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| 1 | +\def\ftmagnification{1200} |
| 2 | +\def\spacingNumerator{1} |
| 3 | +\def\spacingDenominator{1} |
| 4 | +\input jmacros |
| 5 | +%\input figMac |
| 6 | +\def\figdir{fig/} |
| 7 | +\def\showintremarks{n} |
| 8 | + |
| 9 | +\titlea{The Check Column} |
| 10 | + |
| 11 | +Ordinary arithmetic errors are a big problem when you do row operations |
| 12 | +by hand. There is a technique called ``the check column'' (that is modeled |
| 13 | +after the ``parity bit'' in computer hardware design) which provides a |
| 14 | +very effective way to catch mechanical errors. Here is an example which |
| 15 | +illustrates the technique |
| 16 | + |
| 17 | + |
| 18 | + |
| 19 | +\example{}{ |
| 20 | +The augmented matrix for the system of equations |
| 21 | +$$\deqalign{ |
| 22 | + 2&x_1&+&&x_2&+3x_3&=\,&&1\cr |
| 23 | + 4&x_1&+&5&x_2&+7x_3&=\,&&7\cr |
| 24 | + 2&x_1&-&5&x_2&+5x_3&=\,&-&7\cr |
| 25 | +}$$ |
| 26 | +is |
| 27 | +$$ |
| 28 | +\left[\left.\matrix{2&1&3\cr 4&5&7\cr 2&-5 &5\cr} |
| 29 | +\right|\matrix{1\cr7\cr-7\cr}\right] |
| 30 | +$$ |
| 31 | +To implement a ``check column'' you tack onto the right hand side of the |
| 32 | +augmented matrix an additional column. Each entry in this check column is |
| 33 | +the sum of all the entries in the row of the augmented matrix that is to |
| 34 | +the left of the check column entry. For example, the top entry in the check |
| 35 | +column is $2+1+3+1=7$. |
| 36 | +$$ |
| 37 | +\left[\left.\matrix{2&1&3\cr 4&5&7\cr 2&-5 &5\cr} |
| 38 | +\right|\matrix{1\cr7\cr-7\cr}\right]\matrix{7\cr 23\cr -5} |
| 39 | +$$ |
| 40 | + To use the check column you just perform the same row |
| 41 | +operations on the check column as you do on the augmented matrix. After |
| 42 | +each row operation you check that each entry in the check column is |
| 43 | +still the sum of all the entries in the corresponding row of the augmented |
| 44 | +matrix. |
| 45 | + |
| 46 | +We now want to eliminate the $x_1$'s from equations (2) and (3). That is, |
| 47 | +we want to make the first entries in rows 2 and 3 of the augmented matrix |
| 48 | +zero. We can achieve this by subtracting two times row (1) from row (2) and |
| 49 | +subtracting row (1) from row (3). |
| 50 | +$$ |
| 51 | +\matrix{(1)\cr (2)-2(1)\cr (3)-(1)} |
| 52 | +\left[\left.\matrix{2&1&3\cr 0&3&1\cr 0&-6 &2\cr} |
| 53 | +\right|\matrix{1\cr 5\cr-8\cr}\right]\matrix{7\cr 9\cr -12} |
| 54 | +$$ |
| 55 | +Observe that the check column entry $9$ is the sum $0+3+1+5$ |
| 56 | +of the entries in the second row of the augmented matrix. If this were |
| 57 | +not the case, it would mean that we made a mechanical error. Similarly |
| 58 | +the check column entry $-12$ is the sum $0-6+2-8$. |
| 59 | + |
| 60 | +We have now succeeded in eliminating all of the $x_1$'s from equations |
| 61 | +(2) and (3). For example, row 2 now stands for the equation |
| 62 | +$$ |
| 63 | +3x_2+x_3=5 |
| 64 | +$$ |
| 65 | +We next use equation (2) to eliminate all $x_2$'s from equation |
| 66 | +(3). |
| 67 | +$$ |
| 68 | +\matrix{(1)\cr (2)\cr (3)+2(2)} |
| 69 | +\left[\left.\matrix{2&1&3\cr 0&3&1\cr 0&0 &4\cr} |
| 70 | +\right|\matrix{1\cr5\cr2\cr}\right]\matrix{7\cr 9\cr 6} |
| 71 | +$$ |
| 72 | +We can now easily solve (3) for $x_3$, substitute the result back into (2) and |
| 73 | +solve for $x_2$ and so on: |
| 74 | +$$\meqalign{ |
| 75 | +(3)&\implies && 4x_3& =2 && &\implies && x_3&=\half\cr |
| 76 | +(2)&\implies && 3x_2+\half&=5 && &\implies && x_2&=\sfrac{3}{2}\cr |
| 77 | +(1)&\implies && 2x_1+\sfrac{3}{2}+3\times\half&=1 && &\implies && x_1&=-1\cr |
| 78 | +}$$ |
| 79 | +This last step is called ``backsolving''. |
| 80 | + |
| 81 | +Note that there is an easy way to make sure that we have not made any mechanical |
| 82 | +errors in deriving this solution --- just substitute the purported solution |
| 83 | +$(-1,3/2,1/2)$ back into the original system: |
| 84 | +$$\deqalign{ |
| 85 | + 2&(-1)&\,+\,&&\sfrac{3}{2}&+3\times\half&=\,&&1\cr |
| 86 | + 4&(-1)&\,+&5\times &\sfrac{3}{2}&+7\times\half&=\,&&7\cr |
| 87 | + 2&(-1)&\,-&5\times &\sfrac{3}{2}&+5\times\half&=\,&-&7\cr |
| 88 | +}$$ |
| 89 | +and verify that each left hand side really is equal to its corresponding |
| 90 | +right hand side. |
| 91 | +} |
| 92 | + |
| 93 | + |
| 94 | + |
| 95 | + |
| 96 | + |
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| 111 | + |
| 112 | + |
| 113 | + |
| 114 | +\end |
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