Explanation of solution to given question (subspace)

[negation]

Homework Statement

-4x₁ +2x₂ =2
4x₁ -3x₂ -2x₃ =-3
2x₁ -x₂ +(k - k²)x₃ =-k

Find the values of k for which the system has 1) unique solution, 2) infinitely many solution and 3) no solution

The Attempt at a Solution

In REF: the matrix is
-4 2 0 | 2
0 -1 -2 | -1
0 0 k(1-k) | 1-k


1) I do not understand why the solution sets x₃ = y to give a single parameter solution set.


2) for a system to have infinitely solution k must be 1 because 1(1-1) = 0 and 1-1 = 0
0 = 0 implies infinitely many solution

3) for a system to have no solution, the solution must be inconsistent
and so, k(1-k) must equal to 0 for some value k and 1-k gives a value that is an element of real number with the exclusion of 0. For this to be true, k = 0.
0(1-0) = 0 and 1-0 = 1
we see that 0 = 1 but this produces a contradiction.

 

[ehild]

Homework Statement

-4x₁ +2x₂ =2
4x₁ -3x₂ -2x₃ =-3
2x₁ -x₂ +(k - k²)x₃ =-k

Find the values of k for which the system has 1) unique solution, 2) infinitely many solution and 3) no solution

The Attempt at a Solution

In REF: the matrix is
-4 2 0 | 2
0 -1 -2 | -1
0 0 k(1-k) | 1-k


1) I do not understand why the solution sets x₃ = y to give a single parameter solution set.

If k=1 x₃ is arbitrary. y means just an arbitrary number. And you can write the solution set in terms of y.

2) for a system to have infinitely solution k must be 1 because 1(1-1) = 0 and 1-1 = 0
0 = 0 implies infinitely many solution

3) for a system to have no solution, the solution must be inconsistent
and so, k(1-k) must equal to 0 for some value k and 1-k gives a value that is an element of real number with the exclusion of 0. For this to be true, k = 0.
0(1-0) = 0 and 1-0 = 1
we see that 0 = 1 but this produces a contradiction.

Correct. But what happens if k≠1 and k≠0? What is (are) the solution(s)?


ehild

 

[negation]

If k=1 x₃ is arbitrary. y means just an arbitrary number. And you can write the solution set in terms of y.



Correct. But what happens if k≠1 and k≠0? What is (are) the solution(s)?


ehild

I'm still not capturing the full essence of part (1). Maybe some simple examples?

part(3): if k ≠ 1 and k ≠ 0 then we can pick any values of k? But what is the corollary from this?

 

[ehild]

Part (1) asks about unique solution (for a given k) K can be anything, but 1 and zero. What is the solution set in terms of k?

Part(2) : infinite many solution exist if k=1, as x3 can be arbitrary. What are x1 and x2 in terms of x3?

Part 3 : no solution exist when k=0.

(1) Remember that a row of your matrix equation means an equation for x1,x2,x3. The last row of the reduced matrix means the equation k(1-k) x3=1-k. If k≠1 and k≠0 you get x3=1/k.
The second row means equation -x2-2x3=-1. Substitute 1/k for x3: -x2-2/k=-1, x2=1-2/k.
Continue with the first row and find x1.
Example: k=2. x3=0.5, x2=0.

You see that the solution set is unique for a given value of k.
In case (2) k=1 and x3 is arbitrary. x3=y
The second raw means -x2-2x3=-1→-x2-2y=-1→x2=1-2y.
Do the same with the first row.
If y=0, for example, x3=y=0, x2=1 and x1=0.

ehild

 

[negation]

Part (1) asks about unique solution (for a given k) K can be anything, but 1 and zero. What is the solution set in terms of k?

Part(2) : infinite many solution exist if k=1, as x3 can be arbitrary. What are x1 and x2 in terms of x3?

Part 3 : no solution exist when k=0.

(1) Remember that a row of your matrix equation means an equation for x1,x2,x3. The last row of the reduced matrix means the equation k(1-k) x3=1-k. If k≠1 and k≠0 you get x3=1/k.
The second row means equation -x2-2x3=-1. Substitute 1/k for x3: -x2-2/k=-1, x2=1-2/k.
Continue with the first row and find x1.
Example: k=2. x3=0.5, x2=0.

You see that the solution set is unique for a given value of k.
In case (2) k=1 and x3 is arbitrary. x3=y
The second raw means -x2-2x3=-1→-x2-2y=-1→x2=1-2y.
Do the same with the first row.
If y=0, for example, x3=y=0, x2=1 and x1=0.

ehild

It is clear now. I was earlier able to work to the part where k(1-k)x₃ = (-k+1)
simplifying produces x₃ = 1/k but was confused as to what I should do with what I had especially since it conflicted with what my answer key showed.

In response to your question, x₃ = 1/k. If x₃ = y, 1/k = y.

∴ x₃ = y

-x₂ -2x₃ = -1
-x₂ -2y = -1
∴ x₂ = 1-2y

-4x₁ +2x₂ =2
-4x₁ +2(1-2y) 2
∴x₁ =-y

1/k can be substituted into y (preferences)

 

[ehild]

It is clear now. I was earlier able to work to the part where k(1-k)x₃ = (-k+1)
simplifying produces x₃ = 1/k but was confused as to what I should do with what I had especially since it conflicted with what my answer key showed.

In response to your question, x₃ = 1/k. If x₃ = y, 1/k = y.

No, y comes in when k=1.

If k is neither 1 nor zero, there is a unique solution , and that is case (1)

∴ x₃ = [STRIKE]y[/STRIKE]=1/k

-x₂ -2x₃ = -1
-x₂ -2y = -1
∴ x₂ = 1-[STRIKE]2y[/STRIKE]2/k

-4x₁ +2x₂ =2
-4x₁ +2(1-2y) 2
∴x₁ =-[STRIKE]y[/STRIKE]1/k
[STRIKE]
1/k can be substituted into y (preferences)[/STRIKE]

If k=1 you have a free parameter - y - in the solution. That is case (2)

x3=y; x2=1-2y, x1=? ehild

 

[negation]

No, y comes in when k=1.

If k is neither 1 nor zero, there is a unique solution , and that is case (1)



If k=1 you have a free parameter - y - in the solution. That is case (2)

x3=y; x2=1-2y, x1=?


ehild

I should have known not to complicate the answer by setting 1/k = y.

But I have this one.

If k = 1, we have 1(1-1) | (-1+1) and this produces 0 = 0.
We have an infinite solution set.
0x₃ = 0
setting x₃ = y we see that via back substituition

x₂ = 1-2y
x₁ = [2+2(1-2y)]/-4 (lazy to simplify but I have the conceptual idea)

 

[ehild]

x₂ = 1-2y
x₁ = [2+2(1-2y)]/-4 (lazy to simplify but I have the conceptual idea)

Good!

ehild