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Inverse of 5x5 matrix

Syeda Nadiah Nahri さんによって質問されました 2019 年 10 月 11 日
最新アクティビティ Steven Lord
さんによって コメントされました 2019 年 10 月 11 日
Hi,
I am trying to obtain inverse of the following 5x5 matrix:
[ k_rot - C_M*s - C_rot*s + J_M*s^2, C_rot*s - k_rot, 0, 0, 0]
[ C_rot*s - k_rot, k_rot - C_S*s - C_rot*s + J_S*s^2 + i^2*k_n - C_n*i^2*s, 0, i*k_n - C_n*i*s, C_n*i*s - i*k_n]
[ 0, 0, k_B + k_ax - C_B*s - C_ax*s + M_B*s^2, C_ax*s - k_ax, 0]
[ 0, i*k_n - C_n*i*s, C_ax*s - k_ax, k_ax + k_n - C_ax*s - C_n*s + M_S*s^2, C_n*s - k_n]
[ 0, C_n*i*s - i*k_n, 0, C_n*s - k_n, k_n - C_g*s - C_n*s + M_T*s^2]
I used the following code in Matlab:
syms s k_rot C_M C_rot J_M J_S C_n i k_n M_B k_B k_ax C_B C_ax M_S M_T C_g C_S
A1 = ((s^2)*J_M)+(k_rot)-(s*C_M)-(s*C_rot);
A2 = (s*C_rot)-(k_rot);
A3 = (-k_rot)+(s*C_rot);
A4 = (s^2*J_S)+(k_rot)+(i^2*k_n)-(s*C_S)-(s*C_rot)-(i^2*s*C_n);
A5 = (i*k_n)-(i*s*C_n);
A6 = -(i*k_n)+(i*s*C_n);
A7 = (s^2*M_B)+(k_B)+(k_ax)-(s*C_B)-(s*C_ax);
A8 = -(k_ax)+(s*C_ax);
A9 = (i*k_n)-(i*s*C_n);
A10 = -(k_ax)+(s*C_ax);
A11 = (s^2*M_S)+(k_ax)+(k_n)-(s*C_ax)-(s*C_n);
A12 = -(k_n)+(s*C_n);
A13 = -(i*k_n)+(i*s*C_n);
A14 = -(k_n)+(s*C_n);
A15 = (s^2*M_T)+(k_n)-(s*C_n)-(s*C_g);
Arow1 = [A1 A2 0 0 0];
% Arow1 = [ k_rot - C_M*s - C_rot*s + J_M*s^2, C_rot*s - k_rot, 0, 0, 0];
Arow2 = [ A3 A4 0 A5 A6];
%Arow2 = [ C_rot*s - k_rot, k_rot - C_S*s - C_rot*s + J_S*s^2 + i^2*k_n - C_n*i^2*s, 0, i*k_n - C_n*i*s, C_n*i*s - i*k_n]
Arow3 = [0 0 A7 A8 0];
%Arow3 =[ 0, 0, k_B + k_ax - C_B*s - C_ax*s + M_B*s^2, C_ax*s - k_ax, 0]
Arow4 = [0 A9 A10 A11 A12]
% Arow4 = [ 0, i*k_n - C_n*i*s, C_ax*s - k_ax, k_ax + k_n - C_ax*s - C_n*s + M_S*s^2, C_n*s - k_n]
Arow5 = [0 A13 0 A14 A15]
%Arow5 = [ 0, C_n*i*s - i*k_n, 0, C_n*s - k_n, k_n - C_g*s - C_n*s + M_T*s^2]
A = [Arow1; Arow2; Arow3; Arow4; Arow5]
Answer = inv(A)
I am getting the following message after running the above code: "Output truncated. Text exceeds maximum line length for Command Window display."
Is there any other method to obtain the inverse of the above 5 by 5 matrix on Matlab?
Would appreciate the help. Thanks so much!

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1 件の回答

Steven Lord
回答者: Steven Lord
2019 年 10 月 11 日

The answer, written out as text, is about 731,000 characters long.
>> length(char(Answer))
ans =
731073
A quick Google search yielded the rule of thumb that an average page contains about 3,000 characters. So your expression, if printed as a book, would be a little longer than the original UK printing of the first Harry Potter novel at about 243 pages. Even calling simplify on the answer doesn't shorten it that much; it's down to around 707,000 characters (about 235 pages.)
What do you want to actually do with this result? To substitute values into that expression, use subs. I tried substituting 1 for each of the symbolic variables individually, but the results were still really long. Substituting s = 1 had the largest impact, decreasing the length of the answer to about 629,000.

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Syeda Nadiah Nahri 2019 年 10 月 11 日
Dear Sir,
I need to first obtain the inverse matrix, and then obtain the relation between X_T(s) and T_M,
where, X_T(s) is handle position of my project i.e. the output.
T_M is the torque from DC motor fed as input to the system.
Untitled1.png
Steven Lord
2019 年 10 月 11 日
You're trying to find the result of multiplying the inverse by a vector? In that case, don't use inv. Use the backslash operator (\) to solve A*x = b for x. [I assume by T_M in your picture you mean the symbolic variable M_T used in constructing your A matrix.]
V = A\[M_T; 0; 0; 0 ;0];
sol5 = V(5);
The expression sol5 is still pretty long, at a little over 27,000 characters or about 9 pages. But that's a lot better than 700,000+. pretty printing sol5 still requires some scrolling to see the whole thing, but you can see the whole thing. It also makes an impressively wide line of output when I ran this code in a Live Script in the Live Editor.

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