i want to solve this equation...
| 1 1 1 | |b0| |exp(t) |
| 0 1 2 | |b1|=|exp(t) |
| 1 1 1 | |b2| |exp(2*t)|
i like the answer be like this:
for example:
b0=2*exp(t)+exp(2*t)
b1=exp(t)+1
b2=exp(
Asked
Active
Viewed 1,004 times
0
-
You can't solve that linear system, it's dependent (many solutions). – Ben Voigt Apr 29 '12 at 16:55
-
Actually, there are many solutions when `t=0` and no solutions otherwise. – Ben Voigt Apr 29 '12 at 17:31
1 Answers
1
That matrix is singular, so there's no unique solution (depending on t, there may be zero or infinitely many solutions). I will replace it with an invertible matrix to demonstrate the method:
>> A = [1,1,1;0,1,2;1,1,0]
A =
1 1 1
0 1 2
1 1 0
After that, solving is a straightforward use of the symbolic capability:
>> t = sym('t');
>> rhs = [exp(t);exp(t);exp(2*t)]
rhs =
exp(t)
exp(t)
exp(2*t)
>> b = A\rhs
b =
exp(t) - exp(2*t)
2*exp(2*t) - exp(t)
exp(t) - exp(2*t)
Ben Voigt
- 277,958
- 43
- 419
- 720
-
Hi mr. Ben Voigt...thank you for your help...but i hace matrix like this |0 -2 1 | % a= |-1 -2 2 | % |0 -2 1 | that i wanted to solve it in hamilton style(e^at) and the ansert must be like this form(parametric) |2e^t-e^2t 0 2e^t-2e^2t| | 0 e^t 0 | |e^2t 0 2e^2t-e^t | but thank you for replying anyway..be goodluck. – mjjv Apr 30 '12 at 13:55