Determine for which values of a is the matrix A(a) invertible?

Question

I have a $4\times 4$ matrix and I am supposed to find for which values of $a$ is the matrix $A(a)$ invertible? The matrix is as follows: $$ A(a) = \begin{bmatrix} 1 & 0 & -1 & a \\ 2 & 3 & 1 & 4 \\ 0 & 0 & a & 0 \\ 5 & 0 & 1 & 5 \end{bmatrix} $$

I tried Cramer's rule by expanding through the second column because it has three zeros in it, which made it easier. After solving I got:

$3[5a-5a^2] \implies 15a-15a^2$

How am I supposed to go further now? how can I find the values of $a$ from here?

Any help would be highly appreciated!

Answer 1

Hint : For which $a$ do we have $15a-15a^2=15a(1-a)=0$ ?

Answer 2

The method is certainly sound. The determinant is indeed $$ \det A(a)=3\det \begin{bmatrix} 1 & -1 & a \\ 0 & a & 0 \\ 5 & 1 & 5 \end{bmatrix} =3a\det\begin{bmatrix} 1 & a \\ 5 & 5 \end{bmatrix} =3a(5-5a) $$ which is different from zero if and only if $a\ne 0$ and $a\ne 1$.