\magnification 2200
\parindent 0pt
\parskip 10pt
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\hsize 7truein
\hoffset -0.35truein
\vsize 9.2truein
\def\u{\vskip -5pt}
\def\v{\vskip 10pt}



Linear algebra pre-requisites you must know.


${\bf b_1, ..., b_n}$ are linearly independent if 
\u

$c_1{\bf b_1} + c_2{\bf b_2} + ... + c_n{\bf b_n}  = 
d_1{\bf b_1} + d_2{\bf b_2} + ... + d_n{\bf b_n}$ 

\centerline{implies
$c_1 = d_1$, $c_2 = d_2$..., $c_n = d_n$.}

or equivalently,
 
${\bf b_1, ..., b_n}$ are linearly independent if
\u

$c_1{\bf b_1} + c_2{\bf b_2} + ... + c_n{\bf b_n}  = 0$ 
implies $c_1 = c_2 = ... c_n$.

\v\v

Example 1:
${\bf b_1} = (1, 0, 0)$, ${\bf b_2} = (0, 1, 0)$, ${\bf b_3} = (0, 0, 1)$.
\v

\centerline{$(1, 2, 3) \not= (1, 2, 4)$.}

\v
\centerline{If $(a, b, c) = (1, 2, 3)$ then $a = 1$, $b = 2$, $c = 3$.}

\v\v
Example 2:
${\bf b_1} = 1$, ${\bf b_2} = t$, ${\bf b_3} = t^2$.
\v
\centerline{$1 + 2t +  3t^2 \not= 1 + 2t +  4t^2$.}
\v

\centerline{If $a + bt +  ct^2 = 1 + 2t + 3t^2$ then $a = 1$, $b = 2$, $c = 3$.}

\eject
Application:  Partial Fractions

${ 4\over (x^2 + 1)(x - 3)} = {Ax + B \over x^2 + 1} + {C \over x - 3}$


$\hskip .7in = {(Ax + B)(x-3) + C(x^2 + 1) \over (x^2 + 1)(x - 3)}$

Hence ${ 4\over (x^2 + 1)(x - 3)} = 
{(Ax + B)(x-3) + C(x^2 + 1)  \over (x^2 + 1)(x - 3)}$
\v
\centerline{Thus $4 = (Ax + B)(x-3) + C(x^2 + 1)$}
\v
\centerline{$ {4 = Ax^2 + Bx - 3Ax - 3B + Cx^2 + C}$}
\v
\centerline{${4 = (A + C)x^2 + (B - 3A)x - 3B + C }$}



I.e., $0x^2 + 0x + 4 = {(A + C)x^2 + (B - 3A)x - 3B + C }$

Thus $0 = A + C$, $0 = B - 3A$, $4 = -3B + C$.  

$C = -A$, $B = 3A$,\hfil \break
 $4 = -3(3A) + -A$ implies $4 = -10A$. \hfil \break  Hence $A = -{2 
\over 5}$, $B = 3(-{2 
\over 5}) = -{6 \over 5}$, $C = {2 \over 5}$.
\v
\centerline{Thus,
${ 4\over (x^2 + 1)(x - 3)} 
= {-{2 \over 5}x  - {6 \over 5} \over x^2 + 1} + {{2 \over 5} \over 
x - 3}$}

\hskip 1.65in = ${-{2}x - {6} \over 5(x^2 + 1)} + {{2} \over 
5(x - 3)}$}

\end