# Math Help - linear systems help

1. ## linear systems help

Hello,

Can anyone help me with this problem.

Find the λ for which the system has a unique solution
x + y + λz =1
x +λy + z = 1
λx +y + z = 1

Thanks

2. The system will have a unique solution for any value of lambda for
which the 3x3 determinant of the coefficients of the left hand side
is not 0.

In your case all lambda but 1 or -2

3. Originally Posted by Papi3
Hello,

Can anyone help me with this problem.

Find the λ for which the system has a unique solution
x + y + λz =1
x +λy + z = 1
λx +y + z = 1

Thanks
There is no single solution. Any value of $\lambda$ such that the determinant $\left|\begin{array}{ccc}1 & 1 & \lambda \\ 1 & \lambda & 1 \\ \lambda & 1 & 1 \end{array}\right|$ is NOT 0 gives a unique solution.

4. in previous post I meant to say the determinant of the left hand side

5. 1 1 lambda
1 lambda 1
lambda 1 1

This matrix has a determinant:

lambda + lambda + lambda - 1 -1 - lambda3

This has to be different from 0, to make the matrix non-singular, and thus guarantee that it has only one solution.

This third-order polynomial has the following solutions:

-2 and 1

Any other lambda will give you a system that has a unique solution.

The determinant gives 3λ-λ^3-2
How can we find the solutions for λ?

Thanks again

7. You can find one solution and then divide the polynomial.

For example, it is easy to see that 1 is a solution.

3l - l^3 - 2 = 0

This means that 3l - l^3 - 2 = p(l ) * (l - 1)

where p(l ) is a second order polynomial. You can find p(l ) by dividing:

(3l - l^3 - 2) / (l -1) = -l^2 - l + 2

and using the solving formula for 2nd order polynomials

p(l ) = 0 <=> l = -2 V l = 1