Find X,Y,Z of this linear

Question

tiara 2023-8-24

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https://ww2.mathworks.cn/matlabcentral/answers/2012197-find-x-y-z-of-this-linear

移动：Stephen23 2023-8-24

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

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Stephen23 2023-8-24

移动：Stephen23 2023-8-24

在 MATLAB Online 中打开

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

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Answer 1

Sam Chak 2023-8-24

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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

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Answer 2

Nathan Hardenberg 2023-8-24

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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

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Stephen23 2023-8-24

"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

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Answer 3

Ramtej 2023-8-24

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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!

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Answer 4

Paul Bower 2023-8-24

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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