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Section A.4 System of Linear Equations
Solving systems of linear equations is a central topic of linear algebra. In this appendix we only illustrate hands-on methods: substitution and elimination, with no matrix notation.
These methods appear in the main part of the book when we compare coefficients in partial fractions; see
Subsection 3.4.1 and especially
Example 3.43 .
Example A.11 .
Solve
\begin{align*}
x+y+z &= 6\\
2x-y+z &= 3\\
x+2y-z &= 2.
\end{align*}
From the first equation, \(x=6-y-z\text{.}\) Substitute into the other two equations:
\begin{align*}
2(6-y-z)-y+z &= 3 \Rightarrow 3y+z=9\\
(6-y-z)+2y-z &= 2 \Rightarrow y-2z=-4.
\end{align*}
Now eliminate: from \(y-2z=-4\) we get \(y=2z-4\text{.}\) Substitute into \(3y+z=9\text{:}\)
\begin{equation*}
3(2z-4)+z=9 \Rightarrow 7z=21 \Rightarrow z=3.
\end{equation*}
Then \(y=2\) and \(x=1\text{.}\)
Example A.12 .
In
Example 3.43 , comparing coefficients gives the system for
\(A_{11},B_{11},C_{11},B_{12},C_{12}\text{:}\)
\begin{align*}
A_{11}+B_{11} &= 1\\
-B_{11}+C_{11} &= -1\\
2A_{11}+B_{11}-C_{11}+B_{12} &= 4\\
-B_{11}+C_{11}-B_{12}+C_{12} &= -3\\
A_{11}-C_{11}-C_{12} &= 3.
\end{align*}
Use substitution first:
\begin{equation*}
B_{11}=1-A_{11},
\qquad
C_{11}=-1+B_{11}=-A_{11},
\qquad
C_{12}=A_{11}-C_{11}-3=2A_{11}-3.
\end{equation*}
Next, from the fourth equation and \(-B_{11}+C_{11}=-1\text{,}\)
\begin{equation*}
-1-B_{12}+C_{12}=-3 \Rightarrow B_{12}=C_{12}+2=2A_{11}-1.
\end{equation*}
From the third equation and \(B_{11}-C_{11}=1\text{,}\)
\begin{equation*}
2A_{11}+1+B_{12}=4 \Rightarrow B_{12}=3-2A_{11}.
\end{equation*}
Therefore
\begin{equation*}
2A_{11}-1=3-2A_{11} \Rightarrow A_{11}=1.
\end{equation*}
Back-substitute:
\begin{equation*}
B_{11}=0,\quad C_{11}=-1,\quad B_{12}=1,\quad C_{12}=-1.
\end{equation*}
\begin{equation*}
\frac{2x^5-x^4+3x^3-x+1}{(x-1)(x^2+1)^2}
=2+\frac{1}{x-1}-\frac{1}{x^2+1}+\frac{x-1}{(x^2+1)^2}.
\end{equation*}
