I Solving a complex linear system with parameters

[iopz]

TL;DR Summary

Study the solutions of a complex linear system of three equations and three variables upon variation of two parameters.

I have to study the solutions of the following system of three equations and three unknowns upon variation of parameters k and h.

ix₁+kx₂-x₃ = 1+i
(k+i)x₁+(1-i)x₂-(ik-1)x₃ = h
kx₁+(4+2i)x₂-(k-3-3i)x₃ = 1-i

Obviously i is the imaginary unit.
And as stated k and h are the parameters .

I can't find a way to solve these kinds of system in a "clean" way w/out tons of computations and huge resulting equations.
I've thought about various methods (Cramer's rule, gauss reduction & Rouche-Capelli theorem, just straight up substitution) but all of them still require lots of computation and give insane results.

These kinds of system are present in all the tests of my linear algebra faculty and it seems strange to me that there wouldn't be a "cleaner" and faster way to solve them.
Cause all the other exercises are much easier (in terms of computations) with the right method.

How would you go to solve this? Is there any quicker way?

Thanks in advance.

 

[pasmith]

You can say this:
- If the determinant is non-zero, there is exactly one solution for any value of h.
- If the determinant is zero, there are either no solutions or infinitely many solutions, depending on whether the right hand side is in the image of the map or not.

Calculating the determinant will give you conditions on k for the determinant will vanish. Whether or not a solution then exists depends on the value of h.

 

[graphking]

because k and h are parameters so it could be hard to calculate. maybe you know what field is defined in algebra? actually k,h are regarded as the "x" in polynomial. you are sovling this linear equations in a bigger field called C(k,h), which means all the fractions of the polynomials (ring)
C[k,h]

 

[Mark44]

because k and h are parameters so it could be hard to calculate. maybe you know what field is defined in algebra? actually k,h are regarded as the "x" in polynomial. you are sovling this linear equations in a bigger field called C(k,h), which means all the fractions of the polynomials (ring)
C[k,h]

IMO, the above is not helpful. @pasmith's advice to calculate the determinant is the more reasonable approach.

 

[graphking]

IMO, the above is not helpful. @pasmith's advice to calculate the determinant is the more reasonable approach.

i
thought that the questioner already know those basic linear algebra method. he is confusing in the sophisticated calculations, which actually was done in the field C(k,h)

 

[Mark44]

i
thought that the questioner already know those basic linear algebra method. he is confusing in the sophisticated calculations, which actually was done in the field C(k,h)

The field is just the complex numbers. It seems obvious to me that the OP understood what the field was.

 

[graphking]

The field is just the complex numbers. It seems obvious to me that the OP understood what the field was.

i thought the field is not C, because k and h could be regarded as variables like "x,y" in the polynomials. maybe that is called "rational function field", sorry for my bad english i can't find the exact word. and the symbol of this kind of field is like C(k,h), as you may know the symbol of polynomial ring is like C[x]

 

[Mark44]

thought the field is not C, because k and h could be regarded as variables like "x,y" in the polynomials.

No, k and h are parameters, unknown, but fixed, unlike the variables $x_1, x_2$, and $x_3$. From the problem summary:

Study the solutions of a complex linear system of three equations and three variables upon variation of two parameters.

The problem has nothing to do with rational function fields or polynomial rings, which is why I said that your post was not helpful.

 

[graphking]

No, k and h are parameters, unknown, but fixed, unlike the variables $x_1, x_2$, and $x_3$. From the problem summary:

The problem has nothing to do with rational function fields or polynomial rings, which is why I said that your post was not helpful.

ok i got it, you mean discussing whether the det is zero. actually i mean why the gauss reduction is hard to do

 

[Mark44]

actually i mean why the gauss reduction is hard to do

The reason that Gauss reduction is hard to do is that it involves complex numbers. It would also be difficult to do if the coefficients were real and not rational.

 

[graphking]

The reason that Gauss reduction is hard to do is that it involves complex numbers. It would also be difficult to do if the coefficients were real and not rational.

i could agree that, but i want to add: the calculate of polynomials is also hard, even you are doing in some filed like Q(x). just because there is too many monomials. and i thought the calculate of a*b in C is just as hard as the calculate of (ax+b)*(cx+d) in Q...

 

[Stephen Tashi]

These kinds of system are present in all the tests of my linear algebra faculty and it seems strange to me that there wouldn't be a "cleaner" and faster way to solve them.

Can you provide a link to an online publication of such a test?

Since this type of question appears on all tests, have you asked other students how they solve this type of problem?