Matrix one-parameter solution help

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In summary, the conversation discusses a system of equations with variables a and b and determines the conditions for the system to have no solution, a unique solution, a one-parameter solution, or a two-parameter solution. The conversation includes the steps of subtracting equations, dividing by a variable, and setting the determinant equal to zero to solve the system. However, it is noted that simply setting the determinant equal to zero is not enough to distinguish between the different solution cases. Additional steps and considerations are needed to accurately determine the solution set.
  • #1
sara_87
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hello people
i was wondering if anyone could help me with this question:

For what values of a and b does the system

ax +bz = 2
ax +ay +4z = 4
ay + 2z = b

have: (i) no solution
(ii) a unique solution
(iii) a one-parameter solution
(iv) a two-parameter solution?

this is what i did, i wrote it in a matrix form and i did gussian elimination, and across the diagonal i got a, a, and 1
i multiplied the digits on he main diagonal and i got a^2, i made that equal to 0... then i made this post
(basically I'm just doing random things with the numbers and I'm so stuck i don't know what to do! :frown: )

any tips would be really appreciated
cheers in advance!

(tomorrow is christmas :biggrin: )
 
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  • #2
Subtract the first from the second equation, and get:
ay+(4-b)z=2 (2*)

Subtract (2*) from your third equation, and get:
(b-2)z=(b-2)

Thus, you have the following system of equations to work with:
ax+bz=2
ay+(4-b)z=2
(b-2)z=(b-2)
 
  • #3
Thus, you have the following system of equations to work with:
ax+bz=2
ay+(4-b)z=2
(b-2)z=(b-2) (3*)

then i divided 3* by (b-2)

and i got
ax+bz=2
ay+(4-b)z=2
z=1

to work with...should i not do that last step?

and (i think you know this but) to get a no solution or a unique solution etc. we should use the determinant of the matrix which is the multiplication of the numbers on the main diagonal...i bet I'm talking rubbish...then we make the multiplication of the numbers on the main daigonal equal to zero, then i get stuck for real!

i just don't know what to do! lol
 
  • #4
You get 0,1 or infinitely many solutions if there are, well, 0, 1 or infinitely many solutions. So have a look at the solution set and let's see if when we can have 0,1 or infinitely many solutions.

We know (b-2)z=b-2. Now, you divided by b-2. That's fine *when it's allowed*. When is it allowed?
 
  • #5
When is it allowed?
'clueless'
 
  • #6
Suppose we have the equation ax=4. Then this gives the solution x=4/a iff a is not equal to zero. Does this help?
 
  • #7
sara:
It is perfectly true that if the determinant of a quadratic matrix is non-zero, then we'll have a unique solution of the associated system of equations.

Thus, the "no solution" or "infinite solution" cases will have the value of the determinant equal to zero.
But do you see the problem here?

The problem is that if we just set the determinant equal to zero, gaining a relation between a and b, we haven't really DISTINGUISHED between the "no solution"/"infinite number of solutions" cases!
But to make that distinction is precisely one of your tasks..
Thus, just setting the determinant equal to zero is insufficient to answer your particular problem.
 
  • #8
thanx all for your help!
i really really appreciated it, i think i know what to do now...you know i went on this other site to ask a question and by the time i got a response i had worked out the answer my self!
cheers!
 

1. What is a matrix one-parameter solution?

A matrix one-parameter solution is a mathematical technique used to find a solution to a system of linear equations with one variable. It involves representing the system as a matrix and using matrix operations to solve for the variable.

2. How is a matrix one-parameter solution different from other methods?

A matrix one-parameter solution is different from other methods because it uses matrix operations, which make it more efficient and easier to solve for complex systems of equations. It also provides a more general solution that can be applied to different types of equations.

3. When is a matrix one-parameter solution useful?

A matrix one-parameter solution is useful when dealing with large systems of equations, as it simplifies the process and allows for the general solution to be easily obtained. It is also useful for solving systems of equations with multiple variables and constraints.

4. What are the limitations of a matrix one-parameter solution?

One limitation of a matrix one-parameter solution is that it can only be used for linear equations. It also requires knowledge of matrix operations and may not be suitable for solving complex systems with non-linear relationships.

5. How can I learn more about matrix one-parameter solutions?

There are many online resources and textbooks available that provide detailed explanations and examples of matrix one-parameter solutions. You can also consult with a math tutor or take a course in linear algebra to learn more about this technique.

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