I just started my second year of university and am taking Linear Algebra. I'm a bit rusty on all this systems of equations stuff, so I just want some confirmation, really. If don't have this right, I'd like some guidance
Anyways, the question is worded:
Find all solutions for the following systems of linear equations by writing the solution in parametric form. How many parameters are in the solution?
a.) x + y - z = 2
x - y = 0
Where there are 2 parameters with s and t arbitrary.
As I said, I'm feeling very rusty with this stuff and I don't remember our prof going over in class how to write SYSTEMS in parametric form, only 1 equation in parametric form. My book isn't helping much either.
Unfortunately, the question was given as is. I think she just requires that all Parametric form solutions be given, since the other questions in the section are such like:
b.) x + 2y - 3z = 4
x - 2y = 1
C.) x + 2y - 3z + u = 4
x + z = 0
d.) x + y + z + u + v + 10 = 0
Our profs first language is Russian, so the wording might just be a little wonky. As I said, I think she just needs the parametric form written and I need help with how to properly do that with a system of equations.
we set up a matrix containing the coefficients of the variables with the solution set in the last column, so we have:
.... .... .....
putting this in reduced row-echelon form, we have:
so now we see that the z-column is the only one without a leading 1, therefore, we replace z with a parameter. so our solution set is:
the rest are done in a similar fashion
I actually am just learning this (I just started first year.)
Like jhevon said, you need to first make an augmented matrix using the linear equations you're given. Then just work your way to reduced row echelon form.
Taken from my textbook...
A matrix is said to be in row-echelon form if it satisfies the following three conditions:
- 1. All zero rows (consisting entirely of zeros) are at the bottom
- 2. The first nonzero entry from the left in each nonzero row is 1, called the leading 1 for that row.
- 3. Each leading 1 is to the right of all leading 1's in the rows above it
A row-echelon matrix is said to be in reduced row echelon form if, in addition, it satisfies the following condition:
- 4. Each leading 1 is the only nonzero entry in its column.
As you can see from jhevon's post, you have one parameter, not two.
Do you remember the techniques to get the matrix to reduced row echelon form?
Step 1. If the matrix consists entirely of zeros, stop - it is already in row-echelon form.
Step 2. Otherwise, find the first column from the left containing a nonzero entry (call it a,) and move that row containing that entry to the top position.
Step 3. Now multiply that row by 1/a to create a leading 1.
Step 4. By subtracting multiples of that row from rows below it, make each entry below the leading 1 zero.
Step 5. Repeat steps 1-4 on the matrix consisting of the remaining rows.