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I've been trying to figure out how to reduce this matrix without introducing fractions in the intermediate stages but can't figure out how to do it. $$\begin{bmatrix}2&1&3\\ 0&-2&-29\\ 3&4&5\end{bmatrix}$$

Parcly Taxel
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Caroline E. Drummond
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  • A naive approach is to create leading ones by finding columns whose greatest common divisor is one, so that adding some multiple of one row to another allows (possibly after several steps) a one entry in that column to be created. If a one entry is created, then that can be swapped into the appropriate row, and the remaining entries in the column brought to zero (again by adding/subtracting an appropriate multiple of one row to the others). For small matrices such as above, little more needs to be said. But for larger matrices the growth of intermediate results is cause for caution. – hardmath Feb 27 '18 at 03:59
  • Note that this would not be possible if all elements in, for example, the left-hand column, were divisible by a common prime ( but not all zero) – Will Jagy Feb 27 '18 at 04:29

2 Answers2

2

Here's one way.

$\begin{bmatrix}2&1&3\\ 0&-2&-29\\ 3&4&5\end{bmatrix} \xrightarrow {R3-R1} \begin{bmatrix}2&1&3\\ 0&-2&-29\\ 1&3&2 \end{bmatrix} \xrightarrow {R1-2R3} \begin{bmatrix}0&-5&-1\\ 0&-2&-29\\ 1&3&2 \end{bmatrix} \xrightarrow {-R1} \begin{bmatrix}0&5&1\\ 0&-2&-29\\ 1&3&2 \end{bmatrix} \xrightarrow {R1+2R2} \begin{bmatrix}0&1&-57\\ 0&-2&-29\\ 1&3&2 \end{bmatrix} \xrightarrow {R2+2R1} \begin{bmatrix}0&1&-57\\ 0&0&-143\\ 1&3&2 \end{bmatrix} \xrightarrow{-R2/143} \begin{bmatrix}0&1&-57\\ 0&0&1\\ 1&3&2 \end{bmatrix} \xrightarrow{R1+57R2} \begin{bmatrix}0&1&0\\ 0&0&1\\ 1&3&2 \end{bmatrix} \xrightarrow[R3-2R2]{R3-3R1} \begin{bmatrix}0&1&0\\ 0&0&1\\ 1&0&0 \end{bmatrix} \xrightarrow[R2\leftrightarrow R3]{R2\leftrightarrow R3} \begin{bmatrix}1&0&0\\ 0&1&0\\ 0&0&1 \end{bmatrix} $

Martin Argerami
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Subtract the first from the last line. This gives a $1$. Use this $1$ to eliminate evrything else on the first column. For the second column, the same trick, subtract first to get a $1$.

Rene Schipperus
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