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Rezz5Best ResponseYou've already chosen the best response.0
\[\left[\begin{matrix}1 & 1 &2\\ 3 & 3&6\\2&2&4\end{matrix}\right]\]
 one year ago

shubhamsrgBest ResponseYou've already chosen the best response.0
even i want to learn this.. so *bookmark*
 one year ago

slaaibakBest ResponseYou've already chosen the best response.1
Do the gauss jordan elimination method and check how many leading ones there are
 one year ago

amistre64Best 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
 one year ago

amistre64Best ResponseYou've already chosen the best response.1
if the row and col have different dimensions, then rank (A) is equal or smaller than the smaller value
 one year ago

slaaibakBest ResponseYou've already chosen the best response.1
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
 one year ago

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

Rezz5Best ResponseYou've already chosen the best response.0
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?
 one year ago

slaaibakBest ResponseYou've already chosen the best response.1
no. the third one is incorrect
 one year ago

amistre64Best ResponseYou've already chosen the best response.1
given an nxm matrix rankA + nullA = m
 one year ago

slaaibakBest ResponseYou've already chosen the best response.1
soz third one is correct. it is 3
 one year ago

slaaibakBest ResponseYou've already chosen the best response.1
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
 one year ago

Rezz5Best ResponseYou've already chosen the best response.0
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?
 one year ago

Rezz5Best ResponseYou've already chosen the best response.0
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
 one year ago

slaaibakBest ResponseYou've already chosen the best response.1
counting leading ones would suffice. yes, above is correct
 one year ago
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