how do i multiply n number of matrix for 2x2 with the symboic output

hi how do i use my symbolc output results and do a matrix multiplication for a 2x2 matrix?thank you in advance

[Walter Roberson]

ABCD=[A B;C D]
ABCD*M

[grace lim]

how do i do a for loop for n=10?

[Walter Roberson]

E = ABCD
for n=1:10
   E = E*M;
end

However if that were the operation you wanted to do then instead

E = ABCD * M^10;

[grace lim]

im usng symbolic to simplify symbols ,theres error,'M ' is not a variable

is there another way to solve?

what does M means?

[Walter Roberson]

M is the 2x2 matrix that you want to multiply by.

Maybe you want

E = ABCD
for n=1:9
   E = E*ABCD;
end

Answer 1

[grace lim]

i tried this way but there is no output

E = ABCD

for n=1:9

E = E*ABCD;

end

[Walter Roberson]

I am not going to type in your equations from your image, so I will illustrate with some examples:

syms h0 h1 h2 h3
A = 1/(h0+h1)

A = 

B = h0/(h0+h2)

B = 

C = h2*h3/h0

C = 

D = h3 - h1

D = 

ABCD = [A B; C D]

ABCD = 

E = ABCD

E = 

for n = 1:9
   E = E*ABCD;
end

E

E = 

If you need to see the results of each multiplication in turn, then change E = E*ABCD; to become E = E*ABCD without the semi-colon

Answer 2

[grace lim]

thank you,

how about if i wnt to do 2 matrix multiplication for n number of times?

do i do it this way?

E=[A B;CD]

F={A B;C D]

EF=E*F

for n = 1:9

E = E*ABCD;

end

[Walter Roberson]

My understanding is that you have symbols A, B, C, and D.

As I showed you earlier, ABCD = [A, B; C D] would arrange the symbols in a 2 x 2 array, in the order

A B
C D

Now you have

E=[A B;CD]

but we do not know what the symbol CD is . CD is not the same as C D

F={A B;C D]

That is not valid syntax. The { starts a cell array, but the ] ends a list .

If you intended for E =[A B;C D] then what is the difference between that and using the existing ABCD ?

[grace lim]

sorry ,what i meant was:

syms ....

h11..

h12..

h21...

h22...

E=[A B;C D]

...

F=[A B;C D]

..

EF=E*F

the 2 matrix ABCD are different

then,

for n = 1:9

E = E*ABCD;

end

[Walter Roberson]

How does ABCD differ from the [A B;C D] you assign to E, and how does that differ from the [A B;C D] that you assign to F?

Or are you changing A B C D between those points?

Anyhow... go ahead with your code. Just make sure you get right how many multiplications you want to do.

If we call the new A B C D as nA, nB, nC, nD then do you want [nA, nB; nC, nD] * ABCD^9 or do you want [nA, nB; nC, nD] * ABCD^10 ? Your current code would give [nA, nB; nC, nD] * ABCD^9 . In the special case where nA, nB, nC, nD are the same as A, B, C, D, then nA, nB; nC, nD] * ABCD^9 would be the same as ABCD^10