System of Linear Equations

1. Jul 26, 2012

AGNuke

If the system of Linear Equations
$$x+y+z=6$$
$$x+2y+3z=14$$
$$2x+5y+\lambda z=\mu$$
has infinite number of solution in x, y, z

I need to find out two things
1. The value of λ
2. Maximum value of $$(\mu x+\lambda y-20z)sin^2\theta+(\lambda x+\mu y+64z)cos2\theta, \theta \in \mathbb{R}$$ is 272

I used the Matrix method of AX=B to find out λ by solving for A=0; I got the answer 8, and it is correct.

Now my catch is to validate the second question. It is given true, I just need to validate. I tried to solve it with the three existing equations but I was unable to get answer.

Last edited: Jul 26, 2012
2. Jul 26, 2012

HallsofIvy

Staff Emeritus
That isn't quite correct- first you haven't told us what "A" is! More important you mean det(A)= 0, not A= 0.

3. Jul 27, 2012

AGNuke

A is the coefficient matrix which is present when we try to solve the system of equations using matrix.

$$A=\begin{bmatrix} 1 &1 &1 \\ 1 & 2 &3 \\ 2 &5 &\lambda \end{bmatrix}$$

and yes, I meant |A|=0, my bad.

Whatever that may be, I found out the value of λ and I need to find answer to my second question. I am on it, but haven't made progress.

4. Jul 27, 2012

Ray Vickson

You also need to determine the value of μ (because if you don't have the correct value the system will have no solutions at all). Once you know λ and μ you have an optimization problem in the 4 variables x,y,z,θ, subject to linear restrictions on x,y,z. This can be tackled via Lagrange multiplier methods, or in some other way that handles constraints. At that point the problem is more suitable for the "Calculus and Beyond" Forum.

RGV

Last edited: Jul 27, 2012
5. Jul 28, 2012

AGNuke

μ can be determined by solving the three given equations, at which point, it returns the value 36.

We know λ and we know μ, I think we are good to go and find the answer.

UPDATE: I got my answer. I just found out y=4-2x and z=x+2; since the system has infinite solutions. I substituted y and z in the asked equation, got -8sin2θ + 272cos2θ. Then made sinθ=0 and cos2θ=1. Proved the statement right.

Last edited: Jul 28, 2012