Find the solutions of the system, for all λ

[Michael_0039]

Thread moved from technical math section, so there is no homework template.
(∀λ∃ℝ)

-x + y - z = 1
-2x + 10y + (2λ + 6) = 6
3x + 11y + (λ²+6)z = 5λ - 1

after gaussian elimination I have this:

-1 4 -2 | 1
0 1 λ | 2
0 0 λ(λ-1) | 5λ

So, for λ=0 ⇒ ∞ solutions, for λ=1 system is impossible and for λ≠0 and λ≠1 only one solution.

Am I right?

 

[Mark44]

∈(∀λ∃ℝ)

∃ means "there exists". The symbol you want is ∈.

-x + y - z = 1
-2x + 10y + (2λ + 6) = 6
3x + 11y + (λ²+6)z = 5λ - 1

In the 2nd equation is there supposed to be a z on the left side?

after gaussian elimination I have this:

-1 4 -2 | 1
0 1 λ | 2
0 0 λ(λ-1) | 5λ

So, for λ=0 ⇒ ∞ solutions, for λ=1 system is impossible and for λ≠0 and λ≠1 only one solution.

Am I right?

I'm not getting the final augmented matrix you show, but then I'm not sure what the starting system is.

 

[Michael_0039]

∃ means "there exists". The symbol you want is ∈.
In the 2nd equation is there supposed to be a z on the left side?
I'm not getting the final augmented matrix you show, but then I'm not sure what the starting system is.

Yes, my mistake I mean "∈" and also for the 2nd equation I forgot the z, you are right.
The final augnebted matrix it came after calculations between rows, as shown bellow:

1) R₂ → R₂ - 2R₁
2) R₃ → R₃ + 3R₁
3) R₃ → R₃ - R₂

 

[Mark44]

1) R₂ → R₂ - 2R₁
2) R₃ → R₃ + 3R₁
3) R₃ → R₃ - R₂

After the first two row operations above, I get this augmented matrix:
$\begin{bmatrix}-1 & 1 & -1 & | & 1 \\ 0 & 8 & 2\lambda + 8 & | & 4 \\ 0 & 14 & \lambda^2 + 3 & | & 5\lambda - 4 \end{bmatrix}$
I don't see how your third row operation is useful, as it doesn't eliminate the leading entry of row 3.
Also, none of your three row operations modified row 1, so I don't see how you ended up with an augmented matrix whose first row was [-1 4 -2 | 1].

 

[Michael_0039]

After the first two row operations above, I get this augmented matrix:
$\begin{bmatrix}-1 & 1 & -1 & | & 1 \\ 0 & 8 & 2\lambda + 8 & | & 4 \\ 0 & 14 & \lambda^2 + 3 & | & 5\lambda - 4 \end{bmatrix}$
I don't see how your third row operation is useful, as it doesn't eliminate the leading entry of row 3.
Also, none of your three row operations modified row 1, so I don't see how you ended up with an augmented matrix whose first row was [-1 4 -2 | 1].

I am very sorry, I was careless on this post. Correction: -x + 4y - 2z = 1
-2x + 10y + (2λ - 4)z = 6
3x - 11y + (λ²+6)z = 5λ - 1

 

[Mark44]

I am very sorry, I was careless on this post. Correction:

-x + 4y - 2z = 1
-2x + 10y + (2λ - 4)z = 6
3x - 11y + (λ²+6)z = 5λ - 1

Is the system above the correct one? The original system of post #1, copied below, was quite different.

-x + y - z = 1
-2x + 10y + (2λ + 6) = 6
3x + 11y + (λ2+6)z = 5λ - 1

 

[Michael_0039]

Is the system above the correct one? The original system of post #1, copied below, was quite different.

Yes, my apologies.

 

[Mark44]

Post #5:

-x + 4y - 2z = 1
-2x + 10y + (2λ - 4)z = 6
3x - 11y + (λ2+6)z = 5λ - 1

Post #1:

after gaussian elimination I have this:

-1 4 -2 | 1
0 1 λ | 2
0 0 λ(λ-1) | 5λ

So, for λ=0 ⇒ ∞ solutions, for λ=1 system is impossible and for λ≠0 and λ≠1 only one solution.

With the corrected system shown in post #5, I agree with the solution you posted in the OP.

Geometrically, if $\lambda = 0$ you have two planes that intersect in a line, thus there are an infinite number of solutions. If $\lambda = 1$, you have an inconsistent system, thus no solutions. In that case, the third equation becomes $0z = 5$, which can't happen.