How Can I solve this equations system?

2014 2 25
2 个回答
更新时间：2021 8 20
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0 个投票

K.a+M.b=L.b
P.a+ Q .b=R.a
I have the equations above and I am trying to find a and b. All capital letters are constants. How can I solve this problem in Matlab?
Thanks

回答（2 个）

Walter Roberson
Walter Roberson 2014-2-25

0 个投票

Do the "." represent multiplication? Are the elements all scalars or are they matrices?
If everything is scalar and "." represents ordinary multiplication, then the solution is a = 0, b = 0.

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Walter Roberson
Walter Roberson 2014-2-25
I seem to find that there are potential multiple solutions,
a = b * (L-M)/K
for all real-valued b
Provided that Q has a particular ratio in K, L, M, P, R (independent of "a" and "b"). And when it does not have that particular ratio, a = 0, b = 0 looks to be the only solution.
Mehmet
Mehmet 2014-2-26
编辑：Mehmet 2014-2-26
Actually I should tell you my problem with a different way;
My equations are; (Ty25=Ty and Tx25=Tx so I am trying to find alpha and beta values)
Ty25 =
(551666622438673794225934829796875*beta)/1045347431181122959759486794030391945592832
Tx25 =
(551666622438673794225934829796875*alpha)/1045347431181122959759486794030391945592832
Ty =
(8563063365494341*cos(alpha)^2*((log(sin(alpha)*sin(beta) - cos(alpha) - 225)*(cos(alpha) + 225))/(1000000*cos(alpha)*(cos(alpha)^2 - 1)*(cos(beta)^2 - 1)) - (log(200*sin(alpha)*sin(beta) - cos(alpha) - 225)*(cos(alpha) + 225))/(1000000*cos(alpha)*(cos(alpha)^2 - 1)*(cos(beta)^2 - 1)) + (log(8*sin(alpha)*sin(beta) - 8*cos(alpha) - 9)*(8*cos(alpha) + 9))/(40000*cos(alpha)*(cos(alpha)^2 - 1)*(cos(beta)^2 - 1)) - (log((sin(alpha)*sin(beta))/25 - 8*cos(alpha) - 9)*(8*cos(alpha) + 9))/(40000*cos(alpha)*(cos(alpha)^2 - 1)*(cos(beta)^2 - 1))))/(4835703278458516698824704*(pi - 2*alpha)^2) - (8563063365494341*cos(alpha)^2*((log(- cos(alpha) - sin(alpha)*sin(beta) - 225)*(cos(alpha) + 225))/(1000000*cos(alpha)*(cos(alpha)^2 - 1)*(cos(beta)^2 - 1)) - (log(- cos(alpha) - 200*sin(alpha)*sin(beta) - 225)*(cos(alpha) + 225))/(1000000*cos(alpha)*(cos(alpha)^2 - 1)*(cos(beta)^2 - 1)) + (log(- 8*cos(alpha) - 8*sin(alpha)*sin(beta) - 9)*(8*cos(alpha) + 9))/(40000*cos(alpha)*(cos(alpha)^2 - 1)*(cos(beta)^2 - 1)) - (log(- 8*cos(alpha) - (sin(alpha)*sin(beta))/25 - 9)*(8*cos(alpha) + 9))/(40000*cos(alpha)*(cos(alpha)^2 - 1)*(cos(beta)^2 - 1))))/(4835703278458516698824704*(pi - 2*alpha)^2) - (8563063365494341*(1791/(1000000*cos(beta)^2*(8*cos(beta) - 9)) - 1791/(1000000*cos(beta)^2*(cos(beta)/25 - 9)) - (199*log(8*cos(beta) - 9))/(1000000*cos(beta)^2) + (199*log(cos(beta)/25 - 9))/(1000000*cos(beta)^2)))/(19342813113834066795298816*(sin(beta)^2/(cos(alpha)^2*cos(beta)^2) + 1)*(pi/2 - atan(sin(beta)/(cos(alpha)*cos(beta))))^2) - (8563063365494341*(1791/(1000000*cos(beta)^2*(8*cos(beta) + 9)) - 1791/(1000000*cos(beta)^2*(cos(beta)/25 + 9)) + (199*log(- 8*cos(beta) - 9))/(1000000*cos(beta)^2) - (199*log(- cos(beta)/25 - 9))/(1000000*cos(beta)^2)))/(19342813113834066795298816*(sin(beta)^2/(cos(alpha)^2*cos(beta)^2) + 1)*(pi/2 - atan(sin(beta)/(cos(alpha)*cos(beta))))^2)
Tx =
68160280339725220986141/(967140655691703339764940800000000*(sin(beta)^2/(cos(alpha)^2*cos(beta)^2) + 1)*(pi/2 - atan(sin(beta)/(cos(alpha)*cos(beta))))^2*(8*cos(beta)^2 - 1809*cos(beta) + 2025)) - (8563063365494341*cos(alpha)^2*((log(- cos(alpha) - sin(alpha)*sin(beta) - 225)*(cos(alpha) + 225))/(1000000*cos(alpha)*(cos(alpha)^2 - 1)*(cos(beta)^2 - 1)) - (log(- cos(alpha) - 200*sin(alpha)*sin(beta) - 225)*(cos(alpha) + 225))/(1000000*cos(alpha)*(cos(alpha)^2 - 1)*(cos(beta)^2 - 1)) + (log(- 8*cos(alpha) - 8*sin(alpha)*sin(beta) - 9)*(8*cos(alpha) + 9))/(40000*cos(alpha)*(cos(alpha)^2 - 1)*(cos(beta)^2 - 1)) - (log(- 8*cos(alpha) - (sin(alpha)*sin(beta))/25 - 9)*(8*cos(alpha) + 9))/(40000*cos(alpha)*(cos(alpha)^2 - 1)*(cos(beta)^2 - 1))))/(4835703278458516698824704*(pi - 2*alpha)^2) - (8563063365494341*(1791/(1000000*cos(beta)^2*(8*cos(beta) + 9)) - 1791/(1000000*cos(beta)^2*(cos(beta)/25 + 9)) + (199*log(- 8*cos(beta) - 9))/(1000000*cos(beta)^2) - (199*log(- cos(beta)/25 - 9))/(1000000*cos(beta)^2)))/(19342813113834066795298816*(sin(beta)^2/(cos(alpha)^2*cos(beta)^2) + 1)*(pi/2 - atan(sin(beta)/(cos(alpha)*cos(beta))))^2) + (8563063365494341*cos(alpha)^2*((log(8*sin(alpha)*sin(beta) - 8*cos(alpha) - 9)*(8*sin(alpha)*sin(beta) - 9))/(40000*cos(alpha)^2*sin(alpha)*sin(beta)) - (log(8*sin(alpha)*sin(beta) - cos(alpha)/25 - 9)*(8*sin(alpha)*sin(beta) - 9))/(40000*cos(alpha)^2*sin(alpha)*sin(beta)) + (log(sin(alpha)*sin(beta) - cos(alpha) - 225)*(sin(alpha)*sin(beta) - 225))/(1000000*cos(alpha)^2*sin(alpha)*sin(beta)) - (log(sin(alpha)*sin(beta) - 200*cos(alpha) - 225)*(sin(alpha)*sin(beta) - 225))/(1000000*cos(alpha)^2*sin(alpha)*sin(beta))))/(4835703278458516698824704*(pi - 2*alpha)^2)
Star Strider
Star Strider 2014-2-26
How did you get that from the equations in your original post?
What are you not telling us?
Mehmet
Mehmet 2014-2-27
it is lots of integral calculations (ABOUT 600 HUNDREDS ROW). At the end of all calculatıons I got these equatıons.
Walter Roberson
Walter Roberson 2014-2-27
I ran it through Maple and did some testing. You aren't going to be able to get a symbolic solution, I don't think.
You have a difficulty in the equations. Look carefully at the log() terms. One of them is log(8*cos(beta) - 9). For all real beta, 8*cos(beta)-9 is negative, ranging from -17 to -1. Thus you are introducing a complex term. Getting rid of that requires precise cancellation of all of the complex terms in the equation, which is pretty tricky to achieve.
Now if the alpha and beta are permitted to be complex then getting a solution numerically would be easier.
Mehmet
Mehmet 2014-2-27
nope the main point is that I am trying to get rid of complex term. How can I do that? Is there any way of it? Thanks
Mehmet
Mehmet 2014-2-25

0 个投票

Yes it is multiplication and they are scalar.

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2014-2-25

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