LaPlace Transformation DFQ Help

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[Q1[1]



I am struggling to move beyond the inverse step of taking the LaPlace general solution.



I know we take
$$mathcalLy'' + 9mathcalLy = g(t)$$



I know that $mathcalLy'' = s^2$Y(s) - sf(0) - f'(0)$



So then from simple plug and chug, I get that $mathcalLy'' = s^2Y(s) - 1 - 0.$ This is just directly from the IVP problem.



Likewise I know that $mathcalLy' = sY(s) - y(0)$.



Combining these together into the given equation $y''+9y'= g(t)$



I solve for $Y(s)$ and get $(g(t)+1)/(s^2+9s).$



Up to this point, I am stuck.




differential-equations laplace-transform




share|cite|improve this question





















  • In the right hand side you should have the Laplace transform of $g$. Its computation can be reduced to tabulated transforms using the properties of the Laplace transform, but it is simple enough that a direct computation looks even shorter: $mathcalL(g)=int_0^inftyg(t)e^-stdt=int_0^1e^-stdt+int_1^inftyte^-stdt=-frace^-ss+frac1s-frace^-s (s + 1)s^2$. Now is when you divide by $s^2+9s$. The next task is to compute the inverse Laplace transform of that quotient.
    – spiralstotheleft
    2 days ago











  • Your MathJax code was a mess. See my edits for proper usage.
    – Michael Hardy
    2 days ago










  • @spiralstotheleft from your method of computing the transform, how does one immediately know to divide by s^2+9s. Can this be intuitively seen from the problem?
    – SleepApnea
    2 days ago






  • 1




    That comes from what you yourself did on the left hand side of the equation. You computed that $mathcalL(y''+9y')=(s^2+9s)mathcalL(y)$. The whole goal of the method of applying the Laplace transform is to be able to solve for $mathcalL(y)$.
    – spiralstotheleft
    yesterday










  • the g(t) solution differs greatly from the solution posted by Isham
    – SleepApnea
    yesterday














up vote
2
down vote

favorite












[Q1[1]



I am struggling to move beyond the inverse step of taking the LaPlace general solution.



I know we take
$$mathcalLy'' + 9mathcalLy = g(t)$$



I know that $mathcalLy'' = s^2$Y(s) - sf(0) - f'(0)$



So then from simple plug and chug, I get that $mathcalLy'' = s^2Y(s) - 1 - 0.$ This is just directly from the IVP problem.



Likewise I know that $mathcalLy' = sY(s) - y(0)$.



Combining these together into the given equation $y''+9y'= g(t)$



I solve for $Y(s)$ and get $(g(t)+1)/(s^2+9s).$



Up to this point, I am stuck.




differential-equations laplace-transform




share|cite|improve this question





















  • In the right hand side you should have the Laplace transform of $g$. Its computation can be reduced to tabulated transforms using the properties of the Laplace transform, but it is simple enough that a direct computation looks even shorter: $mathcalL(g)=int_0^inftyg(t)e^-stdt=int_0^1e^-stdt+int_1^inftyte^-stdt=-frace^-ss+frac1s-frace^-s (s + 1)s^2$. Now is when you divide by $s^2+9s$. The next task is to compute the inverse Laplace transform of that quotient.
    – spiralstotheleft
    2 days ago











  • Your MathJax code was a mess. See my edits for proper usage.
    – Michael Hardy
    2 days ago










  • @spiralstotheleft from your method of computing the transform, how does one immediately know to divide by s^2+9s. Can this be intuitively seen from the problem?
    – SleepApnea
    2 days ago






  • 1




    That comes from what you yourself did on the left hand side of the equation. You computed that $mathcalL(y''+9y')=(s^2+9s)mathcalL(y)$. The whole goal of the method of applying the Laplace transform is to be able to solve for $mathcalL(y)$.
    – spiralstotheleft
    yesterday










  • the g(t) solution differs greatly from the solution posted by Isham
    – SleepApnea
    yesterday












up vote
2
down vote

favorite









up vote
2
down vote

favorite











[Q1[1]



I am struggling to move beyond the inverse step of taking the LaPlace general solution.



I know we take
$$mathcalLy'' + 9mathcalLy = g(t)$$



I know that $mathcalLy'' = s^2$Y(s) - sf(0) - f'(0)$



So then from simple plug and chug, I get that $mathcalLy'' = s^2Y(s) - 1 - 0.$ This is just directly from the IVP problem.



Likewise I know that $mathcalLy' = sY(s) - y(0)$.



Combining these together into the given equation $y''+9y'= g(t)$



I solve for $Y(s)$ and get $(g(t)+1)/(s^2+9s).$



Up to this point, I am stuck.




differential-equations laplace-transform




share|cite|improve this question













[Q1[1]



I am struggling to move beyond the inverse step of taking the LaPlace general solution.



I know we take
$$mathcalLy'' + 9mathcalLy = g(t)$$



I know that $mathcalLy'' = s^2$Y(s) - sf(0) - f'(0)$



So then from simple plug and chug, I get that $mathcalLy'' = s^2Y(s) - 1 - 0.$ This is just directly from the IVP problem.



Likewise I know that $mathcalLy' = sY(s) - y(0)$.



Combining these together into the given equation $y''+9y'= g(t)$



I solve for $Y(s)$ and get $(g(t)+1)/(s^2+9s).$



Up to this point, I am stuck.





differential-equations laplace-transform





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited 2 days ago









Michael Hardy

204k23185460




204k23185460









asked 2 days ago









SleepApnea

476




476











  • In the right hand side you should have the Laplace transform of $g$. Its computation can be reduced to tabulated transforms using the properties of the Laplace transform, but it is simple enough that a direct computation looks even shorter: $mathcalL(g)=int_0^inftyg(t)e^-stdt=int_0^1e^-stdt+int_1^inftyte^-stdt=-frace^-ss+frac1s-frace^-s (s + 1)s^2$. Now is when you divide by $s^2+9s$. The next task is to compute the inverse Laplace transform of that quotient.
    – spiralstotheleft
    2 days ago











  • Your MathJax code was a mess. See my edits for proper usage.
    – Michael Hardy
    2 days ago










  • @spiralstotheleft from your method of computing the transform, how does one immediately know to divide by s^2+9s. Can this be intuitively seen from the problem?
    – SleepApnea
    2 days ago






  • 1




    That comes from what you yourself did on the left hand side of the equation. You computed that $mathcalL(y''+9y')=(s^2+9s)mathcalL(y)$. The whole goal of the method of applying the Laplace transform is to be able to solve for $mathcalL(y)$.
    – spiralstotheleft
    yesterday










  • the g(t) solution differs greatly from the solution posted by Isham
    – SleepApnea
    yesterday
















  • In the right hand side you should have the Laplace transform of $g$. Its computation can be reduced to tabulated transforms using the properties of the Laplace transform, but it is simple enough that a direct computation looks even shorter: $mathcalL(g)=int_0^inftyg(t)e^-stdt=int_0^1e^-stdt+int_1^inftyte^-stdt=-frace^-ss+frac1s-frace^-s (s + 1)s^2$. Now is when you divide by $s^2+9s$. The next task is to compute the inverse Laplace transform of that quotient.
    – spiralstotheleft
    2 days ago











  • Your MathJax code was a mess. See my edits for proper usage.
    – Michael Hardy
    2 days ago










  • @spiralstotheleft from your method of computing the transform, how does one immediately know to divide by s^2+9s. Can this be intuitively seen from the problem?
    – SleepApnea
    2 days ago






  • 1




    That comes from what you yourself did on the left hand side of the equation. You computed that $mathcalL(y''+9y')=(s^2+9s)mathcalL(y)$. The whole goal of the method of applying the Laplace transform is to be able to solve for $mathcalL(y)$.
    – spiralstotheleft
    yesterday










  • the g(t) solution differs greatly from the solution posted by Isham
    – SleepApnea
    yesterday















In the right hand side you should have the Laplace transform of $g$. Its computation can be reduced to tabulated transforms using the properties of the Laplace transform, but it is simple enough that a direct computation looks even shorter: $mathcalL(g)=int_0^inftyg(t)e^-stdt=int_0^1e^-stdt+int_1^inftyte^-stdt=-frace^-ss+frac1s-frace^-s (s + 1)s^2$. Now is when you divide by $s^2+9s$. The next task is to compute the inverse Laplace transform of that quotient.
– spiralstotheleft
2 days ago





In the right hand side you should have the Laplace transform of $g$. Its computation can be reduced to tabulated transforms using the properties of the Laplace transform, but it is simple enough that a direct computation looks even shorter: $mathcalL(g)=int_0^inftyg(t)e^-stdt=int_0^1e^-stdt+int_1^inftyte^-stdt=-frace^-ss+frac1s-frace^-s (s + 1)s^2$. Now is when you divide by $s^2+9s$. The next task is to compute the inverse Laplace transform of that quotient.
– spiralstotheleft
2 days ago













Your MathJax code was a mess. See my edits for proper usage.
– Michael Hardy
2 days ago




Your MathJax code was a mess. See my edits for proper usage.
– Michael Hardy
2 days ago












@spiralstotheleft from your method of computing the transform, how does one immediately know to divide by s^2+9s. Can this be intuitively seen from the problem?
– SleepApnea
2 days ago




@spiralstotheleft from your method of computing the transform, how does one immediately know to divide by s^2+9s. Can this be intuitively seen from the problem?
– SleepApnea
2 days ago




1




1




That comes from what you yourself did on the left hand side of the equation. You computed that $mathcalL(y''+9y')=(s^2+9s)mathcalL(y)$. The whole goal of the method of applying the Laplace transform is to be able to solve for $mathcalL(y)$.
– spiralstotheleft
yesterday




That comes from what you yourself did on the left hand side of the equation. You computed that $mathcalL(y''+9y')=(s^2+9s)mathcalL(y)$. The whole goal of the method of applying the Laplace transform is to be able to solve for $mathcalL(y)$.
– spiralstotheleft
yesterday












the g(t) solution differs greatly from the solution posted by Isham
– SleepApnea
yesterday




the g(t) solution differs greatly from the solution posted by Isham
– SleepApnea
yesterday










1 Answer
1






active

oldest

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up vote
0
down vote



accepted










$$y''+9y'=g(t)$$
Take the Laplace transform of both sides
$$s^2F(s)-sy(0)-y'(0)+9(sF(s)-y(0))=mathcal L(g(t))$$
$$s^2F(s)-1+9sF(s)=mathcal L(g(t))$$
$$F(s)(s^2+9s)=1+mathcal L(g(t))$$
You can use the heaviside function
$$g(t)=1(H(t)-H(t-1))+t(H(t-1))$$
$$g(t)=1H(t)+(t-1)(H(t-1))$$
You have also that the laplace transform of
$$mathcal L(H(t-a)f(t-a))=e^-asF(s)$$



Therefore
$$mathcal L(g(t))=frac 1s+frac e^-ss^2$$
you have now that
$$F(s)(s^2+9s)=1+frac 1s+frac e^-ss^2$$
$$F(s)=frac 1 s(s+9)+frac 1s^2(s+9)+frac e^-ss^3(s+9)$$
You can use fractions decompositions ..



For the first fraction we decompose this way
$$h(s)=frac 1 s(s+9)=frac 19(frac 1s-frac 1s+9)$$
then with the laplace transform table you see that
$$mathcalL(h(s))=frac 19 (1-e^-9t)$$



Edit2
for the last fraction dont pay attention to the exponential put it outside
$$f(s)=frac e^-ss^3(s+9)=e^-sleft (frac 1s^3(s+9) right)$$
then perform fraction decomposition...the exponential is just a shift






share|cite|improve this answer























  • Isham. Thank you for your response. Why don't we take the inverse of the LaPlace transform?
    – SleepApnea
    yesterday










  • decompose the fraction so you can take the inverse Laplace transform thats the last step ...@SleepApnea
    – Isham
    yesterday










  • the first fraction would thus decompose into sin(3t), the second fraction will decompose into sin^-1(3t). What transform contains e^s in its numerator? Thank you for your help
    – SleepApnea
    yesterday






  • 1




    You are right. Sorry. I made a mistake. You are absolutely right. Mr. Isham. If it is at all convenient for you, can you please post some insights into this question as well? math.stackexchange.com/questions/2872881/…
    – SleepApnea
    yesterday






  • 1




    Thank you kind sir
    – SleepApnea
    yesterday










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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
0
down vote



accepted










$$y''+9y'=g(t)$$
Take the Laplace transform of both sides
$$s^2F(s)-sy(0)-y'(0)+9(sF(s)-y(0))=mathcal L(g(t))$$
$$s^2F(s)-1+9sF(s)=mathcal L(g(t))$$
$$F(s)(s^2+9s)=1+mathcal L(g(t))$$
You can use the heaviside function
$$g(t)=1(H(t)-H(t-1))+t(H(t-1))$$
$$g(t)=1H(t)+(t-1)(H(t-1))$$
You have also that the laplace transform of
$$mathcal L(H(t-a)f(t-a))=e^-asF(s)$$



Therefore
$$mathcal L(g(t))=frac 1s+frac e^-ss^2$$
you have now that
$$F(s)(s^2+9s)=1+frac 1s+frac e^-ss^2$$
$$F(s)=frac 1 s(s+9)+frac 1s^2(s+9)+frac e^-ss^3(s+9)$$
You can use fractions decompositions ..



For the first fraction we decompose this way
$$h(s)=frac 1 s(s+9)=frac 19(frac 1s-frac 1s+9)$$
then with the laplace transform table you see that
$$mathcalL(h(s))=frac 19 (1-e^-9t)$$



Edit2
for the last fraction dont pay attention to the exponential put it outside
$$f(s)=frac e^-ss^3(s+9)=e^-sleft (frac 1s^3(s+9) right)$$
then perform fraction decomposition...the exponential is just a shift






share|cite|improve this answer























  • Isham. Thank you for your response. Why don't we take the inverse of the LaPlace transform?
    – SleepApnea
    yesterday










  • decompose the fraction so you can take the inverse Laplace transform thats the last step ...@SleepApnea
    – Isham
    yesterday










  • the first fraction would thus decompose into sin(3t), the second fraction will decompose into sin^-1(3t). What transform contains e^s in its numerator? Thank you for your help
    – SleepApnea
    yesterday






  • 1




    You are right. Sorry. I made a mistake. You are absolutely right. Mr. Isham. If it is at all convenient for you, can you please post some insights into this question as well? math.stackexchange.com/questions/2872881/…
    – SleepApnea
    yesterday






  • 1




    Thank you kind sir
    – SleepApnea
    yesterday














up vote
0
down vote



accepted










$$y''+9y'=g(t)$$
Take the Laplace transform of both sides
$$s^2F(s)-sy(0)-y'(0)+9(sF(s)-y(0))=mathcal L(g(t))$$
$$s^2F(s)-1+9sF(s)=mathcal L(g(t))$$
$$F(s)(s^2+9s)=1+mathcal L(g(t))$$
You can use the heaviside function
$$g(t)=1(H(t)-H(t-1))+t(H(t-1))$$
$$g(t)=1H(t)+(t-1)(H(t-1))$$
You have also that the laplace transform of
$$mathcal L(H(t-a)f(t-a))=e^-asF(s)$$



Therefore
$$mathcal L(g(t))=frac 1s+frac e^-ss^2$$
you have now that
$$F(s)(s^2+9s)=1+frac 1s+frac e^-ss^2$$
$$F(s)=frac 1 s(s+9)+frac 1s^2(s+9)+frac e^-ss^3(s+9)$$
You can use fractions decompositions ..



For the first fraction we decompose this way
$$h(s)=frac 1 s(s+9)=frac 19(frac 1s-frac 1s+9)$$
then with the laplace transform table you see that
$$mathcalL(h(s))=frac 19 (1-e^-9t)$$



Edit2
for the last fraction dont pay attention to the exponential put it outside
$$f(s)=frac e^-ss^3(s+9)=e^-sleft (frac 1s^3(s+9) right)$$
then perform fraction decomposition...the exponential is just a shift






share|cite|improve this answer























  • Isham. Thank you for your response. Why don't we take the inverse of the LaPlace transform?
    – SleepApnea
    yesterday










  • decompose the fraction so you can take the inverse Laplace transform thats the last step ...@SleepApnea
    – Isham
    yesterday










  • the first fraction would thus decompose into sin(3t), the second fraction will decompose into sin^-1(3t). What transform contains e^s in its numerator? Thank you for your help
    – SleepApnea
    yesterday






  • 1




    You are right. Sorry. I made a mistake. You are absolutely right. Mr. Isham. If it is at all convenient for you, can you please post some insights into this question as well? math.stackexchange.com/questions/2872881/…
    – SleepApnea
    yesterday






  • 1




    Thank you kind sir
    – SleepApnea
    yesterday












up vote
0
down vote



accepted







up vote
0
down vote



accepted






$$y''+9y'=g(t)$$
Take the Laplace transform of both sides
$$s^2F(s)-sy(0)-y'(0)+9(sF(s)-y(0))=mathcal L(g(t))$$
$$s^2F(s)-1+9sF(s)=mathcal L(g(t))$$
$$F(s)(s^2+9s)=1+mathcal L(g(t))$$
You can use the heaviside function
$$g(t)=1(H(t)-H(t-1))+t(H(t-1))$$
$$g(t)=1H(t)+(t-1)(H(t-1))$$
You have also that the laplace transform of
$$mathcal L(H(t-a)f(t-a))=e^-asF(s)$$



Therefore
$$mathcal L(g(t))=frac 1s+frac e^-ss^2$$
you have now that
$$F(s)(s^2+9s)=1+frac 1s+frac e^-ss^2$$
$$F(s)=frac 1 s(s+9)+frac 1s^2(s+9)+frac e^-ss^3(s+9)$$
You can use fractions decompositions ..



For the first fraction we decompose this way
$$h(s)=frac 1 s(s+9)=frac 19(frac 1s-frac 1s+9)$$
then with the laplace transform table you see that
$$mathcalL(h(s))=frac 19 (1-e^-9t)$$



Edit2
for the last fraction dont pay attention to the exponential put it outside
$$f(s)=frac e^-ss^3(s+9)=e^-sleft (frac 1s^3(s+9) right)$$
then perform fraction decomposition...the exponential is just a shift






share|cite|improve this answer















$$y''+9y'=g(t)$$
Take the Laplace transform of both sides
$$s^2F(s)-sy(0)-y'(0)+9(sF(s)-y(0))=mathcal L(g(t))$$
$$s^2F(s)-1+9sF(s)=mathcal L(g(t))$$
$$F(s)(s^2+9s)=1+mathcal L(g(t))$$
You can use the heaviside function
$$g(t)=1(H(t)-H(t-1))+t(H(t-1))$$
$$g(t)=1H(t)+(t-1)(H(t-1))$$
You have also that the laplace transform of
$$mathcal L(H(t-a)f(t-a))=e^-asF(s)$$



Therefore
$$mathcal L(g(t))=frac 1s+frac e^-ss^2$$
you have now that
$$F(s)(s^2+9s)=1+frac 1s+frac e^-ss^2$$
$$F(s)=frac 1 s(s+9)+frac 1s^2(s+9)+frac e^-ss^3(s+9)$$
You can use fractions decompositions ..



For the first fraction we decompose this way
$$h(s)=frac 1 s(s+9)=frac 19(frac 1s-frac 1s+9)$$
then with the laplace transform table you see that
$$mathcalL(h(s))=frac 19 (1-e^-9t)$$



Edit2
for the last fraction dont pay attention to the exponential put it outside
$$f(s)=frac e^-ss^3(s+9)=e^-sleft (frac 1s^3(s+9) right)$$
then perform fraction decomposition...the exponential is just a shift







share|cite|improve this answer















share|cite|improve this answer



share|cite|improve this answer








edited yesterday


























answered 2 days ago









Isham

10.4k3829




10.4k3829











  • Isham. Thank you for your response. Why don't we take the inverse of the LaPlace transform?
    – SleepApnea
    yesterday










  • decompose the fraction so you can take the inverse Laplace transform thats the last step ...@SleepApnea
    – Isham
    yesterday










  • the first fraction would thus decompose into sin(3t), the second fraction will decompose into sin^-1(3t). What transform contains e^s in its numerator? Thank you for your help
    – SleepApnea
    yesterday






  • 1




    You are right. Sorry. I made a mistake. You are absolutely right. Mr. Isham. If it is at all convenient for you, can you please post some insights into this question as well? math.stackexchange.com/questions/2872881/…
    – SleepApnea
    yesterday






  • 1




    Thank you kind sir
    – SleepApnea
    yesterday
















  • Isham. Thank you for your response. Why don't we take the inverse of the LaPlace transform?
    – SleepApnea
    yesterday










  • decompose the fraction so you can take the inverse Laplace transform thats the last step ...@SleepApnea
    – Isham
    yesterday










  • the first fraction would thus decompose into sin(3t), the second fraction will decompose into sin^-1(3t). What transform contains e^s in its numerator? Thank you for your help
    – SleepApnea
    yesterday






  • 1




    You are right. Sorry. I made a mistake. You are absolutely right. Mr. Isham. If it is at all convenient for you, can you please post some insights into this question as well? math.stackexchange.com/questions/2872881/…
    – SleepApnea
    yesterday






  • 1




    Thank you kind sir
    – SleepApnea
    yesterday















Isham. Thank you for your response. Why don't we take the inverse of the LaPlace transform?
– SleepApnea
yesterday




Isham. Thank you for your response. Why don't we take the inverse of the LaPlace transform?
– SleepApnea
yesterday












decompose the fraction so you can take the inverse Laplace transform thats the last step ...@SleepApnea
– Isham
yesterday




decompose the fraction so you can take the inverse Laplace transform thats the last step ...@SleepApnea
– Isham
yesterday












the first fraction would thus decompose into sin(3t), the second fraction will decompose into sin^-1(3t). What transform contains e^s in its numerator? Thank you for your help
– SleepApnea
yesterday




the first fraction would thus decompose into sin(3t), the second fraction will decompose into sin^-1(3t). What transform contains e^s in its numerator? Thank you for your help
– SleepApnea
yesterday




1




1




You are right. Sorry. I made a mistake. You are absolutely right. Mr. Isham. If it is at all convenient for you, can you please post some insights into this question as well? math.stackexchange.com/questions/2872881/…
– SleepApnea
yesterday




You are right. Sorry. I made a mistake. You are absolutely right. Mr. Isham. If it is at all convenient for you, can you please post some insights into this question as well? math.stackexchange.com/questions/2872881/…
– SleepApnea
yesterday




1




1




Thank you kind sir
– SleepApnea
yesterday




Thank you kind sir
– SleepApnea
yesterday












 

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What is the equation of a 3D cone with generalised tilt?

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Clash Royale CLAN TAG #URR8PPP up vote 2 down vote favorite What is the equation of a 3D cone with generalized tilt? I've noticed that in most equations given to represent a cone, there is no parameter which defines the tilt of the cone in 3D space and that most of them have their point at the origin $(0,0,0)$ - I was wondering if anyone could give me a more generalized cone equation for a cone in any position in 3D space 3d share | cite | improve this question edited Jul 25 '16 at 0:05 ervx 9,425 3 13 36 asked Jul 24 '16 at 15:38 Charlene 11 2 2 Welcome to Math.SE! Could you please explain a few things to help people give an answer useful to you: 1. What do you mean by "generalized tilt"? 2. Are you asking about a right circular cone? Does the angle at the vertex matter? 3. What form of answer (implicit, parametric...?) are you looking for? 4. If this is homework, what tools do you have available, an...
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Color the edges and diagonals of a regular polygon

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Clash Royale CLAN TAG #URR8PPP up vote 5 down vote favorite 1 Here is the problem: For what $n$ is it possible to color the edges and diagonals of an $n$-side regular polygon with $dfracbinomn23$ colors, such that you use every color exactly three times and for every color the three segments (edges or diagonals) with that color form a triangle? Trivially $nequiv 0 text or 1 pmod3$. I can also prove that if the statement is true for $k$ than it is true for $3k$ as well. How to finish? Please help! Thanks combinatorial-geometry share | cite | improve this question edited Aug 6 at 21:57 alcana 118 4 asked Aug 6 at 21:15 Leo Gardner 356 11 Even assuming "polyhpn" in the title is a typo for polygon , it is unclear what you mean by "the edges and sides". Aren't the side of a regular polygon the same as the edges? A regular $n$-sided polygon has $n$ edges. – hardmath Aug 6 at 21:20 3 $n$ ne...
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Relationship between determinant of matrix and determinant of adjoint?

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Clash Royale CLAN TAG #URR8PPP up vote 2 down vote favorite We are studying adjoints in class, and I was curious if there is a relationship between the determinant of matrix A, and the determinant of the adjoint of matrix A? I assume there would be a relationship because finding the adjoint requires creating a cofactor matrix and then transposing it. linear-algebra determinant adjoint-operators share | cite | improve this question asked Nov 17 '17 at 17:32 CluelessCoder 32 5 For a square matrix $A$ of order $n$, we have $A(operatornameadj A)=(operatornameadj A)A=|A|I_n$ where $I_n$ is the identity matrix of order $n$. This should tell you about the determinant of the adjoint in terms of that of $A$ (use multiplicative property and other elementary properties of determinants). – Prasun Biswas Nov 17 '17 at 17:35 1 @CluelessCoder: see here: math.stackexchange.com/questions/516127/…...
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