MHB How Do You Invert This 2x2 Matrix?

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(-2e^2t)(sin(4t)) , (-2e^4t)(cos(4t))

(-2e^2t)(cos(4t)) , (2e^4t)(sin(4t))

Please and Thank you!
 
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Can you show us what you have tried? Our helpers will be better able to provide you with relevant help if they can see where you are stuck and/or where you may be making mistakes.
 
MarkFL said:
Can you show us what you have tried? Our helpers will be better able to provide you with relevant help if they can see where you are stuck and/or where you may be making mistakes.

So since it's a 2x2 matrix, it's easier to use the equation A(INVERSE) = (1/ad-bc)(d , -b
-c , a)

I get stuck here, I don't really know how to apply this formula when it's in a more complex form like this.
 
ends said:
So since it's a 2x2 matrix, it's easier to use the equation A(INVERSE) = (1/ad-bc)(d , -b
-c , a)

I get stuck here, I don't really know how to apply this formula when it's in a more complex form like this.

Hi ends!

I don't see why that formula wouldn't work here. Try calculating $ad$ and $bc$ first. What is $(-2e^{2t}\sin(4t))*(2e^{4t}\sin(4t))$ for example?
 
Jameson said:
Hi ends!

I don't see why that formula wouldn't work here. Try calculating $ad$ and $bc$ first. What is $(-2e^{2t}\sin(4t))*(2e^{4t}\sin(4t))$ for example?

Thank you, but can you equate this one for me so I have a general idea of how to multiply these two large terms? I'm not entirely sure how to go about it, and since it's my last chance to submit it online, I don't want to mess it up.
 
ends said:
Thank you, but can you equate this one for me so I have a general idea of how to multiply these two large terms? I'm not entirely sure how to go about it, and since it's my last chance to submit it online, I don't want to mess it up.

Multiply things that are alike. I think of each term as containing a regular number, an "e" expression and a trig term. So $(-2e^{2t}\sin(4t))*(2e^{4t}\sin(4t))=-4e^{6t}\sin^2(4t)$. So there's your $ad$. Try $bc$ and then try combining these two to simplify them somehow.
 
Thanks guys...it simplified nicely, really appreciated the help, cheers.
 
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