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Show matrix is not invertible.
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2 件のコメント
John D'Errico
2023 年 10 月 23 日
編集済み: John D'Errico
2023 年 10 月 23 日
Geez. Was there a reason for everybody to jump in and do this homework assignment? You teach the student nothing more than that there are too many people willing to do their work for them. And that can never be a good thing.
回答 (5 件)
Fabio Freschi
2023 年 10 月 23 日
You can simply show that A is not full rank
syms a b c d e f g h
A = [a 0 0 b 0; 0 c 0 0 d; 0 0 e f 0; 0 0 0 0 g; 0 h 0 0 0]
rank(A)
5 件のコメント
Torsten
2023 年 10 月 23 日
In Mathematica, it is assumed that all symbols used in the matrix representation are independent. In the present case, this means 0 ~= a ~= b ~= c ~=d ~=e ~= f ~=g ~= h.
Matt J
2023 年 10 月 23 日
syms a b c d e f g h
A = [a 0 0 b 0; 0 c 0 0 d; 0 0 e f 0; 0 0 0 0 g; 0 h 0 0 0];
det(A)
Walter Roberson
2023 年 10 月 23 日
When you calculate the terms of the determinant, every component is 0.
syms a b c d e f g h
A = [a 0 0 b 0; 0 c 0 0 d; 0 0 e f 0; 0 0 0 0 g; 0 h 0 0 0];
syms M [5 5]
D = det(M);
DC = children(D)
det_components = cellfun(@(c) subs(c, M, A), DC, 'uniform', 0)
temp = [det_components{:}];
sum(temp)
4 件のコメント
Matt J
2023 年 10 月 23 日
Yes, you've shown a stronger condition than det(A)=0, but why did you need to?
Walter Roberson
2023 年 10 月 23 日
What was needed was to show that the sum of the components was 0. Along the way it turned out that for this particular matrix, that every component was 0 . (Which saved having to do any kind of proof that the sum of the non-zero components would definitely be 0.)
Sam Chak
2023 年 10 月 24 日
Hi @L
I believe this is an exercise in the Linear Algebra course. You have likely learned how to find the determinant of a matrix and calculate the inverse of a matrix using formulas. Additionally, you may know that if the determinant of a matrix is zero, the matrix is considered singular and not invertible.
However, I doubt your professor expects you to manually calculate the symbolic determinant of a 5x5 matrix using pen and paper. There are likely properties of matrix A that allow you to demonstrate that its determinant is zero with pen and paper, similar to the approach shown by @Matt J, but perhaps not as theoretically detailed as in @Walter Roberson's proof.
One of these properties is the fact that "if matrix A is invertible, then its reduced row-echelon form is an identity matrix". You can perform Gaussian elimination by hand to obtain the reduced row-echelon form. In MATLAB, you can achieve this using the "rref(A)" command. By inspecting the reduced row-echelon form, you can determine whether it resembles an identity matrix or not.
syms a b c d e f g h
% Original matrix
A = [a 0 0 b 0; 0 c 0 0 d; 0 0 e f 0; 0 0 0 0 g; 0 h 0 0 0]
% Perform Gaussian elimination to obtain the reduced row echelon form
R = rref(A)
Essentially, this result also implies that a row or column of matrix A is a linear combination of two or more rows or columns, respectively; thus, 
. This is why you can't achieve full rank, as demonstrated by @Fabio Freschi.
. This is why you can't achieve full rank, as demonstrated by @Fabio Freschi.rank(R)
0 件のコメント
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