Inverse Of A Matrix Puns

I create matrix A following an algorithm that theoretically guarantees it to be inverse negative. In other words, A is always invertible and all elements of A^-1 are negative. When A is small, I can easily calculate its inverse in Matlab. When it is large (15,000 by 15,000, for example), Matlab fails to calculate the inverse correctly. I have uploaded an example of A here on my Google Drive. I use the following line of code to calculate the inverse:inverse = sparse(A) \ speye(size(A));The inverse I get has positive elements (in the order of 10^12, which theoretically is wrong) and even A *A^-1 is far from I. It seems the miscalculations occur due to IEEE double-precision arithmetic used in Matlab. However, I am not an expert in this field and don't know how to handle this issue. I was wondering if someone could help me with finding a package/tool that could do the calculation correctly for me.

Background: Let Ω be the state space of an absorbing Markov chain with Ωa being the set of absorbing states, and its complement Ωt, being the set of transient states. The element ij of the matrix A that I have is the transition rate between states i and j, where both i and j belong to Ωt. The element ij of the matrix B = -A^{-1} is the expected time the system spends in transient state j before absorption, given it started from the transient state i. I need to calculate the inverse, because I need the expected time spent in some specific transient states before absorption, given the system started from some other specific transient states.

I am trying to calculate the inverse of an upper triangular matrix (R from a QR decomposition) using back substitution and equating U = I. U being an upper triangular matrix, X being the matrix I want, and I being an identity matrix.

I am programming in R and have everything set up. Right now, all that is left is programming a back substitution function in order to get the inverse of a triangular matrix, as we cannot use the R Base function.

However, all programming examples I have found online only use vectors as the outcome. My matrix algebra is not super sharp, so I am having trouble translating these examples to my problem.

Here is one implementation in Fortran that I have managed to translate to R (see below). Now, this code obviously doesn't work if b is a matrix (and x, consequently also a matrix)

# b = outcome matrix (in this case an identity matrix)
# A = the input matrix
# x = A'

# Solve last element
x[n, n] <- b[n, n] / A[n, n]
    
# Do backward substitution
for (i in (n - 1):1) {
    sum = b[i]
    for (j in (i + 1):n) {
        sum <- sum - (A[i, j] %*% x[j])
    }
    x[i] <- sum / A[i, i]
}

return (A %*% x)

This is the full text of the problem.

This is what I have.

I know the answer is 17, but I have -17. Am I on the right track?

Is it possible to prove a matrix is invertible when in modulo 2?

https://hbfs.wordpress.com/2017/01/17/choosing-the-right-pseudoinverse/

I saw the answer here but I don't understand why you can pre-multiply right hand side with A transpose but not do the same on the left hand side (for wide matrix, as explained in the link)

You can find it here.

I would love any feedback you have as well! I'm curious where to go in order to increase the speed and performance, so if you have any ideas on that please share!

Q:

https://gyazo.com/8a8feea31fd86f7973ae6ea27b5b417d

Given Elementary Matrix x Matrix = Matrix.

In this situation, I've been thaught that here is a short cut to find the inverse of an elemntary matrix. If you perform a multiplication operaito on a row, you can find the inverse by converting the elemntary matrix into an identity matrix and perorm a division on the row of the identity matrix. Not sure what it's called. But in this case, I already found the elemntary matrix, shouldn't the inverse be the identity matrix? I don't want to solve this by the 'ad-bc' formula.

Given an equation

AX^tr = 5B

where A, X, and B, are invertible matrices, find an expression for X.

the first step, according to the solution, is to say X^tr = A^(-1)5B.

Why can't I say X^tr = 5BA^-1 ?

I guess more generally, why does

AX = B imply A=X^(-1)B and not A=BX^(-1) ?

I know AB != BA when it comes to matrices.

Let A=

4 -2 3

8 -3 5

7 -2 4

Find the 2nd column of A^-1 without finding the entire A^-1

I know how to find the inverse by attaching the identity matrix to A and completing row operations until I get the Identity matrix on the left side but Im not sure how to find just the 2nd column of A^-1

Any Insight would be appreciated. Ive tried a couple things but can never seem to get the right answer.

Hello! I have a trouble with MINVERSE function. Excel calculated inverse matrix, but it's wrong: Inverse matrix * Original one does not result in E. At a loss, what to do. Could anyone explain, if i did something wrong? https://imgur.com/tDta9Qs

I'm reading about the graphical lasso and one the interpretations of this method is to assume a sparse prior on the inverse covariance (precision) matrix, i.e. that the distribution of the inverse covariance is Laplace which has a high probability around zero. I'm wondering if this is a reasonable assumption or if it's just something to make our life a lot easier.

Thanks

I'm new to CUDA and want to implement matrix inversion in CUDA for large matrices. I do not know how to verify the results. Can anyone please guide me how to go about this?

Hi all,

I want to do inverse projective transformation of each pixel from one image to another, given the intrinsics of both matrix, the relative projection matrix P (R, T) and the depth values of each pixel in image 1.

The ask is that I have to create a single projective transformation matrix for all pixels. i.e.

x' = Hx, where x' = [u', v', d'] and x = [u, v, d]

Generally the transformation takes place as:

x'' = [u, v, 1]

x''' = D * K^-1 * x', where K is intrinsic matrix, and D is the depth vector of all pixels.

X''' = Px'''

x''' = x''' / x'''[2], divides by depth

x'''' = K * x''', this denotes final step.

Now doing this using single matrix would involve removing the operations involving multiplication by D and divide by D.

So one way is as below: (Note here d is inverse depth (not depth, not disparity))

https://preview.redd.it/f17j2gkh8qo41.png?width=596&format=png&auto=webp&s=03f7ed725760830c11326410eec7319652d2f862

Can I do this task in some other way using depth instead of inverse depth ?

Would really appreciate any help. [Studying perspective geometry] Thanks !!

I am just curious looking for real examples(which can be very basic) that are analogous to matrix multiplication and matrix inverse.

Your answers mean a lot to me.Thanks in Advance <3

I want to find the least squares solution to the matrix equation y = mx + b via an inverse / pseudo-inverse approach. In my case, I have the inputs x and outputs y, so I want to find the coefficient m and constant b. This in in an over-determined situation.

There's a million articles that talk about this (such as here: https://courses.lumenlearning.com/ivytech-collegealgebra/chapter/solving-a-system-of-linear-equations-using-the-inverse-of-a-matrix/) but they never seem to include the constant. I.e, its always just y = mx. How do I recover both the weights m and constant b?

Thanks...I'm sure its probably something stupid.

I'm trying to find the camera's position from an aruco marker.

(https://i.stack.imgur.com/3FyQ4.png) is what the camera sees, and (https://i.stack.imgur.com/068VO.png) is also what the camera sees. The Z-axis of the marker flips back and forth. I only captured Rvec and Tvec for one of the above situations.

A. These are what Aruco returned:

rotation vector Rvec: [1.98895, 1.67426, -0.570106] translation vector Tvec: [0.00876591, 0.100794, 0.630018]

B. This is rotation matrix R, converted from Rvec:

0.16674686138210792, 0.9859319291786623, -0.011563530830037305, 0.7881094090430687, -0.14031913821875497, 0.5993280394135843, -0.5925192347907373, -0.09082274206159546, -0.8004200059515079

C. This is the inverse transformation matrix [R|t]^-1, the R part is the rotation matrix R above, the T part is the Tvec mentioned above:

0.5532041267642692, 2.62162273212768, 1.9549923883934333, -1.5007735737997756, 0.9146872306386324, -0.46552676124283743, -0.36178537886158296, 0.2668355392503998, -0.513303622140676, -1.8878607068233544, -2.6554961620678954, 1.8677949864716032, 0, 0, 0, 1

And the final result is incorrect. The final result looks like: i.stack.imgur.com/RPDoo.png The dummyeye(or eye1) is the camera, it's way off. eye3 and eye4 are in the correct position.

So I was wondering what's wrong. Is Aruco lib giving noise? Or is the rotation matrix in B calculated wrong? Or is the inverse wrong?

Thanks!