
#1
Mar1708, 09:03 PM

P: 4

1. The problem statement, all variables and given/known data
Find all possible values of rank(A) as a varies. A=[1, 2, a] [2, 4a, 2] [a, 2, 1] A is 3x3, just merge the 3 row vectors. 3. The attempt at a solution I have the solution but I create this thread to find the most effective/efficient procedure for me to solve these kinds of problems. What are the possible dimensions of the row/column space as a varies? When do the columns/ranks become linearly dependent based on different values for a? I don't want to guess or plug in arbitrary numbers; I want a generalized, systematic approach that works every time. 



#2
Mar1808, 04:20 AM

Sci Advisor
HW Helper
P: 4,301

There are the following possibilities:
Now for each case, you can write down an equation and solve it for a. For example,let me do the third case (first and third are dependent, but independent of the second). If the third column is a multiple n of the first one, you must have 1 = n a 2 = 2 n a = n 1 From the second equation you see that there is just one possibility for n. Then you get a solution for a from one of the others. Finally, use the remaining equation to see if this value indeed satisfies all of them. Then plug this value into the 4a in the second column, and check that it is indeed independent of the first (and/or third) 



#3
Mar1808, 09:00 AM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,877

The simplest way to determine the rank of a matrix is to "rowreduce". The rank is the number of rows that contain nonzero entries.
In the case you give [tex]A= \left(\begin{array}{ccc}1 & 2 & a \\2 & 4a & 2\\ a & 2 & 1\end{array}\right)[/tex] Add twice the first row to the second and subtract a times the first row from the third to get [tex]A= \left(\begin{array}{ccc}1 & 2 & a \\0 & 4a+ 4 & 2+ 2a\\ 0 & 22a & 1a^2\end{array}\right)[/tex] Now add half the second row to the third to get [tex]A= \left(\begin{array}{ccc}1 & 2 & a \\0 & 4a+ 4 & 2+ 2a\\ 0 & 0 & 2+ 2a a^2\end{array}\right)[/tex] If a= 1 that has only 2 nonzero rows and so the rank of A is 2. If a= [itex]1\pm\sqrt{3} the last row is 0 and again the rank of A is 2. For any other value of a, the rank is 3. 



#4
Mar1808, 10:17 PM

P: 4

Rank of a matrix
Thanks a lot; that was very helpful.



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