Chapter 7.3, Problem 68E

Solution

To calculate : The partial fraction decomposition of $\frac{x-3}{x\left(x+3\right)}=\frac{A}{x+3}+\frac{B}{x}$ using inverse matrices.

The partial fraction decomposition of $\frac{x-3}{x\left(x+3\right)}=\frac{A}{x+3}+\frac{B}{x}$ using inverse matrices is $X=\left[\begin{array}{c}2\\ -1\end{array}\right]$ .

Given information :

The fraction $\frac{x-3}{x\left(x+3\right)}=\frac{A}{x+3}+\frac{B}{x}$ .

Formula used :

For $A$ to be coefficient matrix of a system of $n$ linear equations in $n$ variables given by $AX=B$ . If $A^{-1}$ exists, then system of equations has unique solution $X=A^{-1}B$ .

Calculation :

Consider the fraction $\frac{x-3}{x\left(x+3\right)}=\frac{A}{x+3}+\frac{B}{x}$ ,

Multiply both sides with $x\left(x+3\right)$ .

  $x-3=Ax+B\left(x+3\right)$

Expand the like terms.

  $\begin{array}{c}x-3=Ax+Bx+3B\\ =\left(A+B\right)x+3B\end{array}$

Using the coefficient and constants, write the system of equations.

  $\begin{array}{c}A+B=1\\ 3B=-3\end{array}$

Write the system as matrix equation $CX=D$ where $C=\left[\begin{array}{cc}1& 1\\ 0& 3\end{array}\right]$ , $X=\left[\begin{array}{c}A\\ B\end{array}\right]$ and $D=\left[\begin{array}{c}1\\ -3\end{array}\right]$ .

Solve for $X$ by multiplying both sides $C^{-1}$ that is $X=C^{-1}D$ . Use graphing calculator to find the solution of the system as shown below.

  

Thus, the partial fraction decomposition of $\frac{x-3}{x\left(x+3\right)}=\frac{A}{x+3}+\frac{B}{x}$ using inverse matrices is $X=\left[\begin{array}{c}2\\ -1\end{array}\right]$ .