Solving Multivariable Linear Systems:

3x-5y+5z = 1
5x-2y+3z = 0
7x-y+z = 5

I am confused on how to solve this. I'm supposed to use the Row-Echelon form.

Thanks.

argghhh! I don't like that method, however this page shows you step by step what to do

http://www.math.odu.edu/~bogacki/cgi-bin/lat.cgi?c=rref

enter: 3 rows, 4 columns, then

3 -5 5 1
5 -2 3 0
7 -1 1 5

Wow, thanks.

To solve this system of linear equations using the row-echelon form, follow these steps:

Step 1: Create an augmented matrix
Write down the coefficients of the variables and the constant terms as a matrix. For the given system, the augmented matrix would be:

[3 -5 5 | 1]
[5 -2 3 | 0]
[7 -1 1 | 5]

Step 2: Perform row operations to obtain zeros below the main diagonal
The goal is to transform the matrix into an upper triangular form, also known as the row-echelon form. Start by focusing on the first column and make the first entry (3 in this case) equal to 1.

To make the first entry 1, divide the first row by 3 (R1 = R1/3). The new matrix is:

[1 -5/3 5/3 | 1/3]
[5 -2 3 | 0]
[7 -1 1 | 5]

Next, create zeros below the first entry by eliminating the first element in the second and third rows. To do this, multiply the first row by -5 and add it to the second row. Similarly, multiply the first row by -7 and add it to the third row. After this step, the matrix becomes:

[1 -5/3 5/3 | 1/3]
[0 7 4 | -5/3]
[0 28/3 -2/3 | 32/3]

Step 3: Proceed column by column
Move to the second column and make the second entry (7) equal to 1. Divide the second row by 7 (R2 = R2/7). The matrix now looks like:

[1 -5/3 5/3 | 1/3]
[0 1 4/7 | -5/21]
[0 28/3 -2/3 | 32/3]

Create zeros below the second entry by eliminating the second element in the first and third rows. Multiply the second row by 5/3 and add it to the first row. Multiply the second row by -28/3 and add it to the third row. The matrix becomes:

[1 0 -11/7 | 5/7]
[0 1 4/7 | -5/21]
[0 0 -110/21 | 49/3]

Step 4: Solve for the variables
The matrix is now in row-echelon form. To solve for the variables, go from the bottom row to the top row, back-substituting each variable's value.

From the last row, we can conclude that -110/21z = 49/3. Solving for z:

z = (49/3) / (-110/21)
z = -7/10

Moving up to the second row, we now know that y + (4/7)(-7/10) = -5/21. Solving for y:

y = -5/21 - 4/7*(-7/10)
y = 1/5

Finally, from the first row, we have x + (5/3)(1/5) - (5/3)(-7/10) = 1/3. Solving for x:

x = 1/3 - 5/15 + 7/15
x = 3/5

So the solution to the multivariable linear system is:
x = 3/5
y = 1/5
z = -7/10