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Find rank [[8,-6,2],[-6,7,-4],[2,-4,3]]

Solution:
Your problem `->` rank [[8,-6,2],[-6,7,-4],[2,-4,3]]


 Rank 
`8``-6``2`
`-6``7``-4`
`2``-4``3`


Now, reduce this matrix
`R_1 larr R_1-:8`

 = 
 `1` `1=8-:8`
`R_1 larr R_1-:8`
 `-3/4` `-3/4=-6-:8`
`R_1 larr R_1-:8`
 `1/4` `1/4=2-:8`
`R_1 larr R_1-:8`
`-6``7``-4`
`2``-4``3`


`R_2 larr R_2+6xx R_1`

 = 
`1``-3/4``1/4`
 `0` `0=-6+6xx1`
`R_2 larr R_2+6xx R_1`
 `5/2` `5/2=7+6xx-3/4`
`R_2 larr R_2+6xx R_1`
 `-5/2` `-5/2=-4+6xx1/4`
`R_2 larr R_2+6xx R_1`
`2``-4``3`


`R_3 larr R_3-2xx R_1`

 = 
`1``-3/4``1/4`
`0``5/2``-5/2`
 `0` `0=2-2xx1`
`R_3 larr R_3-2xx R_1`
 `-5/2` `-5/2=-4-2xx-3/4`
`R_3 larr R_3-2xx R_1`
 `5/2` `5/2=3-2xx1/4`
`R_3 larr R_3-2xx R_1`


`R_2 larr R_2-:5/2`

 = 
`1``-3/4``1/4`
 `0` `0=0-:5/2`
`R_2 larr R_2-:5/2`
 `1` `1=5/2-:5/2`
`R_2 larr R_2-:5/2`
 `-1` `-1=-5/2-:5/2`
`R_2 larr R_2-:5/2`
`0``-5/2``5/2`


`R_3 larr R_3+(5/2)xx R_2`

 = 
`1``-3/4``1/4`
`0``1``-1`
 `0` `0=0+(5/2)xx0`
`R_3 larr R_3+(5/2)xx R_2`
 `0` `0=-5/2+(5/2)xx1`
`R_3 larr R_3+(5/2)xx R_2`
 `0` `0=5/2+(5/2)xx-1`
`R_3 larr R_3+(5/2)xx R_2`


The rank of a matrix is the number of non all-zeros rows
`:. Rank = 2`






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