## Rezz5 Group Title how to find the rank of a matrix? one year ago one year ago

1. Rezz5 Group Title

$\left[\begin{matrix}1 & -1 &2\\ 3 & -3&6\\-2&2&4\end{matrix}\right]$

2. shubhamsrg Group Title

even i want to learn this.. so *bookmark*

3. slaaibak Group Title

Do the gauss jordan elimination method and check how many leading ones there are

4. amistre64 Group Title

""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

5. amistre64 Group Title

if the row and col have different dimensions, then rank (A) is equal or smaller than the smaller value

6. slaaibak Group Title

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

7. Servet Group Title

As people have suggested, get it down to reduced form and see how many leading 1's there is :) that is the rank.

8. Rezz5 Group Title

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?

9. slaaibak Group Title

no. the third one is incorrect

10. amistre64 Group Title

given an nxm matrix rankA + nullA = m

11. Rezz5 Group Title

third would be 2?

12. slaaibak Group Title

soz third one is correct. it is 3

13. slaaibak Group Title

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

14. Rezz5 Group Title

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?

15. Rezz5 Group Title

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

16. slaaibak Group Title

counting leading ones would suffice. yes, above is correct

17. Rezz5 Group Title

thanks all!!