Find X,Y,Z of this linear

26.720X + 472.274Y + 791.273Z = 408.162.173.304

42.259X + 567.904Y + 1.358.438Z = 369.152.982.902

49.763X + 593.542Y + 1.405.099Z = 420.198.149.812

[Stephen23]

There is no need to use the symbolic toolbox, MATLAB offers basic and efficient numeric solvers for this task, e.g.:

https://www.mathworks.com/help/matlab/ref/mldivide.html

format long G
M = [26720,472274,791273; 42259,567904,1358438;49763,593542,1405099]

M = 3×3

26720 472274 791273 42259 567904 1358438 49763 593542 1405099

V = [408162173304; 369152982902; 420198149812]

V = 3×1

1.0e+00 * 408162173304 369152982902 420198149812

XYZ = M\V % simpler and faster than slow symbolic operations

XYZ = 3×1

1.0e+00 * 5031741.88514984 1290280.22776427 -424192.123520388

Checking the first equation:

M(1,:)*XYZ

ans = 
              408162173304

Answer 1

[Sam Chak]

Alternative approach. This should produce the same results as @Nathan Hardenberg.

A = [26 472 791;
     42 567 358;
     49 593 405]

A = 3×3

26 472 791 42 567 358 49 593 405

B = [ 408;
      369;
     1420]

B = 3×1

408 369 1420

X = A\B

X = 3×1

182.5581 -15.0968 3.5236

Answer 2

[Nathan Hardenberg]

This is very simple to do. I choose diffenent numbers, since I don't know how your "."-delimiters are meant to be.

syms x y z

equations = [
    26*x + 472*y + 791*z == 408;
    42*x + 567*y + 358*z == 369;
    49*x + 593*y + 405*z == 1420;
    ]

equations = 

solution = solve(equations, [x y z]);
solution.x

ans = 

solution.y

ans = 

solution.z

ans = 

% To evaluate to a double/float you can do
double(solution.x)

ans = 182.5581

eval(solution.x)  % or this

ans = 182.5581

[Stephen23]

"eval(solution.x) % or this"

Best not. EVAL is not part of the symbolic toolbox and its behavior with symbolic variables is not defined.

https://www.mathworks.com/help/symbolic/referencelist.html?type=function&s_tid=CRUX_topnav

Answer 3

[Ramtej]

Hi Tiara,

Please refer to Solve System of Linear Equations for the detailed instructions on how to solve system of linear equations using the Symbolic Math Toolbox.

Hope that helps!

Answer 4

[Paul Bower]

Hi, You could convert the three strings given into a system of equations and solve it using a linear solver. Some code which shows how to do it is given below:

Based on this code:

This is the matrix A:

26720 472274 791273

42259 567904 1358438

49763 593542 1405099

This is the vector b:

1.0e+11 *

4.081621733040000

3.691529829020000

4.201981498120000

This is the solution x:

1.0e+06 *

5.031741885149838

1.290280227764273

-0.424192123520388

I hope this helps. Have a good day. Regards, Paul Bower

% System of equations expressed as strings
str1 = '26.720X + 472.274Y + 791.273Z = 408.162.173.304';
str2 = '42.259X + 567.904Y + 1.358.438Z = 369.152.982.902';
str3 = '49.763X + 593.542Y + 1.405.099Z = 420.198.149.812';

% Initialize matrix A and vector b
A = [];
b = [];

% Convert first string to numbers and write into A & b
[numsArray, lastNum] = extractNumbers(str1);
A = [A; numsArray];
b = [b;lastNum];

% Convert second string to numbers and write into A & b
[numsArray, lastNum] = extractNumbers(str2);
A = [A; numsArray];
b = [b;lastNum];

% Convert third string to numbers and write into A & b
[numsArray, lastNum] = extractNumbers(str3);
A = [A; numsArray];
b = [b;lastNum];

% Display A
disp('This is the matrix A:');
disp(A);
disp('This is the vector b:');
disp(b);

% Solve the system of equations
x = A\b; 

disp('This is the solution, x:');
disp(x);

function [numsArray, lastNum] = extractNumbers(inputString)
    
    % Replace the periods (which represent commas or thousands) with empty spaces
    cleanedString = strrep(inputString, '.', '');

    % Use regular expressions to extract numbers from the string
    nums = regexp(cleanedString, '(-?\d+\.?\d*)', 'match');

    % Convert the extracted numbers into doubles and store them in an array
    numsArray = cellfun(@str2double, nums(1:end-1));
    
    % Extract the last number
    lastNum = str2double(nums{end});
end