## Rezz5 2 years ago how to find the rank of a matrix?

1. Rezz5

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

2. shubhamsrg

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

3. slaaibak

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

4. amistre64

""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

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

6. slaaibak

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

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

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

no. the third one is incorrect

10. amistre64

given an nxm matrix rankA + nullA = m

11. Rezz5

third would be 2?

12. slaaibak

soz third one is correct. it is 3

13. slaaibak

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

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

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

counting leading ones would suffice. yes, above is correct

17. Rezz5

thanks all!!