Linear Algebra and Partial Fractions Review
- Page ID
- 156412
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Linear Algebra
We say \(\mathrm{p}\) is a linear combination of \(\left\{\mathrm{b}_1, \mathrm{~b}_2, \cdots, \mathrm{b}_{\mathrm{n}}\right\}\) if and only if there exists \(c_i\) such that\( \mathbf{p}=c_1 \mathbf{b}_1+c_2 \mathbf{b}_{\mathbf{2}}+\ldots+c_n \mathbf{b}_{\mathbf{n}}. \)
Let \(\mathbf{b}_1=(1,0,0), \mathbf{b}_2=(0,1,0), \mathbf{b}_3=(0,0,1)\). Then \((1,2,3)\) is linear combination of \(\{(1,0,0),(0,1,0),(0,0,1)\}\)
\[\notag
(1,2,3)=1((1,0,0)+2(0,1,0)+3(0,0,1)
\]
We say that \(\mathbf{p}\) is in \(\operatorname{span}\left\{\mathbf{b}_{\mathbf{1}}, \mathbf{b}_{\mathbf{2}}, \cdots, \mathbf{b}_{\mathbf{n}}\right\}\) if and only if there exists \(c_i\) such that
\[ \notag
\mathbf{p}=c_1 \mathbf{b}_1+c_2 \mathbf{b}_2+\ldots+c_n \mathbf{b}_{\mathbf{n}}
\]
Let \(\operatorname{span}\left\{1, t, t^2\right\}=\) polynomials of degree at most 2 .
A polynomial \(p(t)\) is in the span of \(\left\{1, t, t^2\right\}\) if and only if there exists a solution for \(a, b, c\) to the equation
\[ \notag
p(t)=a+b t+c t^2
\]
We say \(\left\{\mathbf{b}_1, \mathbf{b}_2, \cdots, \mathbf{b}_{\mathbf{n}}\right\}\) is a basis for the vector space \(V\) if
1.) \(\operatorname{span}\left\{\mathbf{b}_1, \mathbf{b}_2, \cdots, \mathbf{b}_{\mathbf{n}}\right\}=V\) and
2.) \(\left\{b_1, b_2, \cdots, b_n\right\}\) is a linearly independent set.
In other words if \(\mathbf{p} \in V\), then there exists a solution for \(c_i\) for the following equation and that solution is unique:
\[ \notag
\mathbf{p}=c_1 \mathbf{b}_{\mathbf{1}}+c_2 \mathbf{b}_{\mathbf{2}}+\ldots+c_n \mathbf{b}_{\mathbf{n}}
\]
1) \(\{(1,0,0),(0,1,0),(0,0,1)\}\) is a basis for \(\mathbb{R}^3\).
2) \(\left\{1, t, t^2\right\}=\) is a basis for the set of polynomials of degree at most 2 .
Solve the following systems of equations:
\[
\left[\begin{array}{lll}
1 & 2 & 3 \\
4 & 5 & 6 \\
7 & 8 & 9
\end{array}\right]\left[\begin{array}{l}
x_1 \\
x_2 \\
x_3
\end{array}\right]=\left[\begin{array}{l}
0 \\
0 \\
0
\end{array}\right]
\]
\[\left[\begin{array}{lll}
1 & 2 & 3 \\
4 & 5 & 6 \\
7 & 8 & 9
\end{array}\right]\left[\begin{array}{l}
x_1 \\
x_2 \\
x_3
\end{array}\right]=\left[\begin{array}{l}
0 \\
3 \\
0
\end{array}\right]\left[\begin{array}{lll}
1 & 2 & 3 \\
4 & 5 & 6 \\
7 & 8 & 9
\end{array}\right]\left[\begin{array}{l}
x_1 \\
x_2 \\
x_3
\end{array}\right]=\left[\begin{array}{l}
2 \\
5 \\
8
\end{array}\right]\]
Solution
\begin{array}{l}
{\left[\begin{array}{llllll}
1 & 2 & 3 & 0 & 0 & 2 \\
4 & 5 & 6 & 0 & 3 & 5 \\
7 & 8 & 9 & 0 & 0 & 8
\end{array}\right]} \\
\downarrow R_2-4 R_1 \rightarrow R_2, \quad R_3-7 R_1 \rightarrow R_3
\end{array}
\[
\begin{array}{l}
{\left[\begin{array}{rrrrrr}
1 & 2 & 3 & 0 & 0 & 2 \\
0 & -3 & -6 & 0 & 3 & -3 \\
0 & -6 & -12 & -7 & 0 & -6
\end{array}\right]} \\
\downarrow R_3-2 R_1 \rightarrow R_3 \\
{\left[\begin{array}{rrrrrr}
1 & 2 & 3 & 0 & 0 & 2 \\
0 & -3 & -6 & 0 & 3 & -3 \\
0 & 0 & 0 & 0 & -6 & 0
\end{array}\right]}
\end{array}
\]
\(\downarrow\) already know sol'n to system b.
\[
\begin{array}{l}
{\left[\begin{array}{rrrrr}
1 & 2 & 3 & 0 & 2 \\
0 & -3 & -6 & 0 & -3 \\
0 & 0 & 0 & 0 & 0
\end{array}\right]} \\
\downarrow-\frac{1}{3} R_2 \rightarrow R_2
\end{array}
\]
\[\left[\begin{array}{lllll}
1 & 2 & 3 & 0 & 2 \\
0 & 1 & 2 & 0 & 1 \\
0 & 0 & 0 & 0 & 0
\end{array}\right] \quad R_1-\overrightarrow{2 R_2} \rightarrow R_1\left[\begin{array}{rrrrr}
1 & 0 & -1 & 0 & 0 \\
0 & 1 & 2 & 0 & 1 \\
0 & 0 & 0 & 0 & 0
\end{array}\right]\]
\[\begin{array}{c}
{\left[\begin{array}{lll}
1 & 2 & 3 \\
4 & 5 & 6 \\
7 & 8 & 9
\end{array}\right]\left[\begin{array}{l}
x_1 \\
x_2 \\
x_3
\end{array}\right]=\left[\begin{array}{l}
0 \\
0 \\
0
\end{array}\right]} \\
{\left[\begin{array}{l}
x_1 \\
x_2 \\
x_3
\end{array}\right]=\left[\begin{array}{c}
x_3 \\
-2 x_3 \\
x_3
\end{array}\right]=x_3\left[\begin{array}{r}
1 \\
-2 \\
1
\end{array}\right]}
\end{array}\]
\[\left[\begin{array}{lll}
1 & 2 & 3 \\
4 & 5 & 6 \\
7 & 8 & 9
\end{array}\right]\left[\begin{array}{l}
x_1 \\
x_2 \\
x_3
\end{array}\right]=\left[\begin{array}{l}
0 \\
3 \\
0
\end{array}\right] \quad \text { no solution }\]
\[\begin{array}{c}
{\left[\begin{array}{lll}
1 & 2 & 3 \\
4 & 5 & 6 \\
7 & 8 & 9
\end{array}\right]\left[\begin{array}{l}
x_1 \\
x_2 \\
x_3
\end{array}\right]=\left[\begin{array}{l}
2 \\
5 \\
8
\end{array}\right]} \\
{\left[\begin{array}{l}
x_1 \\
x_2 \\
x_3
\end{array}\right]=x_3\left[\begin{array}{r}
1 \\
-2 \\
1
\end{array}\right]+\left[\begin{array}{l}
0 \\
1 \\
0
\end{array}\right]}
\end{array}\]
Check:
\[
\left[\begin{array}{lll}
1 & 2 & 3 \\
4 & 5 & 6 \\
7 & 8 & 9
\end{array}\right]\left[\begin{array}{r}
1 \\
-2 \\
1
\end{array}\right]=\left[\begin{array}{l}
0 \\
0 \\
0
\end{array}\right] \&\left[\begin{array}{lll}
1 & 2 & 3 \\
4 & 5 & 6 \\
7 & 8 & 9
\end{array}\right]\left[\begin{array}{l}
0 \\
1 \\
0
\end{array}\right]=\left[\begin{array}{l}
2 \\
5 \\
8
\end{array}\right]
\]
Partial Fractions
Observe the following examples to help remind us how to apply partial fraction
\[ \notag
\begin{aligned}
\frac{x+1}{(x+2)(x-3)} & =\frac{A}{x+2}+\frac{B}{x-3} \\
\\
\frac{4}{\left(x^2+1\right)(x-3)} & =\frac{A x+B}{x^2+1}+\frac{C}{x-3}\\
\\
\frac{x^4+1}{\left(x^2+1\right)(x-3)^3}&=\frac{A x+B}{x^2+1}+\frac{C}{x-3}+\frac{D}{(x-3)^2}+\frac{E}{(x-3)^3} \\
\\
\frac{4}{\left(x^2+1\right)^2(x-3)}&=\frac{A x+B}{x^2+1}+\frac{C x+D}{\left(x^2+1\right)^2}+\frac{E}{x-3}
\end{aligned}
\]
Note, it can be helpful to simplify first.
\[ \notag
\begin{array}{c}
\frac{x^2-1}{(x+1)^2}=\frac{(x-1)(x+1)}{(x+1)^2}=\frac{x-1}{x+1}=\frac{x+1-1-1}{x+1} \\
=\frac{x+1-2}{x+1}=\frac{x+1}{x+1}+\frac{-2}{x+1}=1+\frac{-2}{x+1}
\end{array}
\]
For partial fractions, the power in numerator must be less than the power in denominator.
If power in numerator \(\geq\) power in denominator, do long division first (or add a "0" and simplify algebraically).
Also note that if you don't like the denominator, then you can get rid of them:
\[\notag
\begin{array}{c}
\frac{4}{\left(x^2+1\right)(x-3)}=\frac{A x+B}{x^2+1}+\frac{C}{x-3}\\
4=(A x+B)(x-3)+C\left(x^2+1\right) \\
4=A x^2+B x-3 A x-3 B+C x^2+C \\
4=(A+C) x^2+(B-3 A) x-3 B+C\\
\implies 0 x^2+0 x+4=(A+C) x^2+(B-3 A) x-3 B+C
\end{array}
\]
This then allows us to solve for \(A,B\) and \(C\) by the following calculations:
\[\notag
0 x^2+0 x+4=(A+C) x^2+(B-3 A) x-3 B+C
\]
Thus \(0=A+C, \quad 0=B-3 A, \quad 4=-3 B+C\).
\[\notag
\begin{array}{l}
C=-A, B=3 A, \quad 4=-3(3 A)+-A \Rightarrow \notag \\
4=-10 A . \notag
\end{array}
\]
Hence \(A=-\frac{2}{5}, \quad B=3\left(-\frac{2}{5}\right)=-\frac{6}{5}, \quad C=\frac{2}{5}\).
Thus, \(\frac{4}{\left(x^2+1\right)(x-3)}=\frac{-\frac{2}{5} x-\frac{6}{5}}{x^2+1}+\frac{\frac{2}{5}}{x-3}\)
\[\notag
=\frac{-2 x-6}{5\left(x^2+1\right)}+\frac{2}{5(x-3)}
\]
\( 0 x^2+0 x+4=(A+C) x^2+(B-3 A) x-3 B+C \)
Note there are many correct ways to solve for \(A, B, C\). For example, one can plug in \(x=3\) to quickly find \(C\) and then solve for \(A\) and \(B\).
\[\notag
4=(A x+B)(x-3)+C\left(x^2+1\right)
\]
One can also use matrices to solve linear eqns.