# Find the solutions of the system, for all λ

• Michael_0039
In summary: If ##\lambda \neq 0## and ##\lambda \neq 1##, you have three planes that all intersect at a single point, thus one unique solution.
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?

Last edited by a moderator:
Michael_0039 said:
∈(∀λ∃ℝ)
∃ means "there exists". The symbol you want is ∈.
Michael_0039 said:
-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?
Michael_0039 said:
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.

Mark44 said:
∃ 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

Michael_0039 said:
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].

Mark44 said:
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

Michael_0039 said:
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.

Michael_0039 said:
-x + y - z = 1
-2x + 10y + (2λ + 6) = 6
3x + 11y + (λ2+6)z = 5λ - 1

Michael_0039
Mark44 said:
Is the system above the correct one? The original system of post #1, copied below, was quite different.
Yes, my apologies.

Post #5:
Michael_0039 said:
-x + 4y - 2z = 1
-2x + 10y + (2λ - 4)z = 6
3x - 11y + (λ2+6)z = 5λ - 1
Post #1:
Michael_0039 said:
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.

chwala and Michael_0039

## What does λ represent in the system of equations?

λ (lambda) typically represents a parameter or a scalar value that can affect the solutions of the system of equations. It is often used to denote eigenvalues in linear algebra or a parameter in a family of solutions.

## How do you determine if a system of equations has solutions for all values of λ?

To determine if a system of equations has solutions for all values of λ, you can analyze the determinant of the coefficient matrix. If the determinant is non-zero, the system has a unique solution for all λ. If the determinant is zero, you need to check the rank and consistency of the augmented matrix to determine if there are solutions.

## What methods can be used to find solutions of the system for all λ?

Common methods to find solutions of the system for all λ include Gaussian elimination, matrix inversion (if applicable), and eigenvalue/eigenvector analysis. Additionally, for specific types of systems, you might use specialized techniques like Cramer's rule or the characteristic equation.

## What happens if λ is an eigenvalue of the coefficient matrix?

If λ is an eigenvalue of the coefficient matrix, the system may have either no solutions or an infinite number of solutions, depending on the consistency of the system. This is because the coefficient matrix becomes singular (non-invertible) when λ is an eigenvalue, leading to a determinant of zero.

## Can the solutions of the system be expressed in terms of λ?

Yes, the solutions of the system can often be expressed in terms of λ, especially if λ is a parameter that influences the system. The solutions can be written as functions of λ, showing how the solution set changes as λ varies. This is common in parameterized systems and in the study of differential equations and linear algebra.

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