Linear System with Parameter: Solutions and Homogeneity

[Felafel]

Homework Statement

I have done this exercise, but I don't have a file with the solutions. COuld you please check it?
Thank you in advance :)

Given the following system:
$\lambda \in \mathbb{R}$

$x − z = \lambda$
$x + y + 2z + t = 0$
$y + 3z =$
$x + z + t = 0$

1-find $rk(A_{\lambda}) and rk(A_{\lambda}, B_{\lambda})$ according to the different values of $\lambda$

2- for which values of $\lambda$ does tha system have solutions?

3- what is its solutions set?

4- for which values is the system homogeneous ?

The Attempt at a Solution

This is the matrix $(A_{\lambda}, B_{\lambda})$
( 1 0 1 -1 λ)
( λ 1 2 1 0)
( 0 1 3 0 λ)
( 1 0 1 λ 0)

doing some row reduction i get:

( 1 0 -1 0 λ )
( 0 0 2 λ -λ )
(λ-1 0 0 1 -2λ)
( 0 1 3 0 λ)

and i see rk(A)=rk(A,B) for any $\lambda$, so according to Rouchè-Capelli's theorem the system has solutions.

So, answers to 1 and 2 are: 1- rk(A)=rk(A,B) for any $\lambda \in \mathbb{R}$ 2-$\forany \lambda \in \mathbb{R}$

Now, if $\lambda$=1 i get:
(1 0 -1 0 1)
(0 0 2 1 -1)
(0 0 0 1 -2)
(0 1 3 0 1) thus: x=3/2, y= -1/2, z=1/2, t=-2 is the only solution

Doing the same, if $\lambda$=0 i get (x, y, z, t)=(0, 0, 0, 0)
(IS IT ACCEPTABLE AS A SOLUTION?)

If $\lambda$ is different from 0 and 1,

x-z=$\lambda$
2z+λt=-λ
(λ-1)x+t=-2λ
3z+y=λ

with z= $\alpha$
$\Sigma$: (x,y,z,t)= (λ+$\alpha$, λ-3 $\alpha, \alpha, \frac{-λ-2\alpha}{λ}$)

so answer to nu,ber 3: the solutions sets are those above
4- the system is homogeneous for λ=0. for its solutions set, see above.

 

[Felafel]

I know it is a very long exercise, so I'll just ask again for the most urging question: could anyone please tell me if $\lambda=0$ is right?
thank you :)

 

[Mark44]

Homework Statement

I have done this exercise, but I don't have a file with the solutions. COuld you please check it?
Thank you in advance :)

Given the following system:
$\lambda \in \mathbb{R}$

$x − z = \lambda$
$x + y + 2z + t = 0$
$y + 3z =$
$x + z + t = 0$

What's on the right side of the third equation? λ?

1-find $rk(A_{\lambda}) and rk(A_{\lambda}, B_{\lambda})$ according to the different values of $\lambda$

2- for which values of $\lambda$ does tha system have solutions?

3- what is its solutions set?

4- for which values is the system homogeneous ?

The Attempt at a Solution

This is the matrix $(A_{\lambda}, B_{\lambda})$
( 1 0 1 -1 λ)
( λ 1 2 1 0)
( 0 1 3 0 λ)
( 1 0 1 λ 0)

Your matrix doesn't look right to me. I'm assuming this is the augmented matrix that represents your system. If so, the first row in the matrix should be:
1 0 -1 0 λ

You have
1 0 1 -1 λ

doing some row reduction i get:

( 1 0 -1 0 λ )
( 0 0 2 λ -λ )
(λ-1 0 0 1 -2λ)
( 0 1 3 0 λ)

and i see rk(A)=rk(A,B) for any $\lambda$, so according to Rouchè-Capelli's theorem the system has solutions.

So, answers to 1 and 2 are: 1- rk(A)=rk(A,B) for any $\lambda \in \mathbb{R}$ 2-$\forany \lambda \in \mathbb{R}$

Now, if $\lambda$=1 i get:
(1 0 -1 0 1)
(0 0 2 1 -1)
(0 0 0 1 -2)
(0 1 3 0 1) thus: x=3/2, y= -1/2, z=1/2, t=-2 is the only solution

Doing the same, if $\lambda$=0 i get (x, y, z, t)=(0, 0, 0, 0)
(IS IT ACCEPTABLE AS A SOLUTION?)

If $\lambda$ is different from 0 and 1,

x-z=$\lambda$
2z+λt=-λ
(λ-1)x+t=-2λ
3z+y=λ

with z= $\alpha$
$\Sigma$: (x,y,z,t)= (λ+$\alpha$, λ-3 $\alpha, \alpha, \frac{-λ-2\alpha}{λ}$)

so answer to nu,ber 3: the solutions sets are those above
4- the system is homogeneous for λ=0. for its solutions set, see above.

 

[Felafel]

oops, yes, it should be as you said, but I think it's just a typo, because I've solved it on a piece of paper, so the rest of the calculation should be right and done on the correct matrix

 

[Felafel]

and on the right side of the third equation there is a λ, yes

 

[HallsofIvy]

I know it is a very long exercise, so I'll just ask again for the most urging question: could anyone please tell me if $\lambda=0$ is right?
thank you :)

If you mean "is $\lambda= 0$ the correct answer to 'for what values is the system homogeneous,'" then, by the definition of "homogenous system", yes.

 

[Felafel]

thank you!
and was it correct to use rouchè-capelli's thoerem?