Question: I looked up the solution to a math problem in the solution manual, and I'm having trouble understanding how the solution works.

In this problem the lagrange method is used, the part I'm having trouble understanding is how they solved this linear equation:

[![enter image description here](https://i.stack.imgur.com/JXkyL.png)](https://i.stack.imgur.com/JXkyL.png)

Which gives this system of equations:

[![enter image description here](https://i.stack.imgur.com/NyD8Z.png)](https://i.stack.imgur.com/NyD8Z.png)

**The solution and the part I don't understand that they showed in the solution manual:**

*Clearly this has the solution (x, y) = (0, 0) for there to be other solutions the following has to be true:*

[![enter image description here](https://i.stack.imgur.com/cycUF.png)](https://i.stack.imgur.com/cycUF.png)

*Which is fulfilled for λ = 1/3 and λ = -1*

What I don't understand is how does this solution work? Why is the determinant of that matrix 0 when there are other solutions?

Answer: 
Your original equation can be written in the form $$\lambda \begin{bmatrix} 2 & 1\\ 1 & 2 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} y \\ x \end{bmatrix}$$ 
which can be rewritten as 
$$ \begin{bmatrix} 2 \lambda & \lambda \\ \lambda & 2 \lambda \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix}0 & 1 \\ 1 & 0\end{bmatrix} \begin{bmatrix} x \\ y\end{bmatrix}$$
which simplifies to
$$\begin{bmatrix} 2 \lambda & \lambda - 1 \\ \lambda - 1& 2 \lambda \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}.$$

Let $A = \begin{bmatrix} 2 \lambda & \lambda - 1 \\ \lambda - 1& 2 \lambda \end{bmatrix}$. If $\det A \not= 0$ the system is solved uniquely by $$\begin{bmatrix} x \\ y \end{bmatrix} = A^{-1} \begin{bmatrix} 0 \\ 0 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}.$$

Thus if a nonzero solution exists you must have $\det A = 0$.
λ = 1/3 and λ = -1 are just eigen values for the homogeneous system. They are the ones that give rise to non-zero solutions.