Matrix problems with 3D plots

Question

KAO YI 2012-11-29

0
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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/55186-matrix-problems-with-3d-plots

关闭： MATLAB Answer Bot 2021-8-20

Given the following system of linear equations,

-3a + b -c +d + e = 0
b + c + 4e = 0
-3c + 2d + e = 0

Please use MATLAB to do the following:

Use rref command to find the general solution, express all unknowns with two variables s and t;
Plot 3-D figures to show all unknowns as functions of -1<s<1 and -1<t<1;

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Matt Fig 2012-11-29

Sounds like homework. What have you done so far?

KAO YI 2012-11-29

在 MATLAB Online 中打开

>> A= [-3 1 -1 1 1; 0 1 1 0 4; 0 0 -3 2 1]

A =

    -3     1    -1     1     1
     0     1     1     0     4
     0     0    -3     2     1

>> R = rref(A)

R =

    1.0000         0         0    0.1111    1.2222
         0    1.0000         0    0.6667    4.3333
         0         0    1.0000   -0.6667   -0.3333
 this is what I got so far... but I dont quite know how to get t and s then plot into a 3D plot...

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Answer 1

Sara Hafeez 2012-11-30

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链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/55186-matrix-problems-with-3d-plots#answer_67058

在 MATLAB Online 中打开

>>  A= [-3 1 -1 1 1; 0 1 1 0 4; 0 0 -3 2 1]
A =
      -3     1    -1     1     1
       0     1     1     0     4
       0     0    -3     2     1
>> [R,jb] = rref(A)
R =
      1.0000         0         0    0.1111    1.2222
           0    1.0000         0    0.6667    4.3333
           0         0    1.0000   -0.6667   -0.3333
jb =
       1     2     3

[R,jb] = rref(A) also returns a vector jb such that:

    *
      r = length(jb) is this algorithm's idea of the rank of A.
    *
      x(jb) are the pivot variables in a linear system Ax = b.
    *
      A(:,jb) is a basis for the range of A.
    *
      R(1:r,jb) is the r-by-r identity matrix.    if true
      % code
    end