rref matrix with an unkown variable

Question

Nicola 2025-9-10

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/2179887-rref-matrix-with-an-unkown-variable

评论： Paul 2025-9-11

Given the matrix:

M = [1, 2, 3, 5;
    2, 3, 4, 8;
    3, 4, 5, c;
    1, 1, 0, 2];

How does one code this in matlab so that it outputs the proper rref() matrix

I am expecting to get (did this by hand):

rref(M) = [1, 0, 0, 2;

0, 1, 0, 0;

0, 0, 0, c-11;

0, 0, 1, 1];

When solving for rref(M) using my code it always gives me this:

[1, 0, 0, 0]

[0, 1, 0, 0]

[0, 0, 1, 0]

[0, 0, 0, 1]

0 个评论
显示 -2更早的评论隐藏 -2更早的评论

请先登录，再进行评论。

请先登录，再回答此问题。

Answer 1

Matt J 2025-9-10

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/2179887-rref-matrix-with-an-unkown-variable#answer_1570221

编辑：Matt J 2025-9-10

在 MATLAB Online 中打开

I am expecting to get (did this by hand):

Your expected result doesn't look right. For both row-echelon and reduced row-echelon form, the result must be triangular. If you just want (non-reduced) row-echelon form, you could use LU, e.g.,

syms c

M = [1, 2, 3, 5;

2, 3, 4, 8;

3, 4, 5, c;

1, 1, 0, 2];

[L,ref]=lu(M);

ref

ref =

6 个评论
显示 4更早的评论隐藏 4更早的评论

Nicola 2025-9-10

在 MATLAB Online 中打开

My answer by hand for ref is the same just rows 3 and 4 are switched. same for the rref.

I do have c=11

but i have been trying to figure out why it gives me a eye(4) but cant figure it out, i assumed it had to do with the variable.

this here is my code:

All the other matrixes work fine a,c,g and i but I cant get m to work as it should

% Define matrices A, C, G, I, and M
A = [2, -3,  1;
     5,  1,  2];
C = [1,  2,  4];
G = [1,  2,  3,  4;
     5,  6,  7,  8;
     9, 10, 11, 12];
I = [2, -1,  6;
     3,  2,  4;
     1, 10, -12;
     6, 11, -2];
% Defining variable c
syms c; % Define c as a symbolic variable
M = [1, 2, 3, 5;
     2, 3, 4, 8;
     3, 4, 5, c;
     1, 1, 0, 2];
% Calculate the reduced row echelon form
R_A = rref(A);
R_C = rref(C);
R_G = rref(G);
R_I = rref(I);
R_M = rref(M);
% Convert the RREF results to fractions (except M which is symbolic)
fraction_A = rats(R_A);
fraction_C = rats(R_C);
fraction_G = rats(R_G);
fraction_I = rats(R_I);
% Display the fraction representations of the RREF results
disp('Reduced Row Echelon Form of A:');
disp(fraction_A);
disp('Reduced Row Echelon Form of C:');
disp(fraction_C);
disp('Reduced Row Echelon Form of G:');
disp(fraction_G);
disp('Reduced Row Echelon Form of I:');
disp(fraction_I);
disp('Reduced Row Echelon Form of M:'); 
disp(R_M);
% MatLab is compiling rref differently then I did by hand.
% My result was:
% [1 0  0    2 ]
% |0 1  0    0 |
% |0 0  0  c-11|
% {0 0  1    1 ]
% Can only get it to do that if I manualy input all rref steps
% Finding c values that make M singular (determinant = 0)
det_M = det(M);
disp('Determinant of M:');
disp(det_M);
% Solving for c when determinant = 0
c_values = solve(det_M == 0, c);
disp('Values of c that make M singular:');
disp(c_values);

Paul 2025-9-11

编辑：Paul 2025-9-11

在 MATLAB Online 中打开

I'm going to guess that rref does not (and possibly cannot) account for all possible values of the symbolic variables. Somewhere along the way it's generating an expression that cancels out the c, even though such cancellation isn't correct for all possible values of c. For example

syms a b c

M = [a b;0 b]

M =

If I was doing this problem by hand, I might proceed as follows with row operations

R = M;

R(2,:) = R(2,:)/b % 1

R =

R(1,:) = R(1,:)/a % 2

R =

R(1,:) = R(1,:) - R(2,:)*b/a % 3

R =

That happens to be the same result as returned from rref, though I suspect rref got there a different way

rref(M)

ans =

Of course step 1 is invalid if b = 0, but the SMT is happy to do it. We see the same effect with the example from the rref

A = [a b c; b c a; a + b, b + c, c + a]

A =

rref(A)

ans =

That result can't be correct for any combination of (a,b,c) s.t. a*c - b^2 = 0

rref(subs(A,[a,b,c],[1,1,1])) % Case A

ans =

rref(subs(A,[a,b,c],[2,4,8])) % Case B

ans =

Interestingly, assumptions do influence rref

assume(a*c == b^2)

rref(A)

ans =

simplify(ans)

ans =

But I don't see how that result can recover Case A

Paul 2025-9-11

Inspired by @Torsten, here is the Wolfram Alpha result for the example from rref

https://www.wolframalpha.com/input?i=MatrixForm%5BRowReduce%5B%7B%7Ba%2Cb%2Cc%7D%2C%7Bb%2Cc%2Ca%7D%2C%7Ba%2Bb%2Cb%2Bc%2Cc%2Ba%7D%7D%5D%5D

请先登录，再进行评论。

rref matrix with an unkown variable

0 个评论
显示 -2更早的评论隐藏 -2更早的评论

采纳的回答

6 个评论
显示 4更早的评论隐藏 4更早的评论

更多回答（0 个）

另请参阅

类别

标签

产品

Community Treasure Hunt

rref matrix with an unkown variable

0 个评论 显示 -2更早的评论隐藏 -2更早的评论

采纳的回答

6 个评论 显示 4更早的评论隐藏 4更早的评论

更多回答（0 个）

另请参阅

类别

标签

产品

Community Treasure Hunt

0 个评论
显示 -2更早的评论隐藏 -2更早的评论

6 个评论
显示 4更早的评论隐藏 4更早的评论