MHB Creating Linearly Dependent Square Matrix from Cam Mechanism Equation

[Barbudania]

I have an equation that comes from an especific topic of cam mechanisms and it goes like this:

$$
2M[tan(B)-B] - \beta Ntan(B) - 2\pi\sqrt{1 - N^2} = 0 \ \ \ \ \ \ \ \ \ (1)
$$

For this it doesn't matter what each variable means.

I'm trying to create a 3x3 matrix with a determinant equal to zero. The determinant is linearly dependent because I will create a nomogram from it and its vectors have to be on the same plane.

I did come up with this answer:

$$
det \begin{bmatrix} tan(B) & 2\pi\sqrt{1 - N^2} & tan(B) - B \\ 1 & -\beta N & 0 \\ 0 & 2M & 1 \end{bmatrix} = 0\ \ \ \ \ \ \ \ \ (2)
$$

I can manipulate the matrix in order not to change the determinant and it becomes this:

$$
det \begin{bmatrix} Ntan(B) & 2\pi\sqrt{1 - N^2} & tan(B) - B \\ 1 & -\beta & 0 \\ 0 & 2M & 1 \end{bmatrix} = 0 \ \ \ \ \ \ \ \ \ (3)
$$

The way the matrix is written is important because each variable is in a different row. When I create the nomogram, I will have 2 linear axes and a grid. I can have up to two variables in one row. So in this determinant I have B and N in the first row, $\beta$ in the second row and M in the third row.

However this does not solve my problem as I need. What I really need is to create a determinant where variables M and N are in the same row and $\beta$ and B on the other rows.

I've created about 10 different matrices, but I can never create one with this especific condition. I don't know if it's not possible or if I just couldn't find the answer.

It doesn't have to be a determinant straight out of the equation like Eq (2) where you can look at it and see that it will create Eq (1). One could develop a determinant, manipulate it and then arrive at the answer, just like I manipulated Eq (2) into Eq (3).

Can anybody come up with an answer? Or perhaps, is there any methodology, theorem or something else to help me create this apart from trial and error? Is there at least a way for me to know if what I want is impossible?
(actually if someone could find a solution, I would gladly ask the person if he/she wants to be a co-author on a paper I'm working on)

 

[GJA]

Hi Barbudania,

Assuming I've understood your post correctly, I think we can accomplish what you're looking for in the following way:

1. Assuming $\beta\neq 0$, multiply both sides of equation (2) above by $1/\beta$ to get $$\det\begin{bmatrix}\tan(B) & 2\pi\sqrt{1-N^{2}} & \tan(B) - B\\ \frac{1}{\beta} & -N & 0\\ 0 & 2M & 1 \end{bmatrix} = 0.$$
2. Use the fact that the determinant of the transpose of a matrix is equal to the determinant of the matrix; i.e., $\det(A) = \det(A^{T})$. Taking the transpose -- i.e., make the rows of the matrix the columns of the matrix -- of the matrix from step 1, we get: $$\det\begin{bmatrix}\tan(B) & \frac{1}{\beta} & 0\\ 2\pi\sqrt{1-N^{2}} & -N & 2M\\ \tan(B)-B & 0 & 1 \end{bmatrix} = 0.$$

Is this form acceptable for your purposes?

 

[Barbudania]

Hi there, thank you for your help, but it doesn't solve the problem.
In your answer, you have the variable B in the first and third row. The thing is that a variable can't be in two different rows, I don't know if I didn't make it clear in my post.