Question

Use a system of equations to find the partial fraction decomposition of the rational expression. Solve the system using matrices.


`(3x ^(2) + 3x - 2 )/((x + 1)^(2) (x - 1)) = (a)/(x + 1) + (b)/(x - 1) + (c)/((x + 1)^(2) )`
A=
B=
C=​

Answer 1

Combine the fractions on the left with a common denominator:

`\frac a{x+1} + \frac b{x-1} + \frac c{(x+1)^2} = (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) = \frac1{-4} \begin{bmatrix}-3 & -1 & 2 \\ 1 & -1 & -2 \\ 1 & -1 & 2\end{bmatrix}^\top = -\frac14 \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} = -\frac14 \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

`(3x^2+3x-2)/((x+1)^2(x-1)) = \frac 2{x+1} + \frac 1{x-1} + \frac 1{(x+1)^2}`

Answer 2

• A = 2
• B = 1
• C = 1

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}ccc1&1&0&3\\0&2&1&3\\-1&1&-1&-2\end{array}\right]`

Transforming this to reduced row-echelon form using any of a variety of available tools gives ...

`\left[\begin{array}ccc1&0&0&2\\0&1&0&1\\0&0&1&1\end{array}\right]`

which tells you the solution is (A, B, C) = (2, 1, 1).