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Solving a system of equations

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Mitra Taghizadeh
Mitra Taghizadeh 2022 年 8 月 9 日
回答済み: Paul 2022 年 8 月 10 日
Hello everyone.
I have a system of equations in matrix form like this: (This is just an example. In fact the size of matrix A is 13*1 and B is 13*53 and C is 53*13]).
A=B*C
A=[x1,y1,z1,0,0,0,1]'
B=[a1,a2,a3,0,0,0,a4;a5,a6,a7,0,0,0,0;a8,a9,a10,0,0,0,0;0,0,0,a11,a12,0,a13;0,0,0,a14,a15,0,a16;0,0,0,a17,a18,a19,a20;0,0,0,0,0,0,1];
C=[3,4,5,x2,y2,z2,1]'
In this equation x1,x2,y1,y2,z1,z2 are the unknowns and ai (i=1 to 20) are the known values.
Do you know how can I solve this system?
  2 件のコメント
Paul
Paul 2022 年 8 月 9 日
Hi Mitra
Please update the example. B has 6 rows; it should have seven.
Mitra Taghizadeh
Mitra Taghizadeh 2022 年 8 月 10 日
Hi
Yes thank you.

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回答 (1 件)

Paul
Paul 2022 年 8 月 10 日
Hi Mitra
Maybe this will extend to the larger problem.
syms x1 y1 z1 x2 y2 z2
syms a [1 20]
A=[x1,y1,z1,0,0,0,1].';
B=[a1,a2,a3,0,0,0,a4;a5,a6,a7,0,0,0,0;a8,a9,a10,0,0,0,0;0,0,0,a11,a12,0,a13;0,0,0,a14,a15,0,a16;0,0,0,a17,a18,a19,a20;0,0,0,0,0,0,1];
C=[3,4,5,x2,y2,z2,1].';
sol = solve(A==B*C,[x1 y1 z1 x2 y2 z2])
sol = struct with fields:
x1: 3*a1 + 4*a2 + 5*a3 + a4 y1: 3*a5 + 4*a6 + 5*a7 z1: 3*a8 + 4*a9 + 5*a10 x2: (a12*a16 - a13*a15)/(a11*a15 - a12*a14) y2: -(a11*a16 - a13*a14)/(a11*a15 - a12*a14) z2: (a11*a16*a18 - a12*a16*a17 - a13*a14*a18 + a13*a15*a17 - a11*a15*a20 + a12*a14*a20)/(a19*(a11*a15 - a12*a14))

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