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# VBA array inverse

By : Justin Wypack
Date : October 18 2020, 08:10 AM
around this issue You may be trying to invert an ill-conditioned matrix. I tried your code on an easy example:
code :
``````Sub dural()

Dim A As Variant
Dim i As Integer, j As Integer
ReDim A(1 To 3, 1 To 3) As Double

For i = 1 To 3
For j = 1 To 3
A(i, j) = Cells(i, j).Value
Next j
Next i

A = Application.WorksheetFunction.MInverse(A)

For i = 1 To 3
For j = 1 To 3
Cells(i + 5, j + 5).Value = A(i, j)
Next j
Next i
End Sub
``````

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## Algorithm: how calculate INVERSE of bilinear interpolation? INVERSE of mapping on to an arbitrary quadrilateral?

By : Meet Thind
Date : March 29 2020, 07:55 AM
hope this fix your issue To simplify, let's begin by just considering a single intepolated value z.
Assume four values z00, z01, z10, z10, and two weights w0 and w1 applied to the first and second index, giving

## 7 DOF Inverse Kinematic with Jacobian and Pseudo inverse

By : user3616370
Date : March 29 2020, 07:55 AM
With these it helps Hate to keep answering my own questions, but I had been working on this harder than on anything else so far. The problems were at two places:
Instead of calling the .inverse() on the J * J.transpose() I used fullPivLu().solve(deltaP) on the J * J.transpose(). (I had overlooked the fact that we weren't suppose to invert the matrix but use LUD in my class notes.) The line with tP = (float)time * err+currentP; had to be just tP = (float)time*err;

## Python: Calculating the inverse of a pseudo inverse matrix

By : Алексей посевин
Date : March 29 2020, 07:55 AM
this will help Your matrix A does not have full column rank. In consequence helper is singular and not invertible (If you print helper.I you will see some very large numbers).
The solution is to compute the right inverse instead of the left inverse:
code :
``````helper = A * A.T
PI = A.T * helper.I
``````
``````>>> numpy.random.seed(42)
>>> a = mat(numpy.random.random_sample((3, 4)))  # smaller matrix for nicer output
>>> h = a * a.T
>>> h * h.I
matrix([[  1.00000000e+00,   1.33226763e-15,   0.00000000e+00],
[ -1.77635684e-15,   1.00000000e+00,   0.00000000e+00],
[  0.00000000e+00,   1.33226763e-15,   1.00000000e+00]])
``````

## Filter objects array with keys array [inverse]

By : hxseek
Date : March 29 2020, 07:55 AM
I wish this helpful for you I have this following keys array: , You can use filter() and find().
code :
``````var users = [
{id: "333", firstName: "", lastName: "", idCard: "", birthDate: ""},
{id: "334", firstName: "", lastName: "", idCard: "", birthDate: ""},
{id: "335", firstName: "", lastName: "", idCard: "", birthDate: ""},
{id: "336", firstName: "", lastName: "", idCard: "", birthDate: ""}
]
var keys = [{userId: "333"}, {userId: "334"}]

var result = users.filter(function(o) {
return !keys.find(function(e) {
return e.userId == o.id
})
})

console.log(result)``````

## Problem in printing inverse array of an input array

By : Parwan Singh Bisht
Date : March 29 2020, 07:55 AM
it helps some times , Here is one problem:
code :
``````for(j=5;j>=0;--j){
new[i]=arr[j];
``````
``````for(j=4;j>=0;--j){  // 4 instead of 5
new[i]=arr[j];
``````
``````    for(j=4;j>=0;--j){
printf("%d", arr[j]);
printf(" ");
}
printf("\n");
``````
``````    for(j=4;j>=0;--j){
new[4-j] = arr[j];
printf("%d", arr[j]);  // or printf("%d", new[4-j]);
printf(" ");
}
printf("\n");
``````