Find all possible values of rank(A) as a varies

[proxyIP]

Homework Statement

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.

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.

 

[CompuChip]

There are the following possibilities:

• All columns are independent
• The first and the second are dependent, but independent of the third
• The first and the third are dependent, but independent of the second
• The second and the third are dependent, but independent of the first
• All of them are dependent

Or, if you prefer, you can do this with the rows.
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)

 

[HallsofIvy]

The simplest way to determine the rank of a matrix is to "row-reduce". The rank is the number of rows that contain non-zero entries.
In the case you give
A= \left(\begin{array}{ccc}1 & 2 & a \\-2 & 4a & 2\\ a & -2 & 1\end{array}\right)
Add twice the first row to the second and subtract a times the first row from the third to get
A= \left(\begin{array}{ccc}1 & 2 & a \\0 & 4a+ 4 & 2+ 2a\\ 0 & -2-2a & 1-a^2\end{array}\right)
Now add half the second row to the third to get
A= \left(\begin{array}{ccc}1 & 2 & a \\0 & 4a+ 4 & 2+ 2a\\ 0 & 0 & 2+ 2a- a^2\end{array}\right)
If a= -1 that has only 2 non-zero rows and so the rank of A is 2. If a= -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.

 

[proxyIP]

Thanks a lot; that was very helpful.