For an assignment I have to solve the following problem:
Find all possible values of rank(A) as 'a' varies:
\begin{bmatrix}a&2&-1\\3&3&-2\\-2&-1&a\end{bmatrix}
I came across several answers on google and stack (Find all possible values of rank(A) as a varies?, and similar question, different matrix: Finding all possible ranks of matrix $A$ as $a$ varies?) but I need to solve the problem using row reduction instead of calculating the determinant. How do I go about?
I'm stuck here:
swapping rows gives:
\begin{bmatrix}-2&-1&a\\3&3&-2\\a&2&-1\end{bmatrix}
reducing rows 2 and 3 gives:
\begin{bmatrix}-2&-1&a\\0&\frac{3}{2}&-2+\frac{3}{2}a\\\\0&2-\frac{1}{2}a&-1+\frac{1}{2}a^2\end{bmatrix}
How to proceed from here?
according to Find all possible values of rank(A) as a varies? the answer should be: rank = 2 for a = 1 or a = 5/3 for all other values rank = 3.
Best Answer
Your first step is correct. Now multiply the second row by $\frac{2}{3}(\frac{1}{2}a-2)$ and sum to the third line, so you find an upper triangular matrix: $$ \begin {bmatrix} -2&-1&a\\ 0&\frac{3}{2}&-2+\frac{3}{2}a\\ 0&0&a^2-\frac{8}{3}a+\frac{5}{3} \end{bmatrix} $$
If $a^2-\frac{8}{3}a+\frac{5}{3} \ne 0$ the matrix is full rank, and this is done if $a$ is different from 1, and $\frac{5}{3}$ (the roots of $a^2-\frac{8}{3}a+\frac{5}{3} = 0$ ). Substituting these values you see that the other two rows are linearly independent so the rank of the matrix is $2$ for $a=1$ or $a=\frac{5}{3}$.