System of linear equations (Finding Real numbers in a Unique Solution)

[mlk771]

Homework Statement

For which real numbers  does the following system have a unique solution?
$$14x - 6y + 18z = 2\lambda z$$
$$x = \lambda x$$
$$3x - 8y = -\lambda y$$

Homework Equations

The Attempt at a Solution

hi, I rearranged the equations so that becomes:

$$x + 0y + 0z = \lambda x$$
$$3x - 8y + 0z = -\lambda y$$
$$14x - 6y + 18z = 2\lambda z$$
right?
then I put it in Augmented shape and letting lambda to be 1:

[1 0 0 | 1]
[3 -8 0 |-1]
[14 -6 18 | 2]

After that, I kept trying real numbers from -5 to 5 and everyone of them gave me a unique solution, which is what i want.

In this case are the values should be Real numbers?

But I am still confused about letting lambda = 0, in this case x = y = z = 0;

is this looks a unique solution? if no, then obviously I will have R-{0}

Please help me understand this question.

Thanks

 

[LCKurtz]

This must have come from an eigenvalue problem. Your system of equations can be rewritten:

$$\left(\begin{array}{ccc}<br /> 1-\lambda & 0 & 0\\<br /> 3 & -8 + \lambda & 0\\<br /> 14 & -65 & 18-2\lambda <br /> \end{array} \right)<br /> \left(\begin{array}{c}<br /> x\\<br /> y\\<br /> z <br /> \end{array} \right)=<br /> <br /> \left(\begin{array}{c}<br /> 0\\<br /> 0\\<br /> 0 <br /> \end{array} \right)$$

This homogeneous system of equations will have a unique solution if the determinant of coefficients is nonzero and otherwise not. Look at that.

 

[HallsofIvy]

Homework Statement

For which real numbers  does the following system have a unique solution?
$$14x - 6y + 18z = 2\lambda z$$
$$x = \lambda x$$
$$3x - 8y = -\lambda y$$

Homework Equations

The Attempt at a Solution

hi, I rearranged the equations so that becomes:

$$x + 0y + 0z = \lambda x$$
$$3x - 8y + 0z = -\lambda y$$
$$14x - 6y + 18z = 2\lambda z$$
right?
then I put it in Augmented shape and letting lambda to be 1:

[1 0 0 | 1]
[3 -8 0 |-1]
[14 -6 18 | 2]

No, this would be correct only if the equations were x+ 0y+ 0z= 1, 3x- 8y+ 0z= -1, 4x- 6y+ 18z= 2, not with x, y, and z on the right side.

After that, I kept trying real numbers from -5 to 5 and everyone of them gave me a unique solution, which is what i want.

In this case are the values should be Real numbers?

But I am still confused about letting lambda = 0, in this case x = y = z = 0;

is this looks a unique solution? if no, then obviously I will have R-{0}

Please help me understand this question.

Thanks

 

[mlk771]

Sorry all, there was a mistake

1. Homework Statement
For which real numbers λ does the following system have a unique solution?

 

[mlk771]

Please any help here,

 

[A-fil]

If you don't know how to calculate determinants then that's not necessary in this problem.

You can look at the equations one at a time, starting with the one that only contains x.

Now, depending on λ, x may be forced to have certain values. If x,y and z isn't forced to have certain values, then you will have a lot of solutions to the system.

You will want to split up this problem into the case when x is forced to a certain value and the case where x is free to take any value. x taking any value would mean many solutions to the system, so you want to select λ so that this case won't happen. Now, if you write the equation down on a piece of paper and think a bit I think you'll understand how to continue with y and z.

 

[mlk771]

well, thanks

I'm still confused..

I tried to solve it be making λx, λy and λz as parts of right side vector b like: b1, b2, b3..

then found rref of the augmented matrix and I came up with these vaues for λx, λy and λz:

λx = λ
λy = (1/8)λy + (3/8)λx
λz = -2λx -(3/8)λy

I have a similar prob but don't know if it related or no..I attached it