How would I answer the following question about the rank of a matrix?

Question

Let A = \begin{pmatrix}1&-8&a\\ \:\:\:4&8a&-16\\ \:\:\:a&32&16\end{pmatrix}

Then the rank of A is ? for a = ?, ? for a = ?, and ? for all other values of a."

I am struggling to compute the elimination required so that the matrix below turns into row-echelon form. This is mainly because of the variable a.

Any help on how to proceed will be highly appreciated!

Answer 1

Note that $$ \det\left[\begin{array}{rrr} 1 & -8 & a \\ 4 & 8 \, a & -16 \\ a & 32 & 16 \end{array}\right] =-8 \, a^{3} + 384 \, a + 1024 =-8 \, {\left(a + 4\right)}^{2} {\left(a - 8\right)} $$ Thus our matrix $A$ is invertible if and only if $a\neq-4$ and $a\neq8$. That is, $\DeclareMathOperator{rank}{rank}\rank(A)=3$ if and only if $a\neq -4$ and $a\neq 8$.

To compute $\rank(A)$ if $a=-4$, note that $$ \DeclareMathOperator{rref}{rref}\rref \left[\begin{array}{rrr} 1 & -8 & -4 \\ 4 & -32 & -16 \\ -4 & 32 & 16 \end{array}\right] =\left[\begin{array}{rrr} 1 & -8 & -4 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right] $$ Thus $\rank(A)=1$ if $a=-4$.

Can you compute $\rank(A)$ if $a=8$?