Mathematics, 25.11.2021 09:40, bfgnnnbddf6830

Use a system of equations to find the partial fraction decomposition of the rational expression. Solve the system using matrices. $\frac{3x ^{2} + 3x - 2 }{(x + 1)^{2} (x - 1)} = \frac{a}{x + 1} + \frac{b}{x - 1} + \frac{c}{(x + 1)^{2} }$
A=
B=
C=

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answered: kaykardash

Combine the fractions on the left with a common denominator:

$\dfrac a{x+1} + \dfrac b{x-1} + \dfrac c{(x+1)^2} = \dfrac{a(x+1)(x-1) + b(x+1)^2 + c(x-1)}{(x-1)(x+1)^2}$

It follows that

3x^2+3x-2 = a(x+1)(x-1) + b(x+1)^2 + c(x-1)

Expand the right side and collect like powers of x :

3x^2+3x-2 = a(x^2-1) + b(x^2+2x+1) + c(x-1)

3x^2+3x-2 = (a+b)x^2 + (2b + c)x -a +b - c

Then we have the system of equations

$\begin{cases}a+b=3\\2b+c=3\\-a+b-c=-2\end{cases}$

or in matrix form,

$\begin{bmatrix}1&1&0\\0&2&1\\-1&1&-1\end{bmatrix} \begin{bmatrix}a\\b\\c\end{bmatrix} = \begin{bmatrix}3\\3\\-2\end{bmatrix}$

Compute the determinant of the coefficient matrix:

$\det\begin{bmatrix}1&1&0\\0&2&1\\-1&1&-1\end{bmatrix} = -4$

Then the inverse of the coefficient matrix is equal 1/det times the adjugate of the coefficient matrix (a.k.a the transpose of the cofactor matrix):

$\begin{bmatrix}1&1&0\\0&2&1\\-1&1&-1\end{bmatrix}^{-1} = \dfrac1{-4} \begin{bmatrix}-3 & -1 & 2 \\ 1 & -1 & -2 \\ 1 & -1 & 2\end{bmatrix}^\top = -\dfrac14 \begin{bmatrix}3&-1&-1\\1&1&1\\-2&2&-2\end{bmatrix}$

Multiply both sides of the equation by the inverse :

$\begin{bmatrix}a\\b\\c\end{bmatrix} = -\dfrac14 \begin{bmatrix}3&-1&-1\\1&1&1\\-2&2&-2\end{bmatrix} \begin{bmatrix}3\\3\\-2\end{bmatrix} = \begin{bmatrix}2\\1\\1\end{bmatrix}$

So, we have a = 2 and b = c = 1, and the partial fraction decomposition is

$\dfrac{3x^2+3x-2}{(x+1)^2(x-1)} = \dfrac 2{x+1} + \dfrac 1{x-1} + \dfrac 1{(x+1)^2}$

answered: Clervoyantyvonne

A = 2B = 1C = 1

Step-by-step explanation:

One can solve for a, b, c a little more directly than using a system of 3 equations.

If we multiply the rational expression by (x+1)², we get ...

(3x² +3x -2)/(x -1) = (x+1)²(a/(x+1) +b/(x-1)) +c

Evaluating this for x = -1 gives ...

(3(-1)² +3(-1) -2)/(-1 -1) = c

-2/-2 = 1 = c

Similarly, multiplying by (x -1) gives ...

(3x² +3x -2)/(x +1)² = (x -1)(a/(x +1) +c/(x +1)²) + b

Evaluating this for x = 1 gives ...

(3·1² +3·1 -2)/(1 +1)² = b

4/4 = 1 = b

Now, we need to find the value of 'a'. The identity will hold true for any value of x, so we can see what happens when we substitute x=0. We can use the values of 'b' and 'c' that we found above.

(3·0² +3·0 -2)/((0 +1)²(0 -1)) = a/(0 +1) +1/(0 -1) +1/(0 +1)²

-2/-1 = a -1 +1 ⇒ a = 2

_____

System of equations solution

When the terms of the right-side expansion are combined, the numerator of the result is ...

a(x +1)(x -1) +b(x +1)^2 +c(x -1) = (a+b)x² +(2b+c)x +(-a+b-c) ≡ 3x² +3x -2

Equating the coefficients gives the system of equations whose augmented matrix is:

$\left[\begin{array}{ccc|c}1&1&0&3\\0&2&1&3\\-1&1&-1&-2\end{array}\right]$