.

Of course, an Identity matrix means there's only 1 solution, which is preferred =P But if you have the Reduced Row Echelon form of a matrix, and it's not an Identity matrix, then it has infinite solutions instead. You'll never get an Identity matrix in that case.


EDIT: I got the same solution HooKarez found for problem 2. In this case, you can state the solution as:

x = 3z - 10
y = 34 - 3z

Since we're dealing with the real world in this problem, there must be a domain where the set of functions is valid. Usually it's the unique solution or [0,+infinity). Here however, x, y, and z ≥ 0 since I find it extremely hard to have a negative amount of chairs (and this isn't economics where debt exists). Therefore:

3z - 10 ≥ 0
34 - 3z ≥ 0

Or,

10 ≤ 3z ≤ 34

Or,

3 < z < 12

since 3*3 = 9, which is less than 10, and 3*12 = 36, which is greater than 34, and that can't happen.