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Show matrix is not invertible.

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L
L on 23 Oct 2023
Answered: Sam Chak on 24 Oct 2023
  2 Comments
Dyuman Joshi
Dyuman Joshi on 23 Oct 2023
How is your question related to MATLAB?
John D'Errico
John D'Errico on 23 Oct 2023
Edited: John D'Errico on 23 Oct 2023
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.

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Answers (5)

Fabio Freschi
Fabio Freschi on 23 Oct 2023
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]
A = 
rank(A)
ans = 4
  5 Comments
Torsten
Torsten on 23 Oct 2023
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.
Fabio Freschi
Fabio Freschi on 23 Oct 2023
@Torsten: thank you! good to know

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Torsten
Torsten on 23 Oct 2023
Moved: Torsten on 23 Oct 2023
b/a*[a 0 0 0 0] + f/e*[0 0 e 0 0] = [b 0 f 0 0]

Matt J
Matt J on 23 Oct 2023
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)
ans = 
0
  1 Comment
Matt J
Matt J on 23 Oct 2023
Edited: Matt J on 23 Oct 2023
Or, analytically,
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]
A = 
B=A(2:4,3:5)
B = 
By co-factor expansion, clearly,
det(A) = -a*h*det(B)
and clearly det(B)=0.

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Walter Roberson
Walter Roberson on 23 Oct 2023
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)
DC = 1×120 cell array
Columns 1 through 7 {[M1_1*M2_2*M3_3*M…]} {[-M1_1*M2_2*M3_3*…]} {[-M1_1*M2_2*M3_4*…]} {[M1_1*M2_2*M3_4*M…]} {[M1_1*M2_2*M3_5*M…]} {[-M1_1*M2_2*M3_5*…]} {[-M1_1*M2_3*M3_2*…]} Columns 8 through 14 {[M1_1*M2_3*M3_2*M…]} {[M1_1*M2_3*M3_4*M…]} {[-M1_1*M2_3*M3_4*…]} {[-M1_1*M2_3*M3_5*…]} {[M1_1*M2_3*M3_5*M…]} {[M1_1*M2_4*M3_2*M…]} {[-M1_1*M2_4*M3_2*…]} Columns 15 through 21 {[-M1_1*M2_4*M3_3*…]} {[M1_1*M2_4*M3_3*M…]} {[M1_1*M2_4*M3_5*M…]} {[-M1_1*M2_4*M3_5*…]} {[-M1_1*M2_5*M3_2*…]} {[M1_1*M2_5*M3_2*M…]} {[M1_1*M2_5*M3_3*M…]} Columns 22 through 28 {[-M1_1*M2_5*M3_3*…]} {[-M1_1*M2_5*M3_4*…]} {[M1_1*M2_5*M3_4*M…]} {[-M1_2*M2_1*M3_3*…]} {[M1_2*M2_1*M3_3*M…]} {[M1_2*M2_1*M3_4*M…]} {[-M1_2*M2_1*M3_4*…]} Columns 29 through 35 {[-M1_2*M2_1*M3_5*…]} {[M1_2*M2_1*M3_5*M…]} {[M1_2*M2_3*M3_1*M…]} {[-M1_2*M2_3*M3_1*…]} {[-M1_2*M2_3*M3_4*…]} {[M1_2*M2_3*M3_4*M…]} {[M1_2*M2_3*M3_5*M…]} Columns 36 through 42 {[-M1_2*M2_3*M3_5*…]} {[-M1_2*M2_4*M3_1*…]} {[M1_2*M2_4*M3_1*M…]} {[M1_2*M2_4*M3_3*M…]} {[-M1_2*M2_4*M3_3*…]} {[-M1_2*M2_4*M3_5*…]} {[M1_2*M2_4*M3_5*M…]} Columns 43 through 49 {[M1_2*M2_5*M3_1*M…]} {[-M1_2*M2_5*M3_1*…]} {[-M1_2*M2_5*M3_3*…]} {[M1_2*M2_5*M3_3*M…]} {[M1_2*M2_5*M3_4*M…]} {[-M1_2*M2_5*M3_4*…]} {[M1_3*M2_1*M3_2*M…]} Columns 50 through 56 {[-M1_3*M2_1*M3_2*…]} {[-M1_3*M2_1*M3_4*…]} {[M1_3*M2_1*M3_4*M…]} {[M1_3*M2_1*M3_5*M…]} {[-M1_3*M2_1*M3_5*…]} {[-M1_3*M2_2*M3_1*…]} {[M1_3*M2_2*M3_1*M…]} Columns 57 through 63 {[M1_3*M2_2*M3_4*M…]} {[-M1_3*M2_2*M3_4*…]} {[-M1_3*M2_2*M3_5*…]} {[M1_3*M2_2*M3_5*M…]} {[M1_3*M2_4*M3_1*M…]} {[-M1_3*M2_4*M3_1*…]} {[-M1_3*M2_4*M3_2*…]} Columns 64 through 70 {[M1_3*M2_4*M3_2*M…]} {[M1_3*M2_4*M3_5*M…]} {[-M1_3*M2_4*M3_5*…]} {[-M1_3*M2_5*M3_1*…]} {[M1_3*M2_5*M3_1*M…]} {[M1_3*M2_5*M3_2*M…]} {[-M1_3*M2_5*M3_2*…]} Columns 71 through 77 {[-M1_3*M2_5*M3_4*…]} {[M1_3*M2_5*M3_4*M…]} {[-M1_4*M2_1*M3_2*…]} {[M1_4*M2_1*M3_2*M…]} {[M1_4*M2_1*M3_3*M…]} {[-M1_4*M2_1*M3_3*…]} {[-M1_4*M2_1*M3_5*…]} Columns 78 through 84 {[M1_4*M2_1*M3_5*M…]} {[M1_4*M2_2*M3_1*M…]} {[-M1_4*M2_2*M3_1*…]} {[-M1_4*M2_2*M3_3*…]} {[M1_4*M2_2*M3_3*M…]} {[M1_4*M2_2*M3_5*M…]} {[-M1_4*M2_2*M3_5*…]} Columns 85 through 91 {[-M1_4*M2_3*M3_1*…]} {[M1_4*M2_3*M3_1*M…]} {[M1_4*M2_3*M3_2*M…]} {[-M1_4*M2_3*M3_2*…]} {[-M1_4*M2_3*M3_5*…]} {[M1_4*M2_3*M3_5*M…]} {[M1_4*M2_5*M3_1*M…]} Columns 92 through 98 {[-M1_4*M2_5*M3_1*…]} {[-M1_4*M2_5*M3_2*…]} {[M1_4*M2_5*M3_2*M…]} {[M1_4*M2_5*M3_3*M…]} {[-M1_4*M2_5*M3_3*…]} {[M1_5*M2_1*M3_2*M…]} {[-M1_5*M2_1*M3_2*…]} Columns 99 through 105 {[-M1_5*M2_1*M3_3*…]} {[M1_5*M2_1*M3_3*M…]} {[M1_5*M2_1*M3_4*M…]} {[-M1_5*M2_1*M3_4*…]} {[-M1_5*M2_2*M3_1*…]} {[M1_5*M2_2*M3_1*M…]} {[M1_5*M2_2*M3_3*M…]} Columns 106 through 112 {[-M1_5*M2_2*M3_3*…]} {[-M1_5*M2_2*M3_4*…]} {[M1_5*M2_2*M3_4*M…]} {[M1_5*M2_3*M3_1*M…]} {[-M1_5*M2_3*M3_1*…]} {[-M1_5*M2_3*M3_2*…]} {[M1_5*M2_3*M3_2*M…]} Columns 113 through 119 {[M1_5*M2_3*M3_4*M…]} {[-M1_5*M2_3*M3_4*…]} {[-M1_5*M2_4*M3_1*…]} {[M1_5*M2_4*M3_1*M…]} {[M1_5*M2_4*M3_2*M…]} {[-M1_5*M2_4*M3_2*…]} {[-M1_5*M2_4*M3_3*…]} Column 120 {[M1_5*M2_4*M3_3*M…]}
det_components = cellfun(@(c) subs(c, M, A), DC, 'uniform', 0)
det_components = 1×120 cell array
Columns 1 through 22 {[0]} {[0]} {[0]} {[0]} {[0]} {[0]} {[0]} {[0]} {[0]} {[0]} {[0]} {[0]} {[0]} {[0]} {[0]} {[0]} {[0]} {[0]} {[0]} {[0]} {[0]} {[0]} Columns 23 through 44 {[0]} {[0]} {[0]} {[0]} {[0]} {[0]} {[0]} {[0]} {[0]} {[0]} {[0]} {[0]} {[0]} {[0]} {[0]} {[0]} {[0]} {[0]} {[0]} {[0]} {[0]} {[0]} Columns 45 through 66 {[0]} {[0]} {[0]} {[0]} {[0]} {[0]} {[0]} {[0]} {[0]} {[0]} {[0]} {[0]} {[0]} {[0]} {[0]} {[0]} {[0]} {[0]} {[0]} {[0]} {[0]} {[0]} Columns 67 through 88 {[0]} {[0]} {[0]} {[0]} {[0]} {[0]} {[0]} {[0]} {[0]} {[0]} {[0]} {[0]} {[0]} {[0]} {[0]} {[0]} {[0]} {[0]} {[0]} {[0]} {[0]} {[0]} Columns 89 through 110 {[0]} {[0]} {[0]} {[0]} {[0]} {[0]} {[0]} {[0]} {[0]} {[0]} {[0]} {[0]} {[0]} {[0]} {[0]} {[0]} {[0]} {[0]} {[0]} {[0]} {[0]} {[0]} Columns 111 through 120 {[0]} {[0]} {[0]} {[0]} {[0]} {[0]} {[0]} {[0]} {[0]} {[0]}
temp = [det_components{:}];
sum(temp)
ans = 
0
  4 Comments
Matt J
Matt J on 23 Oct 2023
Yes, you've shown a stronger condition than det(A)=0, but why did you need to?
Walter Roberson
Walter Roberson on 23 Oct 2023
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.)

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Sam Chak
Sam Chak on 24 Oct 2023
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]
A = 
% Perform Gaussian elimination to obtain the reduced row echelon form
R = rref(A)
R = 
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.
rank(R)
ans = 4

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