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Rezz5
 2 years ago
Best ResponseYou've already chosen the best response.0\[\left[\begin{matrix}1 & 1 &2\\ 3 & 3&6\\2&2&4\end{matrix}\right]\]

shubhamsrg
 2 years ago
Best ResponseYou've already chosen the best response.0even i want to learn this.. so *bookmark*

slaaibak
 2 years ago
Best ResponseYou've already chosen the best response.1Do the gauss jordan elimination method and check how many leading ones there are

amistre64
 2 years ago
Best ResponseYou've already chosen the best response.1""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

amistre64
 2 years ago
Best ResponseYou've already chosen the best response.1if the row and col have different dimensions, then rank (A) is equal or smaller than the smaller value

slaaibak
 2 years ago
Best ResponseYou've already chosen the best response.1By 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

Servet
 2 years ago
Best ResponseYou've already chosen the best response.0As people have suggested, get it down to reduced form and see how many leading 1's there is :) that is the rank.

Rezz5
 2 years ago
Best ResponseYou've already chosen the best response.0So 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?

slaaibak
 2 years ago
Best ResponseYou've already chosen the best response.1no. the third one is incorrect

amistre64
 2 years ago
Best ResponseYou've already chosen the best response.1given an nxm matrix rankA + nullA = m

slaaibak
 2 years ago
Best ResponseYou've already chosen the best response.1soz third one is correct. it is 3

slaaibak
 2 years ago
Best ResponseYou've already chosen the best response.11 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

Rezz5
 2 years ago
Best ResponseYou've already chosen the best response.0okay, i think i undertand now so we reduce to row echelon form and do leading 1's then count the smallest number of rows?

Rezz5
 2 years ago
Best ResponseYou've already chosen the best response.0so 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

slaaibak
 2 years ago
Best ResponseYou've already chosen the best response.1counting leading ones would suffice. yes, above is correct
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