1. Example `[[8,-6,2],[-6,7,-4],[2,-4,3]]` (Previous example) | 3. Example `[[1,1,1],[-1,-3,-3],[2,4,4]]` (Next example) |
2. Example `[[3,2,4],[2,0,2],[4,2,3]]`
Find Pivots of a Matrix ... `[[3,2,4],[2,0,2],[4,2,3]]`
Solution: First apply Gaussian Elimination method to find Pivots
`A` | = | | `3` | `2` | `4` | | | `2` | `0` | `2` | | | `4` | `2` | `3` | |
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`R_2 larr R_2-2/3xx R_1`
= | | `3` | `2` | `4` | | | `0` `0=2-2/3xx3` `R_2 larr R_2-2/3xx R_1` | `-4/3` `-4/3=0-2/3xx2` `R_2 larr R_2-2/3xx R_1` | `-2/3` `-2/3=2-2/3xx4` `R_2 larr R_2-2/3xx R_1` | | | `4` | `2` | `3` | |
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`R_3 larr R_3-4/3xx R_1`
= | | `3` | `2` | `4` | | | `0` | `-4/3` | `-2/3` | | | `0` `0=4-4/3xx3` `R_3 larr R_3-4/3xx R_1` | `-2/3` `-2/3=2-4/3xx2` `R_3 larr R_3-4/3xx R_1` | `-7/3` `-7/3=3-4/3xx4` `R_3 larr R_3-4/3xx R_1` | |
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`R_3 larr R_3-1/2xx R_2`
= | | `3` | `2` | `4` | | | `0` | `-4/3` | `-2/3` | | | `0` `0=0-1/2xx0` `R_3 larr R_3-1/2xx R_2` | `0` `0=-2/3-1/2xx-4/3` `R_3 larr R_3-1/2xx R_2` | `-2` `-2=-7/3-1/2xx-2/3` `R_3 larr R_3-1/2xx R_2` | |
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Now, Pivots are the first non-zero element in each row of this eliminated matrix. Pivots are `3,-4/3,-2`
This material is intended as a summary. Use your textbook for detail explanation. Any bug, improvement, feedback then
1. Example `[[8,-6,2],[-6,7,-4],[2,-4,3]]` (Previous example) | 3. Example `[[1,1,1],[-1,-3,-3],[2,4,4]]` (Next example) |
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