Solving Linear Math Inverse Homework: A=4e^4txsin(4t)

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The discussion revolves around finding the inverse of a 2x2 matrix defined by A=4e^4t x sin(4t) and related terms. Participants mention using the determinant equation 1/(ad-bc) to solve for the inverse but express confusion over simplifying the multiplication due to the complex variables involved. There is a request for clarification on the determinant expression and the steps taken to arrive at a solution. Additionally, the use of the trigonometric identity sin^2(t) + cos^2(t) = 1 is suggested to aid in simplification. The conversation emphasizes the need for clearer steps in the calculation process.
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Homework Statement


pt)
If A=
4e^4t x sin(4t) 4e^2t x cos(4t)
-3e^4t x cos(4t) 3e^2t x sin(4t)

Then A inverse is?

x= times
^=power

sorry for bad format

Homework Equations



Since it's 2x2 I know you can use the determinant equation 1/(ad-bc) ..etc

The Attempt at a Solution



Because of the weird variables I'm not sure how to simplify the multiplication when doing the determination equation.
 
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I assume you've have managed to get an answer that is correct, and are just looking for help simplifying?

Care to show us what you got? (Or, maybe just the expression you got for the determinant?)
 
Hurkyl said:
I assume you've have managed to get an answer that is correct, and are just looking for help simplifying?

Care to show us what you got? (Or, maybe just the expression you got for the determinant?)

well the answer i got was long..i just subbed the variables in the equation.
 
?? You "subbed the variables" into what equation?
 
use sin^2(t) +cos^2(t) = 1
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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