Solving Equaiton with multplying power

I have a equation given below, I need to solve this equaiton in matlab and find base X,Y,Z . There is solution given.

In matlab, I try to setup base but I cant figure out exactly. How I setup this equation in Matlab ? Thanks for help.

My matlab code for design base of the equation :

clear; clc;
[0 1 1^-1 ] % B*C^(-1)
[0 1 0] % B
[1 1^-3 0] %A*B^-3
[0 1^3 1^-1] %B^3*C^-1

Answer 1

[Walter Roberson]

syms A B C
syms X Y Z integer
eqn = (B*C^-1)^X * (B)^Y * (A*B^-3)^Z * (B^3*C^-1) == A^0*B^0*C^0

eqn = 

seqn = simplify(lhs(eqn)) == simplify(rhs(eqn))

seqn = 

collect(lhs(seqn),C)

ans = 

powers = findSymType(ans, 'power')

powers = 

need_to_solve = arrayfun(@(expression) children(expression,2), powers)

need_to_solve = 

syms X Y Z %remove assumptions
sol = solve(need_to_solve)

sol = struct with fields:
    X: -1
    Y: -2
    Z: 0

[Zafer Sahin]

Thank you very much!

Could you write answer in this form:

Thanks!

[Walter Roberson]

You might be able to get somewhere using displayFormula https://www.mathworks.com/help/symbolic/displayformula.html

[jack london]

Hi again,

When I run code on my matlab R2021a the output is:

Gıves this format, how I convert to numeric values (X=-1 Y=-2 Z=0)

[Walter Roberson]

X = double(sol.X)
Y = double(sol.Y)
Z = double(sol.Z)

Remember though that if you are using displayFormula, then what it prefers is character values -- which would be from char(sol.X) or string(sol.X) rather than double(sol.X)

Answer 2

[Matt J]

LHS=[0 0  1;
    -1 0  0;
     1 1 -3];

RHS=[0;1;-3];

XYZ=LHS\RHS

XYZ = 3×1

-1 -2 0

[jack london]

Thansk for answer.

I dont understant what represents LHS and RHS, could you give more detail please