# Thread: help with partial fractions

1. I got

s = -1

1 = -A + B -5C + 5/2

s = 2

1 = 8A + 4B + 4C + 1

then not sure what to do from here.

2. I don't agree with either one of those equations. Try again, and double-check your work.

3. S = -1

1 = -5a + 5/2 - c + d

s = = 2

1 = 4a + 1 + 8c + 4d

4. There you go. Now you have three linear equations with three unknowns. Pick your favorite solution method and fire away. (I'd recommend Gaussian elimination with back substitution - there is no more efficient exact method known to man.)

5. I have tried cramers rule to solve the simultaneous equation but it doesnt work because you have to multiply by 0.

6. So your system appears to be this:

$\begin{bmatrix}1&1&1\\-5&-1&1\\4&8&4\end{bmatrix}\begin{bmatrix}A\\C\\D\end{ bmatrix}=\begin{bmatrix}1/2\\-3/2\\0\end{bmatrix}.$

Correct? Show me what work you've done.

7. Ive gone through it and all the values = 1/2.

When $\dfrac{1/2s+1/2}{(s-1)^2+1)}.$ I have to put that in laplace transform form so this should = 1/2e^tcost. Just in the answers its says -1/2

8. A=C=D=1/2 doesn't even satisfy A+C+D=1/2. Therefore, your solution is incorrect. Show me some work, here, and I can show you where you're going wrong. If you don't show me work, I can't help you.

9. I got det A = $\begin{bmatrix}1/2&1&1\\-3/2&-1&1\\0&8&4\end{bmatrix}.$= (1/2)(-1)(4)+(1)(1)(0)+(1)(-3/2)(8)-(1)(-1)(0)-(1)(-3/2)(4)-(1/2)(1)(8)
divide by $\begin{bmatrix}1&1&1\\-5&-1&1\\4&8&4\end{bmatrix}.$ = (1)(-1)(4)+(1)(1)(4)+(1)(-5)(8)-(1)(-1)(4)-(1)(-5)(4)-(1)(1)(8) = -12/-24

det C = $\begin{bmatrix}1/2&1&1\\-3/2&-5&1\\0&4&4\end{bmatrix}.$= (1/2)(-5)(4)+(1)(1)(0)+(1)(-3/2)(4)-(1)(-5)(0)-(1)(-3/2)(4)-(1/2)(1)(4)
divide by $\begin{bmatrix}1&1&1\\-5&-1&1\\4&8&4\end{bmatrix}.$= (1)(-1)(4)+(1)(1)(4)+(1)(-5)(8)-(1)(-1)(4)-(1)(-5)(4)-(1)(1)(8) = -12/-24

det D = $\begin{bmatrix}1/2&1&1\\-3/2&-5&-1\\0&4&8\end{bmatrix}.$= (1/2)(-5)(8)+(1)(-1)(0)+(1)(-3/2)(4)-(1)(-5)(0)-(1)(-3/2)(8)-(1/2)(-1)(4)
divide by $\begin{bmatrix}1&1&1\\-5&-1&1\\4&8&4\end{bmatrix}.$= (1)(-1)(4)+(1)(1)(4)+(1)(-5)(8)-(1)(-1)(4)-(1)(-5)(4)-(1)(1)(8) = -12/-24

also the answers say the laplace transforms are -1/2e^tcost + 1/2 + 1/2t

10. You're not computing the correct determinants. You should have

$\displaystyle A=\frac{\left|\begin{matrix}
1/2&1&1\\
-3/2&-1&1\\
0&8&4
\end{matrix}\right|}{\left|\begin{matrix}
1&1&1\\
-5&-1&1\\
4&8&4
\end{matrix}\right|},
C=\frac{\left|\begin{matrix}
1&1/2&1\\
-5&-3/2&1\\
4&0&4
\end{matrix}\right|}{\left|\begin{matrix}
1&1&1\\
-5&-1&1\\
4&8&4
\end{matrix}\right|},
D=\frac{\left|\begin{matrix}
1&1&1/2\\
-5&-1&-3/2\\
4&8&0
\end{matrix}\right|}{\left|\begin{matrix}
1&1&1\\
-5&-1&1\\
4&8&4
\end{matrix}\right|}.
$

What does that give you?

11. cheers mate thats correct.

Moderator edit: New question deleted. Ask new questions in a new thread, as explained in the forum rules stuck at every subforum.

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