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Matrix problems with 3D plots

2 ビュー (過去 30 日間)
KAO YI
KAO YI 2012 年 11 月 29 日
閉鎖済み: MATLAB Answer Bot 2021 年 8 月 20 日
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;
  2 件のコメント
Matt Fig
Matt Fig 2012 年 11 月 29 日
Sounds like homework. What have you done so far?
KAO YI
KAO YI 2012 年 11 月 29 日
>> 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...

回答 (1 件)

Sara Hafeez
Sara Hafeez 2012 年 11 月 30 日
>> 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

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