Partial Fractions help!?

Question

$A + C = 0$

$-4A + B - 8C + D = 1$

$3A + 16C - 8D = -29$

$-12A + 3B + 16D = 5$

How do I equate the coefficients? Please provide steps an an explanation.

Answer 1

Hint: $$\begin{bmatrix}1 & 0 & 1&0\\ -4 &1&-8&1\\3 & 0 &16&-8\\-12&3&0&16 \end{bmatrix}.\begin{bmatrix}A\\B\\C\\D \end{bmatrix}=\begin{bmatrix}0\\1\\-29\\5 \end{bmatrix}$$

This should help you. Find inverse of the matrix and multiply. Note $|P|\ne0$thus it is invertible. Where equation signifies $P$.$X$=$D$

Answer 2

Start with the first (and simplest) equation.

$A+C=0$ can be rearranged to give $C=-A$.

Substitute that into the other equations:

$-4A + B - 8C + D = 1 \Rightarrow -4A + B - 8(-A) + D = 1 \Rightarrow 4A+B+D=1$

$3A + 16C - 8D = -29 \Rightarrow 3A + 16(-A) - 8D = -29 \Rightarrow -13A - 8D = -29$

-12A + 3B + 16D = 5

Answer 3

@tomi provides a quick and clever solution. The general solution process of @TheDeadLegend follows.

Form the augmented matrix. $$ \left[ \begin{array}{c|c} \mathbf{A} & \mathbf{I} \end{array} \right] = \left[ \begin{array}{rrrr|cccc} 1 & 0 & 1 & 0 & 1 & 0 & 0 & 0 \\ -4 & 1 & -8 & 1 & 0 & 1 & 0 & 0 \\ 3 & 0 & 16 & -8 & 0 & 0 & 1 & 0 \\ -12 & 3 & 0 & 16 & 0 & 0 & 0 & 1 \\ \end{array} \right] $$ Gauss-Jordan reduction:

Column 1 $$ \left[ \begin{array}{c|c} \mathbf{A} & \mathbf{I} \end{array} \right]_{1} = \left[ \begin{array}{rccc} 1 & 0 & 0 & 0 \\ 4 & 1 & 0 & 0 \\ -3 & 0 & 1 & 0 \\ 12 & 0 & 0 & 1 \\ \end{array} \right] % \left[ \begin{array}{c|c} \mathbf{A} & \mathbf{I} \end{array} \right]_{0} = \left[ \begin{array}{ccrr|rccc} 1 & 0 & 1 & 0 & 1 & 0 & 0 & 0 \\ 0 & 1 & -4 & 1 & 4 & 1 & 0 & 0 \\ 0 & 0 & 13 & -8 & -3 & 0 & 1 & 0 \\ 0 & 3 & 12 & 16 & 12 & 0 & 0 & 1 \\ \end{array} \right] $$

Column 2 $$ \left[ \begin{array}{c|c} \mathbf{A} & \mathbf{I} \end{array} \right]_{2} = \left[ \begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & \frac{1}{13} & 0 \\ 0 & -3 & 0 & 1 \\ \end{array} \right] % \left[ \begin{array}{c|c} \mathbf{A} & \mathbf{I} \end{array} \right]_{1} = \left[ \begin{array}{ccrr|rrrrc} 1 & 0 & 1 & 0 & 1 & 0 & 0 & 0 \\ 0 & 1 & -4 & 1 & 4 & 1 & 0 & 0 \\ 0 & 0 & 1 & -\frac{8}{13} & -\frac{3}{13} & 0 & \frac{1}{13} & 0 \\ 0 & 0 & 24 & 13 & 0 & -3 & 0 & 1 \\ \end{array} \right] $$

Column 3 $$ \left[ \begin{array}{c|c} \mathbf{A} & \mathbf{I} \end{array} \right]_{3} = \left[ \begin{array}{ccrc} 1 & 0 & -1 & 0 \\ 0 & 1 & 4 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & -24 & 1 \\ \end{array} \right] % \left[ \begin{array}{c|c} \mathbf{A} & \mathbf{I} \end{array} \right]_{2} = \left[ \begin{array}{cccr|rrrr} 1 & 0 & 0 & \frac{8}{13} & \frac{16}{13} & 0 & -\frac{1}{13} & 0 \\ 0 & 1 & 0 & -\frac{19}{13} & \frac{40}{13} & 1 & \frac{4}{13} & 0 \\ 0 & 0 & 1 & -\frac{8}{13} & -\frac{3}{13} & 0 & \frac{1}{13} & 0 \\ 0 & 0 & 0 & 1 & \frac{72}{361} & -\frac{39}{361} & -\frac{24}{361} & \frac{13}{361} \\ \end{array} \right] $$

Column 4 $$ \left[ \begin{array}{c|c} \mathbf{A} & \mathbf{I} \end{array} \right]_{4} = \left[ \begin{array}{cccr} 1 & 0 & 0 & -\frac{8}{13} \\ 0 & 1 & 0 & \frac{19}{13} \\ 0 & 0 & 1 & \frac{8}{13} \\ 0 & 0 & 0 & 1 \\ \end{array} \right] % \left[ \begin{array}{c|c} \mathbf{A} & \mathbf{I} \end{array} \right]_{3} = \left[ \begin{array}{cccc|rrrr} 1 & 0 & 0 & 0 & \frac{400}{361} & \frac{24}{361} & -\frac{13}{361} & -\frac{8}{361} \\ 0 & 1 & 0 & 0 & \frac{64}{19} & \frac{16}{19} & \frac{4}{19} & \frac{1}{19} \\ 0 & 0 & 1 & 0 & -\frac{39}{361} & -\frac{24}{361} & \frac{13}{361} & \frac{8}{361} \\ 0 & 0 & 0 & 1 & \frac{72}{361} & -\frac{39}{361} & -\frac{24}{361} & \frac{13}{361} \\ \end{array} \right] $$

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$$ \color{blue}{\left[ \begin{array}{r} A \\ B \\ C \\ D \end{array} \right] } = \mathbf{A}^{-1} b= \frac{1}{361} \left[ \begin{array}{rrrr} 400 & 24 & -13 & -8 \\ 1216 & 304 & 76 & 19 \\ -39 & -24 & 13 & 8 \\ 72 & -39 & -24 & 13 \\ \end{array} \right] \left[ \begin{array}{r} 0 \\ -1 \\ 29 \\5 \end{array} \right] = \color{blue}{\left[ \begin{array}{r} 1 \\ -1 \\ -5 \\ 12 \end{array} \right] } $$