Maple_Maple

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14 years, 8 days

 

 

Reza2961

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These are answers submitted by Maple_Maple

 I read all answered and each result was correct , but it is`nt my c.

because i want to find a special c.

maybe my question wrong or .........

but i want to solve

and for this i use Q= C.C

then i want to find C,



(1)

(2)

(3)



Download MyMatrix.mw

 

 

Reza

with(LinearAlgebra):
q:= 3: Values:= {a[0]=3, a[1]=2, a[2]=1, a[3] = 0, b[m] = -1};
A:= subs(Values,Matrix(q,q, proc(i,j) if i >= j then a[i-j] else b[m] end if end proc));
B:= subs(Values,Matrix(q,q,proc(i,j) if i > j then b[m] else a[q+i-j] end if end proc));
Q:= Matrix(q,q,symbol=x, shape=symmetric);
eq:= Q + A . B^%T . Q^(-1) . B . A^%T - (A . A^%T + B . B^%T);
S:= fsolve({seq(seq(eq[i,j],j=i..q),i=1..q)});

 I don`t Want symmetric matrix.

Reza

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