Express the solutions of the following systems in terms of the free variables:
x1 + 3*(x2) - 2*(x3) + 2*(x5) = 0
2*(x1) + 6*(x2) - 5*(x3) - 2*(x4) + 4*(x5) - 3*(x6) = -1
5*(x3) + 10*(x4) + 15*(x6) = 5
2*(x1) + 6*(x2) + 8*(x4) + 4*(x5) + 18*(x6) = 6
Hey Linus,
First step: simplify where you can. For example, equation 3 can be divided by 5, so you end up with (x3) + 2(x4) + 3(x6) = 1. Equation 4 can also be simplified. This will make the algebra later a little easier.
Second step: Pick an x (x1 seems to be a good one) to start solving, and begin moving around your equations.
Do you know where to go from here? (Or are you onto matrices already? If so ignore step 2 and build your matrix). If you're stuck show me what you've done so far and I'll help out!
I got:
x1 = -3*(x2) -2*(x5) + 66*(x6) - 22
x2 = free
x3 = 33*(x6) - 11
x4 = -18*(x6) + 6
x5 = free
x6 = free
Is that correct?
To express the solutions of the given system in terms of the free variables, we first need to obtain the row-echelon form or reduced row-echelon form of the augmented matrix associated with the system. Then, by analyzing the resulting rows, we can determine the value of the free variables and express the solutions accordingly.
Let's proceed step by step:
1. Write the augmented matrix for the given system:
[1 3 -2 0 2 0 | 0]
[2 6 -5 -2 4 -3 | -1]
[0 0 5 10 0 15 | 5]
[2 6 0 8 4 18 | 6]
2. Apply row operations to obtain the row-echelon form. Here, we can use Gaussian elimination:
a) Replace R2 with R2 - 2R1:
[1 3 -2 0 2 0 | 0]
[0 0 1 -2 0 -3 | -1]
[0 0 5 10 0 15 | 5]
[2 6 0 8 4 18 | 6]
b) Replace R4 with R4 - 2R1:
[1 3 -2 0 2 0 | 0]
[0 0 1 -2 0 -3 | -1]
[0 0 5 10 0 15 | 5]
[0 0 4 8 0 18 | 6]
c) Replace R4 with R4 - (4/5)R3:
[1 3 -2 0 2 0 | 0]
[0 0 1 -2 0 -3 | -1]
[0 0 5 10 0 15 | 5]
[0 0 0 0 0 -2 | 4]
So, the row-echelon form of the augmented matrix is:
[1 3 -2 0 2 0 | 0]
[0 0 1 -2 0 -3 | -1]
[0 0 5 10 0 15 | 5]
[0 0 0 0 0 -2 | 4]
3. Perform additional row operations to achieve reduced row-echelon form. Here, we'll focus on the last row, which represents an equation with only the free variable x6:
a) Multiply R4 by (-1/2):
[1 3 -2 0 2 0 | 0]
[0 0 1 -2 0 -3 | -1]
[0 0 5 10 0 15 | 5]
[0 0 0 0 0 1 | -2]
b) Replace R3 with R3 - 5R4:
[1 3 -2 0 2 0 | 0]
[0 0 1 -2 0 -3 | -1]
[0 0 0 0 0 10 | 15]
[0 0 0 0 0 1 | -2]
c) Replace R2 with R2 + 2R4:
[1 3 -2 0 2 0 | 0]
[0 0 1 -2 0 -1 | -3]
[0 0 0 0 0 10 | 15]
[0 0 0 0 0 1 | -2]
d) Replace R1 with R1 + 2R4:
[1 3 -2 0 2 -2 | -4]
[0 0 1 -2 0 -1 | -3]
[0 0 0 0 0 10 | 15]
[0 0 0 0 0 1 | -2]
4. Analyzing the final row-echelon form:
From the last row, we have the equation 0*x1 + 0*x2 + 0*x3 + 0*x4 + 0*x5 + 1*x6 = -2, which simplifies to x6 = -2.
The other variables x1, x2, x3, x4, and x5 are the free variables since they do not have pivots in the row-echelon form. Therefore, they can take any value.
So, the solutions of the given system in terms of the free variables are:
x1 = a (where "a" is any real number)
x2 = b (where "b" is any real number)
x3 = c (where "c" is any real number)
x4 = d (where "d" is any real number)
x5 = e (where "e" is any real number)
x6 = -2
Note: The values of x1, x2, x3, x4, and x5 will depend on the chosen values for the free variables.