Answer:
[tex]L^{-1} \begin {pmatrix} \dfrac{s}{((s-20)(s-5)(s-4)} \end {pmatrix}= \dfrac{e^{4t}}{12}\begin {pmatrix}e^{16t} - 4e^t+3 \end {pmatrix}[/tex]
Step-by-step explanation:
From the given information the first step to carry out is to find the partial fraction of the given equation; i.e.
[tex]\dfrac{s}{(s-4)(s-5)(s-20)}= \dfrac{A}{s-4}+\dfrac{B}{s-5}+\dfrac{C}{s-20}[/tex]
After finding the partial fraction; we get:
[tex]s= s^2(A+B+C)+s(-2A-24B-25C)+20A+80B+100C[/tex]
By solving this; we get:
[tex]A =\dfrac{1}{12}\ ; \ B = -\dfrac{1}{3} \ ; \ C= \dfrac{1}{4}[/tex]
Thus; the above equation can be re-written as:
[tex]\dfrac{s}{(s-20)(s-5)(s-4)}= \dfrac{\dfrac{1}{12}}{s-20}+\dfrac{-\dfrac{1}{3} }{s-5}+\dfrac{\dfrac{1}{4}}{s-4}[/tex]
Finally as a function of t ;
[tex]L^{-1} \begin {pmatrix} \dfrac{s}{((s-20)(s-5)(s-4)} \end {pmatrix}= \dfrac{e^{4t}}{12}\begin {pmatrix}e^{16t} - 4e^t+3 \end {pmatrix}[/tex]