# Creating Linearly Dependent Square Matrix from Cam Mechanism Equation

• MHB
• Barbudania
In summary, the conversation discusses an equation that comes from a specific topic of cam mechanisms. The equation involves various variables and the goal is to create a 3x3 matrix with a determinant equal to zero. The determinant is linearly dependent and the vectors must be on the same plane. The conversation explores different methods and matrices to achieve this goal, but the desired solution has not been found yet. The request for help and potential co-authorship on a paper related to this topic is also mentioned.
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)

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?

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.

## 1. How is a linearly dependent square matrix created from a cam mechanism equation?

A linearly dependent square matrix is created by taking the coefficients of the cam mechanism equation and arranging them in a square matrix. This matrix will have a determinant of 0, indicating that the rows or columns are linearly dependent.

## 2. Why is it important to create a linearly dependent square matrix from a cam mechanism equation?

Creating a linearly dependent square matrix allows us to determine the existence of a solution for the cam mechanism equation. If the matrix is linearly dependent, then there are an infinite number of solutions. If it is linearly independent, then there is a unique solution.

## 3. What is the significance of a determinant of 0 in a linearly dependent square matrix?

A determinant of 0 in a linearly dependent square matrix indicates that one or more rows or columns are dependent on each other. This means that the equations represented by these rows or columns are not independent and there are an infinite number of solutions to the system of equations.

## 4. Can a linearly dependent square matrix be solved?

Yes, a linearly dependent square matrix can still be solved using various methods such as Gaussian elimination or Cramer's rule. However, the solution will not be unique and there will be an infinite number of solutions.

## 5. How does creating a linearly dependent square matrix affect the motion of a cam mechanism?

The creation of a linearly dependent square matrix from a cam mechanism equation allows us to analyze the motion of the cam mechanism and determine the different possible positions of the cam. This information is crucial in designing efficient and accurate cam mechanisms for various applications.

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