How do i solve this block matrix equation using MATLAB ?

2018 8 15
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更新时间：2018 8 18
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I have this equation
[A B;C 0]*[G;H]=[G*Am;Cm]
which is needed to be solved for G and H , this equation can be written as
[G;H]=[A B;C 0]*inv([A B;C 0]*[A B;C 0]')*[G*Am;Cm]
and the matrices are given as
A=[0 1;1 -2],B=[0;1],C=[1 0],Am=[0 1;-1 0],Cm=[1 0.5]
dimension of G should be 2X2 and H 1X2
How to write a code to solve for G and H ?

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Walter Roberson
Walter Roberson 2018-8-16

2 个投票

A=[0 1;1 -2]; B=[0;1]; C=[1 0]; Am=[0 1;-1 0]; Cm=[1 0.5];
G = sym('G',[2 2]);
H = sym('H' ,[1 2]);
sol = solve([A B;C 0]*[G;H] == [G*Am;Cm], [G(:); H(:)]);
Gsol = subs(G, sol)
Hsol = subs(H, sol)

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siddhartha ganguly
siddhartha ganguly 2018-8-17
编辑：siddhartha ganguly 2018-8-17
Thanks for the help , but when i run the code it's showing sol=[empty sym],Gsol=[empty sym],Hsol=[empty sym]
Walter Roberson
Walter Roberson 2018-8-17
编辑：Walter Roberson 2018-8-17
Which release are you using? I tested in R2017a and R2018a... Oh wait, I see you marked R2013a. I will test more.
Walter Roberson
Walter Roberson 2018-8-17
A=[0 1;1 -2]; B=[0;1]; C=[1 0]; Am=[0 1;-1 0]; Cm=[1 0.5];
G = sym('G',[2 2]);
H = sym('H' ,[1 2]);
sol = solve([A B;C 0]*[G;H] == [G*Am;Cm]);
Gsol = subs(G, sol)
Hsol = subs(H, sol)
Tested in R2013a.
siddhartha ganguly
siddhartha ganguly 2018-8-17
yes R2013a , thanks for the effort , please let me know if it gets solved , thanks again
siddhartha ganguly
siddhartha ganguly 2018-8-17
hey it's working now,thank you so much
siddhartha ganguly
siddhartha ganguly 2018-8-17
编辑：Walter Roberson 2018-8-17
sir i have another doubt , i was using the same code for another example , but it's showing
"Error using vertcat
Dimensions of matrices being concatenated are not consistent.
Error in Untitled6 (line 7)
sol = solve([A B;C 0]*[G;H] == [G*Am;Cm]);"
Why is that happening ?? dimensions seems to be correct to me ..
clc
clear all
close all
A=[0 1 0;0 -0.87 42.22;0 0.99 -1.34]; B=[0 0;-17.25 -1.58;-0.17 -0.25]; C=[1 0 0]; Am=[0 0.3;-0.3 0]; Cm=[1 0];
G = sym('G',[3 2]);
H = sym('H' ,[2 2]);
sol = solve([A B;C 0]*[G;H] == [G*Am;Cm]);
Gsol = subs(G, sol)
Hsol = subs(H, sol)
Walter Roberson
Walter Roberson 2018-8-17
Your A is 3 columns, and your B is 2 columns, so [A B] is 5 columns.
Your C is 3 columns, and 0 is 1 column, so [C 0] is 4 columns.
So you are trying to put a matrix with 4 columns at the bottom of a matrix with 5 columns.
siddhartha ganguly
siddhartha ganguly 2018-8-17
Thanks,got it , but still not getting the whole matrix , here is the code
clc
clear all
close all
A=[0 1 0;0 -0.87 42.22;0 0.99 -1.34]; B=[0 0;-17.25 -1.58;-0.17 -0.25]; C=[1 0 0]; Am=[0 0.3;-0.3 0]; Cm=[1 0];
G = sym('G',[3 2]);
H = sym('H' ,[2 2]);
sol = solve([A B;C [0 0]]*[G;H] == [G*Am ;Cm]);
Gsol = subs(G, sol)
Hsol = subs(H, sol)
The output is :
Walter Roberson
Walter Roberson 2018-8-17
A and B are 3 rows. You put C on the bottom of [A B] and C has 1 row, so [A B; C 0 0] is 4 rows total, each of which has 5 columns. [G; H] is 5 rows by 2 columns. (4 x 5) * (5 x 2) gives a 4 x 2 array. So you have 8 equations. But you have 10 variables to solve for. You do not have enough equations to solve for 10 variables simultaneously.

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