Question

In order to expand in partial fractions $$ \frac{p x^{2}+q x+r}{(x-a)(x-b)^{2}}=\frac{A}{x-a}+\frac{B}{x-b}+\frac{C}{(x-b)^{2}} $$ we need to solve the following equations for \(A, B, C\) : $$ \begin{aligned} &\mid A+B=p \\ &2 b A+(a+b) B-C=-q \\ &b^{2} A+\quad a b B-a C=r \end{aligned} $$ Use row reduction to find \(A, B, C\) in terms of \(a, b, \rho, q\), and \(r\).

Short answer

Solve the system of equations using row reduction. First, express A, B, C in matrix form and row-reduce step-by-step. A, B, C are found through substitution and elimination.

Step-by-step solution

Write the system of equations in matrix form

Write the given system of equations as a matrix equation. The equations are: \(A + B = p\), \(2bA + (a + b)B - C = -q\), \(b^2A + abB - aC = r\). This can be represented as: \[ \begin{pmatrix} 1 & 1 & 0 \ 2b & (a + b) & -1 \ b^2 & ab & -a \end{pmatrix} \begin{pmatrix} A \ B \ C \end{pmatrix} = \begin{pmatrix} p \ -q \ r \end{pmatrix} \] \

Form the augmented matrix

Create the augmented matrix from the equations. The augmented matrix is: \[ \begin{pmatrix} 1 & 1 & 0 & | & p \ 2b & (a + b) & -1 & | & -q \ b^2 & ab & -a & | & r \end{pmatrix} \] \

Apply row reduction - Step 1

Use the first equation to create a leading 1 in the first column for row 1. No change is necessary for the first row: \[ \begin{pmatrix} 1 & 1 & 0 & | & p \ 2b & (a + b) & -1 & | & -q \ b^2 & ab & -a & | & r \end{pmatrix} \] \

Eliminate the first variable from rows 2 and 3

Subtract 2b times the first row from the second row and \( b^2 \) times the first row from the third row: \[ R2 - 2bR1 \rightarrow R2 \] \[ 2b - 2b(1), (a + b) - 2b(1), -1 - 2b(0), -q - 2b(p) \] This simplifies to: \(0, a - b, -1, -q - 2bp\). Similarly, \[ R3 - b^2R1 \rightarrow R3 \] \[ b^2 - b^2(1), ab - b^2(1), -a - b^2(0), r - b^2(p) \] This simplifies to: \(0, ab - b^2, -a, r - b^2p \). The augmented matrix now becomes: \[ \begin{pmatrix} 1 & 1 & 0 & | & p \ 0 & a - b & -1 & | & -q - 2bp \ 0 & ab - b^2 & -a & | & r - b^2p \end{pmatrix} \] \

Simplify row 2

Make the leading coefficient of the second row 1 (if necessary) by dividing the second row by \(a - b\), if \(a eq b\). \[ \frac{R2}{a - b} \rightarrow R2 \] This yields: \(0, 1, -\frac{1}{a - b}, -\frac{q + 2bp}{a - b} \). The augmented matrix now becomes: \[ \begin{pmatrix} 1 & 1 & 0 & | & p \ 0 & 1 & -\frac{1}{a - b} & | & -\frac{q + 2bp}{a - b} \ 0 & ab - b^2 & -a & | & r - b^2p \end{pmatrix} \] \

Eliminate the second variable from row 3

Subtract \((ab - b^2)\) times the second row from the third row: \[ R3 - (ab - b^2)R2 \rightarrow R3 \] \[ 0 - (ab - b^2)(0), (ab - b^2) - (ab - b^2)(1), -a - (ab - b^2)(-\frac{1}{a - b}), (r - b^2p) - (ab - b^2)(-\frac{q + 2bp}{a - b}) \] Simplifies to: \[0, 0, -(a - b), -a - \frac{(ab - b^2)(q + 2bp)}{a - b} + r - b^2p \] The augmented matrix now becomes: \[ \begin{pmatrix} 1 & 1 & 0 & | & p \ 0 & 1 & -\frac{1}{a - b} & | & -\frac{q + 2bp}{a - b} \ 0 & 0 & -(a - b) & | & -a - \frac{(ab - b^2)(q + 2bp)}{a - b} + r - b^2p \end{pmatrix} \] \

Solve for C

Solve the third row for C: \[ C = \frac{ -a - (ab - b^2)(q + 2bp)/(a - b) + r - b^2p }{-(a - b)} \] Simplifying the expression under the fraction: \[ C = \frac{ a + b(q + 2bp) - (q + 2bp)(b^2 - ab)/a - b - r + b^2p }{a - b} \] \[ C = \frac{ a + b^2q + 2b^2p - b - \frac{(q + 2bp)(ab - b^2)}{a - b} + b^2p }{a - b} \] \

Solve for B

With C found, solve for B using the second row equation: \( B - \frac{1}{a - b}C = -\frac{q + 2bp}{a - b} \) Substitute C back in: \( B = -\frac{q + 2bp}{a - b} + \frac{1}{a - b} \times \frac{-a - \frac{(ab - b^2)(q + 2bp)}{a - b} + r - b^2p }{a - b} \) Simplifying further for B, substitute values from previous calculations. \

Solve for A

Finally, use the first row to solve for A: \( A = p - B \). Substitute the value of B back in to get A. \

Key concepts

matrix row reduction

Matrix row reduction is a systematic method of solving systems of linear equations. It transforms a matrix into a simpler form, known as row-echelon form or reduced row-echelon form, by performing row operations. These operations include:

• Swapping two rows
• Multiplying a row by a non-zero scalar
• Adding or subtracting one row to/from another

Using these operations, we reduce the matrix step by step until it’s easier to derive the solution for the variables involved. In our exercise, we started by writing the system of equations in matrix form and then forming an augmented matrix. Through row reduction, we isolated terms and systematically solved the matrix, ultimately finding our variables A, B, and C.

system of equations

A system of equations is a set of two or more equations with the same set of variables. The objective is to find values for the variables that satisfy all equations simultaneously. The methods to solve systems of equations include:

• Substitution
• Elimination
• Graphical methods
• Using matrices

In the given problem, we tackled a system of linear equations involving indicators A, B, and C. By converting these equations into a matrix and applying row reduction, we isolated and solved the variables systematically. Each step in row reduction corresponds to balancing and simplifying the equations effectively.

linear algebra

Linear algebra is a branch of mathematics concerning vector spaces and linear mappings between them. It includes the study of vectors, matrices, determinants, and systems of linear equations. Linear algebra is foundational in many areas, such as:

• Engineering
• Physics
• Computer science
• Economics

In our provided exercise, linear algebra principles helped transform a complex fraction problem into systems of equations and matrices. Using matrix row reduction—a standard technique in linear algebra—enabled us to simplify and solve for the unknowns, illustrating practical uses of these abstract concepts. Additionally, the power of linear algebra is highlighted in simplifying and providing precise solutions efficiently.