# Matrix Differential Equation

1. Feb 17, 2012

### zack7

1. The problem statement, all variables and given/known data
The questions are in the image

3. The attempt at a solution
My solutions are
V1=3*(1 -2)e-2t+ (-2) (1 -3)e-3t
V2=1*(1 -2)e-2t+ (-1) (1 -3)e-3t

How do I get the X matrix since my solutions are in exponential still.

Thank you for all the help

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Last edited: Feb 17, 2012
2. Feb 18, 2012

### Some Pig

$$X=(v_1\ v_2)=\begin{pmatrix}3e^{-2t}-2e^{-3t}&-6e^{-2t}+6e^{-3t}\\ e^{-2t}-e^{-3t}&-2e^{-2t}+3e^{-3t}\end{pmatrix}.$$

3. Feb 18, 2012

### zack7

That means I just plug in t=0 to prove that x(0)=(1 0
0 1) and also use the same method to prove that dx/dt =AX

Last edited: Feb 18, 2012
4. Feb 18, 2012

### vela

Staff Emeritus
You have the columns and rows swapped.

5. Feb 18, 2012

### zack7

Even after multiplying it I get this respective solutions, what do I do next?

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6. Feb 18, 2012

### vela

Staff Emeritus
You can write v1 as a single vector:
$$\vec{v}_1 = \begin{pmatrix} 3e^{-2t} - 2e^{-3t} \\ -6e^{-2t} + 6e^{-3t}\end{pmatrix}$$Do the same for $\vec{v}_2$.

7. Feb 18, 2012

### zack7

i get that part but after that what do I do to get just numbers in my 2x2 matrix so that I can prove dx/dt =AX and x(0)= \begin{pmatrix}1 0\\ 0 1\end{pmatrix}

8. Feb 18, 2012

### vela

Staff Emeritus
Like the problem says, the first column of X is v1. Its second column is v2. It's not going to be just numbers. I'm not sure why you think it has to be.

9. Feb 18, 2012

### zack7

Okay but then if I put t=0 into the x equation, I do not get the identity matrix and how would I verify that dx/dt=AX by just differentiating the X matrix?.

Thank you for all the help

10. Feb 18, 2012

### vela

Staff Emeritus
Show us how you're calculating X when t=0.

11. Feb 18, 2012

### zack7

Okay I got just did a arithmetic error, but how do I verify that dx/dt=AX

12. Feb 18, 2012

### vela

Staff Emeritus
Calculate both sides and show they're equal to each other.

13. Feb 18, 2012

### zack7

Okay I will give it a try, thank you very much for all the help.