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Find the inverse of the matrix

  • Thread starter js14
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  • #1
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Find the inverse of the matrix if it exist.
|e^x -e^2x|
|e^2x e^3x|

1/(e^x)(e^3x)-(-e^2x)(-e^2x)

|e^3 -e^2x|
|-e^2 e^x | I think this is right...

This is about as far as I can get.
Can someone help me complete it?
 

Answers and Replies

  • #2
33,636
5,296
Find the inverse of the matrix if it exist.
|e^x -e^2x|
|e^2x e^3x|

1/(e^x)(e^3x)-(-e^2x)(-e^2x)

|e^3 -e^2x|
|-e^2 e^x | I think this is right...
No, it's not right. You could have checked this yourself by multiplying this matrix and the one you started with. If you get the 2 x 2 identity matrix, then you have the inverse. If you don't get the identity matrix, then you made a mistake.

There is a simple formula for getting the inverse of a 2 x 2 matrix, that involves the determinant of the matrix. It looks like you calculated the determinant, but you didn't simplify it, and there is a sign error.
This is about as far as I can get.
Can someone help me complete it?
[tex]\text{If } A = \begin{bmatrix}a& b\\c & d\end{bmatrix}[/tex]
and det(A) [itex]\neq[/itex] 0, then
[tex]A^{-1} = \frac{1}{det(A)}\begin{bmatrix}d& -b\\-c & a\end{bmatrix}[/tex]
 
  • #3
43
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Alright i get that but im confused as to how im supposed to add and subtract the exponential functions.
 
  • #4
33,636
5,296
You first have to multiply them.
det(A) = exe3x - (-e2xe2x)

What do you get for each product?
 
  • #5
43
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e^4x+e^4x?
 
  • #6
33,636
5,296
Right, but that can be simplified. Isn't that just 2e4x?
 
  • #7
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So I should have 1/2e^4x to multiply by the matrix?
 
  • #9
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Ok I understand. But why isnt this what my matrix is supoosed to look like? |e^x -e^2x|
|e^2x e^3x|
 
  • #10
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Is this it |e^3x -e^-2x|
|e^-2x e^x |
Is this it this the right?
 
  • #12
43
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Ok I thought a=e^x b=-e^2x c=e^-2x d=e^3x

and then when you put it in |d -b| it was |e^3x -e^-2x|
|-c a| |e^-2x e^x |

If not then I don't understand...
 
  • #13
33,636
5,296
Here's the matrix you put in your first post:
|e^x -e^2x|
|e^2x e^3x|

c = e^(2x), not e^(-2x) as you have in post 12.
 
  • #14
43
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Right. Originally b=-e^2x but dosen't it change to -e^-2x when you switch everything around?
 
  • #15
43
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Can you show me how to finish the problem? or just give me an example?
 
  • #16
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is this right? 1/(2e^(4x))*|e^3x e^2x|
|-e^2x e^x|
 
  • #18
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I thought this forum was for help. I don't understand the exponentials. I just want an example to help understand what im doing. i am completely confused by exponentials. If this problem was pure numbers then I could solve it. I understand everything, just not the exponentials.
 
  • #19
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5,296
It is for help, but you will learn better by doing the work rather than having someone give you the answers. If you are at the stage that you are finding inverses of matrices, then you should already know how to multiply matrices. If you are given a matrix A, and find that another matrix B is the inverse of A, then you should get into the habit of checking that AB = I.

There isn't anything very complicated about the exponential functions.

ea*eb = ea + b
ea / eb = ea - b

You probably learned this a while back in precalculus or algebra, but you are evidently pretty rusty at it.
 
  • #20
43
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well anyway my final answer after multipliying by the scaler is |1/2e^x 1/2e^2x|
|1/2e^2x 1/2e^3x|
 
  • #21
33,636
5,296
is this right? 1/(2e^(4x))*
|e^3x e^2x|
|-e^2x e^x|
well anyway my final answer after multipliying by the scaler is
|1/2e^x 1/2e^2x|
|1/2e^2x 1/2e^3x|
These answers don't agree.
 
  • #22
33,636
5,296
Try to be more careful with your signs. You have made several sign errors throughout this thread.

Your final answer should be
|1/2e^x 1/2e^2x|
|-1/2e^2x 1/2e^3x|

There's another problem, though - the terms in this matrix are ambiguous. For example, someone could reasonably interpret the term in the upper left corner as (1/2)ex, and not as you intended, which was 1/(2ex). Same for the other three terms.

And instead of having fractions with exponential functions in the denominator, you could write all of the terms with negative exponents, with the idea being that 1/ax = a-x.

If you're checking your answer against the one in the answer book (if there is an answer), it will probably look something like this:
|(1/2)e^(-x) (1/2)e^(-2x)|
|(-1/2)e^(-2x) (1/2)e^(-3x)|
 
  • #23
46
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I appreciate your help. Thank you.
 

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