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Rezz5
how to find the rank of a matrix?
\[\left[\begin{matrix}1 & -1 &2\\ 3 & -3&6\\-2&2&4\end{matrix}\right]\]
even i want to learn this.. so *bookmark*
Do the gauss jordan elimination method and check how many leading ones there are
""Suppose that A is a matrix then the row space of A and the column space of A will have the same dimension. We call this common dimension the rank of A and denote it by rank (A)." http://tutorial.math.lamar.edu/Classes/LinAlg/FundamentalSubspaces.aspx
if the row and col have different dimensions, then rank (A) is equal or smaller than the smaller value
By inspection, it would be 2. Row 2 is a multiple of row one. row 3 is not a multiple or linear combination of row 1 or 2 so therefore, row 1 and 3 are linearly independent. hence, rank = 2
As people have suggested, get it down to reduced form and see how many leading 1's there is :) that is the rank.
So if i get something like 1 4 6 0 3 4 0 0 0 rank is 2? or 2 5 7 0 0 0 0 0 0 rank is 1? Or 1 4 5 1 0 0 1 0 4 rank is 3?
no. the third one is incorrect
given an nxm matrix rankA + nullA = m
soz third one is correct. it is 3
1 4 5 1 0 0 1 0 4 1 0 0 1 4 5 1 0 4 1 0 0 0 4 5 0 0 4
okay, i think i undertand now so we reduce to row echelon form and do leading 1's then count the smallest number of rows?
so 1 0 0 0 4 5 0 0 4 becomes 1 0 0 0 1 5/4 0 0 1 and that has three 1's so rank is 3
counting leading ones would suffice. yes, above is correct