Evaluate the partial derivative of a matrix element

In summary, the determinant a is defined as the sum of the products of the rows of the coefficient matrix and the cofactor matrix, with the kronecker delta used to account for the expansion over columns. To evaluate (∂a / ∂art), where art is a coefficient matrix element in row r and column t, take the derivative of the sum with respect to a single element, a_{rt}. In the second part, it is recommended to use the notation (∂a / ∂u_j) instead of (∂a / ∂u_t) to avoid confusion with the variable t used in the first part.
  • #1
Biffinator87
24
1

Homework Statement



A determinant a is defined in the following manner ar * Ak = Σns=1 ars Aks = δkr a , where a=det(aij), ar , Ak , are rows of the coefficient matrix and cofactor matrix respectively. The second term in the equation is the expansion over the columns of both matrices, δkr is the kronecker delta that is 1, when r=k and 0 otherwise. Or ar * Ar = Σns=1 ars Ars = a, where a=det(aij). Evaluate (∂a / ∂art), where art is the coefficient matrix element in row r and column t. Assume all of the elements of the coefficient matrix depend on the independent variables (u1,u2,u3). Find (∂a / ∂ut) for t=1,2,3. Your answer should depend on your results for (∂a / ∂art) above.

Homework Equations



No equations given. We are going over the basics of curvilinear coordinates at the moment and thus I believe tangent vector equations are need to solve the problem which is what I think he wants us to find when he says evaluate (see above). However, the professor is very unhelpful and not willing to really help the students understand exactly what he is after. To be fair he has had some kindof illness that has left him partially disabled.

The Attempt at a Solution


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My attempt at the solution poses a problem. The professor has not given any direction as to how to approach these matrix problems. My only attempt that the solution so far was to use the determinant to find the matrix element art and then from there right a very generic answer for what that vector would be in terms of the independent variables given: art = u1 i + u2 j + u3 k. the problem I am encountering is I just can't seem to figure out if this is the correct route to take. We are using a schaums outline for the book and I haven't found any examples in there to help. Any direction that can be offered would be greatly appreciated.

If i have violated the rules of the forums please reply and let me know that I have so that I can correct myself appropriately. I understand we are supposed to have a valid attempt that the problem before posting. I hope I explained clearly enough above that I have tried to figure out what I need to do.

Thanks!
 
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  • #2
You have the equation ##\sum_{s=1}^n a_{rs}A^{ks}=\delta_r^ka##. If you set ##k=r## then you get the equation ##a=\sum_{s=1}^n a_{rs}A^{rs}=##. Note that none of the items ##A^{rs}## depend on any coefficient in row ##r##, and only one of the ##a_{rs}## coefficients in that sum will be the item with respect to which you are asked to differentiate: ##a_{rt}##. What do you get when you take the ##\frac{\partial}{\partial a_{rt}}## of the sum?

When you get to the second part, I suggest you take ##\frac{\partial a}{\partial u_j}## rather than ##\frac{\partial a}{\partial u_t}## as ##t## is already used for something completely different, and using the same index for two different things is unnecessarily confusing.
 
  • #3
andrewkirk said:
You have the equation ##\sum_{s=1}^n a_{rs}A^{ks}=\delta_r^ka##. If you set ##k=r## then you get the equation ##a=\sum_{s=1}^n a_{rs}A^{rs}=##. Note that none of the items ##A^{rs}## depend on any coefficient in row ##r##, and only one of the ##a_{rs}## coefficients in that sum will be the item with respect to which you are asked to differentiate: ##a_{rt}##. What do you get when you take the ##\frac{\partial}{\partial a_{rt}}## of the sum?

When you get to the second part, I suggest you take ##\frac{\partial a}{\partial u_j}## rather than ##\frac{\partial a}{\partial u_t}## as ##t## is already used for something completely different, and using the same index for two different things is unnecessarily confusing.

Thank you! I will try it and see what I come up with!
 

Related to Evaluate the partial derivative of a matrix element

What is a partial derivative of a matrix element?

A partial derivative of a matrix element is the rate of change of that element with respect to a specific variable while holding all other variables constant. It is denoted by the symbol ∂

How do you evaluate the partial derivative of a matrix element?

To evaluate the partial derivative of a matrix element, you would take the derivative of the element with respect to the specific variable and hold all other variables constant. This can be done by using the chain rule and the product rule for multivariate functions.

Why is it important to evaluate the partial derivative of a matrix element?

Evaluating the partial derivative of a matrix element is important because it allows us to understand how the value of that element changes as one variable changes while holding all other variables constant. This is useful in many applications, such as optimization problems and sensitivity analysis.

What are some real-life applications of evaluating the partial derivative of a matrix element?

The partial derivative of a matrix element is used in fields such as physics, economics, and engineering. It can be applied to problems such as calculating the elasticity of demand, finding the maximum or minimum values of a function, and determining the stability of a system.

Are there any limitations or assumptions when evaluating the partial derivative of a matrix element?

Yes, there are some limitations and assumptions when evaluating the partial derivative of a matrix element. One limitation is that the element must be a continuous function. Additionally, the assumption is made that the element is differentiable with respect to the specific variable. If these conditions are not met, the partial derivative may not exist or may be undefined.

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