# System of Equations

1. Nov 20, 2015

### says

1. The problem statement, all variables and given/known data
find the solutions to the following system, depending on the value of the parameter l∈ℝ:
lx+y+z=1
x+ly+z=l
x+y+lz=l2

Find the value of λ∈ℝ for which the following sytem of linear equations has infinitely many solutions, for such value of λ write down all solutions:

x - 2z = 8
x - 3y + λz = 4
2x - λy + z = 12

2. Relevant equations

3. The attempt at a solution
I'm not really sure what the questions are asking. Am I supposed to put both of these into an augmented matrix, complete gauss row reduction then rearrange to find the value for l, λ and x,y,z in each question?

2. Nov 20, 2015

### nuuskur

The system is exactly soluble if the system matrix is regular.
Find the value of $\lambda$ such that the determinant of the system matrix is zero, that means at least two rows of the system matrix are linearly dependent, which means you will have at least one free variable, which implies the number of solutions is infinite.

3. Nov 20, 2015

### says

I'm still a bit confused...
Not sure about the process I should use to complete this question

4. Nov 20, 2015

### nuuskur

This is for the $\lambda$-problem
Consider the system matrix
$$\left (\begin{array}{ccc}1 &0&-2\\1&-3&\lambda\\2&-\lambda &1\end{array}\right )$$
For what value of $\lambda$ is the determinant of this matrix zero?

Please create separate topics for each assignment. Short of your post not being formated in TeX, which makes it difficult to follow, there are two problems to be considered.

Last edited: Nov 20, 2015
5. Nov 20, 2015

### says

Ahhhh, so a system of linear equations that has infinite solutions is linearly dependent, and a linearly dependent set of vectors has a determinant of 0!

I think that should make this easier to solve.

6. Nov 20, 2015

### nuuskur

Linear dependence is defined in a vectorspace. Equations, on the other hand, are not vectors. Essentially, what you are saying is right, but your wording is sloppy.
$n$ $n$-dimensional vectors (of the same vectorspace, naturally) can be written in matrix form and then it can be determined whether they are linearly (in)dependent.

Your arguments do not hold for general cases, be careful with that.

7. Nov 20, 2015

### says

Now I'm even more lost! I tried to calculate the determinant but I got 'i' hahaha

8. Nov 20, 2015

### nuuskur

How could you have been able to calculate the determinant? You don't know what $\lambda$ is. The objective is to determine for which value of $\lambda$ the determinant of said matrix is zero. In other words, for which value of $\lambda$ the row vectors of the matrix are linearly dependent.
The polynomial needed to be solved is: $\lambda ^2 +2\lambda -15 = 0$ and it has roots in $\mathbb{R}$, check your calculations.

9. Nov 20, 2015

### says

Determinant = 0 = 1 ((-3*1)-(λ*-λ) - 2((1*-λ)-(-3*2))

= λ2+2λ=-3

10. Nov 20, 2015

### nuuskur

Going by Sarrus rule:
$-3 + 0 +2\lambda -12 +\lambda ^2 + 0 = 0$

E: The rule of sarrus

11. Nov 20, 2015

### says

Sarrus rule?

12. Nov 20, 2015

### says

i forgot to multiply by 2

13. Nov 20, 2015

### says

−3+λ2+2λ-12=0
λ2+2λ = 15

lambda = 3!

not 3 factorial, just excited ;)

14. Nov 20, 2015

### nuuskur

3 is correct, but there is one more.

There are three kinds of systems of linear equations:
1) The ones that can be solved
2) The ones that can be solved exactly
3) The ones that have no solution

Your objective now is to verify that for either $\lambda$ the system in question satisfies 1). It cannot satisfy 2), this is a result in linear algebra. Also try to think when the system would not be soluble.

15. Nov 20, 2015

### says

-5 is the other solution
\However, if we put -5 into the matrix, we no longer have a linearly dependent set.

16. Nov 20, 2015

### nuuskur

You just determined that for $\lambda = -5$ the determinant is 0, therefore the vectors Are linearly dependent. Question now is, when does a system conflict? (no solutions)
Consider as an example
$x+y=0$
$x+y=1$

Why does this system conflict?

Last edited: Nov 20, 2015
17. Nov 20, 2015

### Ray Vickson

For the first question, there are two "easy" cases: (1) $l=0$; and (2) $l = 1$. Write out the equations in detail in both of those cases, and you can essentially complete the solution by "inspection" almost.

That leaves the case $l \neq 0,1$. In that case, I recommend you just avoid echelon forms (and even matrices) altogether and just go ahead and start solving the old-fashioned way. You can use the first equation to get z in terms of x and y, then substitute that expression in place of z in the next two equations. That will leave you with two equations in the two unknowns x and y, which you can tackle in the same way: solve for y (say) in terms of x from the second equation (say), then substitute that expression into the third equation, getting a single equation in terms of x alone. You can carry out these steps as long as you are never dividing by 0 at any stage, and that tells you something about $l$.

You can deal with the second set of equations in the same way, and the "never divide by zero" rule will tell you something about $\lambda$.

18. Nov 20, 2015

### ehild

Yes, that is the usual procedure.