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Merge pull request #6 from UBCMath/extra
Extra material
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.codespell-ignorelines

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# lines that codespell should ignore: whitespace matters!
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# jmacros.tex: "short implies" not simplifies!
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\def\simplies{\;\Rightarrow\;}
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# figMac.tex
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\openin\labeLfile=\figdir#1.lbl
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\openin\labeLfile=\figdir#2.lbl
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extra/CoverPage2018.pdf

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extra/Ex3point3v2.pdf

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extra/check.tex

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\def\ftmagnification{1200}
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\def\spacingNumerator{1}
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\def\spacingDenominator{1}
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\input jmacros
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%\input figMac
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\def\figdir{fig/}
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\def\showintremarks{n}
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\titlea{The Check Column}
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Ordinary arithmetic errors are a big problem when you do row operations
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by hand. There is a technique called ``the check column'' (that is modeled
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after the ``parity bit'' in computer hardware design) which provides a
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very effective way to catch mechanical errors. Here is an example which
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illustrates the technique
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\example{}{
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The augmented matrix for the system of equations
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$$\deqalign{
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2&x_1&+&&x_2&+3x_3&=\,&&1\cr
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4&x_1&+&5&x_2&+7x_3&=\,&&7\cr
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2&x_1&-&5&x_2&+5x_3&=\,&-&7\cr
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}$$
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is
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$$
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\left[\left.\matrix{2&1&3\cr 4&5&7\cr 2&-5 &5\cr}
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\right|\matrix{1\cr7\cr-7\cr}\right]
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$$
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To implement a ``check column'' you tack onto the right hand side of the
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augmented matrix an additional column. Each entry in this check column is
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the sum of all the entries in the row of the augmented matrix that is to
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the left of the check column entry. For example, the top entry in the check
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column is $2+1+3+1=7$.
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$$
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\left[\left.\matrix{2&1&3\cr 4&5&7\cr 2&-5 &5\cr}
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\right|\matrix{1\cr7\cr-7\cr}\right]\matrix{7\cr 23\cr -5}
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$$
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To use the check column you just perform the same row
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operations on the check column as you do on the augmented matrix. After
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each row operation you check that each entry in the check column is
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still the sum of all the entries in the corresponding row of the augmented
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matrix.
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We now want to eliminate the $x_1$'s from equations (2) and (3). That is,
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we want to make the first entries in rows 2 and 3 of the augmented matrix
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zero. We can achieve this by subtracting two times row (1) from row (2) and
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subtracting row (1) from row (3).
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$$
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\matrix{(1)\cr (2)-2(1)\cr (3)-(1)}
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\left[\left.\matrix{2&1&3\cr 0&3&1\cr 0&-6 &2\cr}
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\right|\matrix{1\cr 5\cr-8\cr}\right]\matrix{7\cr 9\cr -12}
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$$
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Observe that the check column entry $9$ is the sum $0+3+1+5$
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of the entries in the second row of the augmented matrix. If this were
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not the case, it would mean that we made a mechanical error. Similarly
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the check column entry $-12$ is the sum $0-6+2-8$.
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We have now succeeded in eliminating all of the $x_1$'s from equations
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(2) and (3). For example, row 2 now stands for the equation
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$$
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3x_2+x_3=5
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$$
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We next use equation (2) to eliminate all $x_2$'s from equation
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(3).
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$$
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\matrix{(1)\cr (2)\cr (3)+2(2)}
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\left[\left.\matrix{2&1&3\cr 0&3&1\cr 0&0 &4\cr}
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\right|\matrix{1\cr5\cr2\cr}\right]\matrix{7\cr 9\cr 6}
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$$
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We can now easily solve (3) for $x_3$, substitute the result back into (2) and
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solve for $x_2$ and so on:
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$$\meqalign{
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(3)&\implies && 4x_3& =2 && &\implies && x_3&=\half\cr
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(2)&\implies && 3x_2+\half&=5 && &\implies && x_2&=\sfrac{3}{2}\cr
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(1)&\implies && 2x_1+\sfrac{3}{2}+3\times\half&=1 && &\implies && x_1&=-1\cr
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}$$
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This last step is called ``backsolving''.
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Note that there is an easy way to make sure that we have not made any mechanical
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errors in deriving this solution --- just substitute the purported solution
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$(-1,3/2,1/2)$ back into the original system:
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$$\deqalign{
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2&(-1)&\,+\,&&\sfrac{3}{2}&+3\times\half&=\,&&1\cr
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4&(-1)&\,+&5\times &\sfrac{3}{2}&+7\times\half&=\,&&7\cr
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2&(-1)&\,-&5\times &\sfrac{3}{2}&+5\times\half&=\,&-&7\cr
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}$$
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and verify that each left hand side really is equal to its corresponding
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right hand side.
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}
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\end

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