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Solution
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Solution provided by AtoZmath.com
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Column Space calculator
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1. `[[8,-6,2],[-6,7,-4],[2,-4,3]]` 2. `[[6,-2,2],[-2,3,-1],[2,-1,3]]` 3. `[[3,2,4],[2,0,2],[4,2,3]]` 4. `[[1,1,1],[-1,-3,-3],[2,4,4]]` 5. `[[2,3],[4,10]]` 6. `[[5,1],[4,2]]`
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Example1. Find Column Space ... `[[1,-2,0,3,-4],[3,2,8,1,4],[2,3,7,2,3],[-1,2,0,4,-3]]`Solution: | `1` | `-2` | `0` | `3` | `-4` | | | `3` | `2` | `8` | `1` | `4` | | | `2` | `3` | `7` | `2` | `3` | | | `-1` | `2` | `0` | `4` | `-3` | |
Now, reduce the matrix to row echelon form `R_2 larr R_2-3xx R_1` = | | `1` | `-2` | `0` | `3` | `-4` | | | `0` | `8` | `8` | `-8` | `16` | | | `2` | `3` | `7` | `2` | `3` | | | `-1` | `2` | `0` | `4` | `-3` | |
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`R_3 larr R_3-2xx R_1` = | | `1` | `-2` | `0` | `3` | `-4` | | | `0` | `8` | `8` | `-8` | `16` | | | `0` | `7` | `7` | `-4` | `11` | | | `-1` | `2` | `0` | `4` | `-3` | |
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`R_4 larr R_4+ R_1` = | | `1` | `-2` | `0` | `3` | `-4` | | | `0` | `8` | `8` | `-8` | `16` | | | `0` | `7` | `7` | `-4` | `11` | | | `0` | `0` | `0` | `7` | `-7` | |
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`R_3 larr R_3-7/8xx R_2` = | | `1` | `-2` | `0` | `3` | `-4` | | | `0` | `8` | `8` | `-8` | `16` | | | `0` | `0` | `0` | `3` | `-3` | | | `0` | `0` | `0` | `7` | `-7` | |
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`R_4 larr R_4-7/3xx R_3` = | | `1` | `-2` | `0` | `3` | `-4` | | | `0` | `8` | `8` | `-8` | `16` | | | `0` | `0` | `0` | `3` | `-3` | | | `0` | `0` | `0` | `0` | `0` | |
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The rank of a matrix is the number of non all-zeros rows `:. Rank = 3` Column Space : The matrix has 3 pivots and Pivots are in the columns 1,2 and 4. We know that these columns in the original matrix define the column space of the matrix. `:.` The Column Space is `[[1],[3],[2],[-1]],[[-2],[2],[3],[2]],[[3],[1],[2],[4]]`
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