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We know determinant of a square matrix A[ij] (i,j ∈ {1,2,3}) is equal to the following expression

det(A) = 1/6 * e[ijk] * e[pqr] * A[ip] * A[jq] * A[kr] 

in which e[ijk] is a third order Tensor (Permutation notation or Levi-Civita symbol) and has a simple form as follows:

e[mnr] = 1/2 * (m-n) * (n-r) * (r-m).

The (i,j) minor of A, denoted Mij, is the determinant of the (n − 1)×(n − 1) matrix that results from deleting row i and column j of A. The cofactor matrix of A is the n×n matrix C whose (i, j) entry is the (i, j) cofactor of A,

C[ij]= -1 i+j * M[ij]


The general form of Levi-Civita symbol is as bellow:


I want to write a program for finding inverse of (NxN) matrix:

N=2 →

N := 2:
f := (1/2)*(sum(sum(sum(sum((m-n)*(p-q)*A[m, p]*A[n, q], q = 1 .. 2), p = 1 .. 2), n = 1 .. 2), m = 1 .. 2)):
A := Matrix(N, N, proc (i, j) options operator, arrow; evalf((37*i^2+j^3)/(2*i+4*j)) end proc):

N=3 →

N := 3:
f := (1/24)*(sum(sum(sum(sum(sum(sum((m-n)*(n-r)*(r-m)*(p-q)*(q-z)*(z-p)*A[m, p]*A[n, q]*A[r, z], m = 1 .. N), n = 1 .. N), r = 1 .. N), p = 1 .. N), q = 1 .. N), z = 1 .. N)):
A := Matrix(N, N, proc (i, j) options operator, arrow; 10*i^2/(20*i+j) end proc):

The results of above programs are equal to 1 and validation of method is observed.

If we can write the general form of determinant then we can find the inverse of any square non-singular matrices.

Now I try to write the mentioned program.

N := 7:
Digits := 40:
e := product(product(signum(a[j]-a[i]), j = i+1 .. N), i = 1 .. N):
ML := product(A[a[k], b[k]], k = 1 .. N):
s[0] := e*subs(`$`(a[q] = b[q], q = 1 .. N), e)*ML:
for i to N do
s[i] := sum(sum(s[i-1], a[i] = 1 .. N), b[i] = 1 .. N)
end do:
A := Matrix(N, N, proc (i, j) options operator, arrow; evalf((3*i+j)/(i+2*j)) end proc): # arbitrary matrix

Therefore det(A)= CN-1 * e[a1,a2,] * e [b1,b2,.., bn] * A[a1,b1] * A[a2,b2] * ... * A[an,bn].

The correction coefficient is CN(for N)/CN(for N-1) = N!/(N-1)! =N.

with(linalg): N := 4: Digits := 20:
e := product(product(signum(a[j]-a[i]), j = i+1 .. N), i = 1 .. N):
ML := product(A[a[k], b[k]], k = 1 .. N):
s[0] := e*subs(`$`(a[q] = b[q], q = 1 .. N), e)*ML:
for r to N do s[r] := sum(sum(s[r-1], a[r] = 1 .. N), b[r] = 1 .. N) end do:
A := Matrix(N, N, proc (i, j) options operator, arrow; evalf((3*i+2*j)/(i+2*j)) end proc):
for x to N do for y to N do
e := product(product(signum(a[j]-a[i]), j = i+1 .. N-1), i = 1 .. N-1):
ML := product(AA[a[k], b[k]], k = 1 .. N-1):
S[0, x, y] := e*subs(`$`(a[q] = b[q], q = 1 .. N-1), e)*ML:
for r to N-1 do S[r, x, y] := sum(sum(S[r-1, x, y], a[r] = 1 .. N-1), b[r] = 1 .. N-1) end do:
f[y, x] := (-1)^(x+y)*subs(seq(seq(AA[t, u] = delrows(delcols(A, y .. y), x .. x)[t, u], t = 1 .. N-1), u = 1 .. N-1), S[N-1, x, y])
end do: end do:
Matrix(N, N, f)/(DET)*(24/6);

CN for N=4 and N=3 is 24 and 6 respectively i.e. CN(4)/CN(3)=24/6.

When I use bellow procedure the error "(in S) bad index into Matrix" is occurred.

Please help me to write this algorithm by using procedure.

Thank you 

restart; with(linalg): Digits := 40: n := 7:
S := proc (N) local e, ML, s, i:
e := product(product(signum(a[j]-a[i]), j = i+1 .. N), i = 1 .. N):
ML := product(A[a[k], b[k]], k = 1 .. N):
s[0] := e*subs(`$`(a[q] = b[q], q = 1 .. N), e)*ML:
for i to N do s[i] := sum(sum(s[i-1], a[i] = 1 .. N), b[i] = 1 .. N) end do
end proc:
A := Matrix(n, n, proc (i, j) options operator, arrow; evalf((3*i+j)/(i+2*j)) end proc): # arbitrary matrix
CN := simplify(S(n)/det(A))

Hi and thanks in advance

I have a function like

w=f(r) on the other hand

z=g(r) wher r is common between these two.

how i could plot w in term of z?


I've tried to find the solution to my problem, but none of my attempts was succesful.

I have a function which is one-to-one in a particular domain which I am interested in. I would like to get the inverse function of it only in this domain. Here is my function and plot for xp=0..10000:

x := xp-> (-1)*720.5668720*sinh(0.2043018094e-3*xp-0.8532729286)+84952.59423+4.014460003*10^5*arcsinh(0.1219272144e-1*sinh(0.2043018094e-3*xp-0.8532729286)-0.2032498888)

I would appreciate any help,



I tried to solve below equation, but it gives me zeros result. Please help me to find their inverse laplace. 

 It will be clearer if was pasted on Maple:



Ps := [P[0], P[1], P[2], P[3], P[4]]:

eqs := [P[0](s) =~ (P[1](s)*mu[1]+P[2](s)*mu[2]+P[3](s)*mu[3]+P[4](s)*mu[4])/(s+lambda[1]+lambda[2]+lambda[3]+lambda[4]), P[1](s) = lambda[1]*P[0](s)/(s+mu[1]), P[2](s) = lambda[2]*P[0](s)/(s+mu[2]), P[3](s) = lambda[3]*P[0](s)/(s+mu[3]), P[4](s) = lambda[4]*P[0](s)/(s+mu[4])];

Ls := solve(eqs, Ps(s))[];

P(t)=~inttrans[invlaplace]~(rhs~(Ls), s, t);


Thank you




I have a complex set of Markov Processes in reliability application. To make them simpler for me, as a beginner in Maplesoft, I solve them manually to reach a point where I need inverse Laplace for a set of equations. For illustration, I used a simple example below. If I get the concepts for below example, I can apply them on more complicated systems, as following:

P0(s) = 1/(s+λ)+υ*P1(s)/(s+λ)


Mannuly I find that:



Please help me step by step to understand how to solve such inverse Laplace. 

Thank you,

Hello please,

Can some one help me find inverse of this fucntion 

f(x) = binomial(x+r-1, x)*((d*x+1)/(1+d*x+(1/2)*b))^x*((1/2)*b/(1+d*x+(1/2)*b))^r/(d*x+1)





I am trying to find the inverse fourier transform given a vector (of real) as input. I get the following error message:

Error, (in DiscreteTransforms:-InverseFourierTransform) entries of input data must all evaluate to complex floats, got '[.7644453211]'

I am unable to understand this message. I would much appreciate help to resolve this issue.



I was trying to find the solution for two theta variables in a couple of simultaneous equations (infact this is an iverse kinematics problem for a two link system pendulum).
The following are the initial inputs/equations to be manipulated:

Then I use the folowing command to rearrange for the theta values which I am after:

which gives me the result:

This is all fine until I give in values for l1, l2, x and y:


I have a RootOf in there with a _Z term poping up here and there. I know that this configuration of the two link mechanism in fact dows have a solution and that these numbers are reasonable. Thus I have three questions:

Why does this happen?
What does the "signum" mean here?
how do I go about getting the nummerical values?

Many thanks,
- pjf


I have 20 by 20 matrix (with only two symbolic components, say a and b), other entries of the matrix are either populated with zeros or real numbers. I am trying to obtain the inverse.


All the command I have tried returned an error message. In particular M^(-1) yields:

Error, (in radnormal/ifactors) too many levels of recursion.


I would appreciate any suggestion that could solve this issue. Thank you


ps: I am essentially trying to solve Ax=b. The commands within LinearAlgebra all yielded similar error messages.

where A^T is the transpose of A and it's given that (A^T).A is not invertible.

I am stuck as to how to arrive at the solution for x in this case. I initially thought I could multiply both sides by the inverse of A^T reducing it to Ax=b but that was obviously wrong since A^T is itself not invertible(it is singular).

Dear All


I have a question:

How can I computet the laplace inverse of a function that is below!

My version of Maple is 13:

It should be noted that r>0, \rho>0.5, y\in R.


> restart;
> with(MTM);
print(??); # input placeholder
> ilaplace(r^(rho-.5)*abs(y)^(rho-1)*exp(-r*y)/GAMMA(rho-.5), s);

Dear all,

I have a question, why is the output of the inverse Laplace transformation if the signal is multiplied by itself not just convoluted in time domain:









assume system of polynomials are above, how to test whether inverse exist

and find inverse of them




I am trying to find the inverse of 8x8 generic symbolic matrix. Everytime I evaluate the program I get the following error:


Error, (in expand/bigprod) Maple was unable to allocate enough memory to complete this computation. Please see ?alloc


Can anyone guide me how to increase the memory allocation? And would that solve the problem for me?


Any help is appreciated.



Dear all,

It's very convenient to define a DE or PDE through Differential Operator D, for example,

((D[1, 1]+D[1, 2]+D[2, 2])(z))(x, y) = exp(x)*sin(y)

Is it possible to realize Inverse Operator Method of Operator D? How to solve the following equation if we rewrite the pde through inverse operator method?

(z)(x, y)=((D[1, 1]+D[1, 2]+D[2, 2])^(-1))exp(x)*sin(y)


Thanks a lot.

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