Regularity and invertibility of two parameterized matrices? 
$$
C=
\begin{bmatrix}
1+a & 2 & 3 & 4 &  5 \\
1 & 2+a & 3 & 4 &  5 \\
1 & 2 & 3+a & 4 &  5 \\
1 & 2 & 3 & 4+a &  5 \\
1 & 2 & 3 & 4 &  5+a \\
\end{bmatrix}
$$
For which values of $a$ is the matrix $C$ regular?



$$
A=
\begin{bmatrix}
3a & a \\
-a & 1 \\
\end{bmatrix}
$$
Determine for which values of $a$ is there an inverse $A^{-1}$ and then solve it?

Please help....
 A: Actually, there is an easier way to find the values for the matrix $C$ for which it is invertible:
Look at the following representation
$$
\begin{pmatrix}
1+a & 2 & 3 & 4 &  5 \\
1 & 2+a & 3 & 4 &  5 \\
1 & 2 & 3+a & 4 &  5 \\
1 & 2 & 3 & 4+a &  5 \\
1 & 2 & 3 & 4 &  5+a \\
\end{pmatrix}=
\begin{pmatrix}
1 & 2 & 3 & 4 &  5 \\
1 & 2 & 3 & 4 &  5 \\
1 & 2 & 3 & 4 &  5 \\
1 & 2 & 3 & 4 &  5 \\
1 & 2 & 3 & 4 &  5 \\
\end{pmatrix}+
a \operatorname{Id}_{5\times5}=B+a \operatorname{Id}_{5\times5}
$$
which is the eigenvalue problem for matrix B in $\lambda=-a$.
That for $\lambda=0$ we have an eigenvalue is clear, the dimension of the eigenspace is $4$, which is clear because with Gauss we get
$$
\begin{pmatrix}
1 & 2 & 3 & 4 &  5 \\
1 & 2 & 3 & 4 &  5 \\
1 & 2 & 3 & 4 &  5 \\
1 & 2 & 3 & 4 &  5 \\
1 & 2 & 3 & 4 &  5 \\
\end{pmatrix}\to
\begin{pmatrix}
1 & 0 & 0 & 0 &  0 \\
1 & 0 & 0 & 0 &  0 \\
1 & 0 & 0 & 0 &  0 \\
1 & 0 & 0 & 0 &  0 \\
1 & 0 & 0 & 0 &  0 
\end{pmatrix}\
$$
so in fact we are looking for one other eigenvalue $\lambda_2$ with eigenspace dimension $1$, so that everything adds up to $5$ eventually.
If we look hard enough, we can easily figure out it has to be $\lambda=15\Leftrightarrow a=-15$. For this to be seen, observe that
$$
\begin{pmatrix}
1+a \\
1  \\
1  \\
1  \\
1  
\end{pmatrix}+
\begin{pmatrix}
2 \\
2+a  \\
2  \\
2  \\
2  
\end{pmatrix}+
\begin{pmatrix}
3 \\
3  \\
3+a  \\
3  \\
3
\end{pmatrix}+
\begin{pmatrix}
4 \\
4  \\
4  \\
4+a  \\
4 
\end{pmatrix}=
\begin{pmatrix}
10+a \\
10+a  \\
10+a  \\
10+a \\
10 
\end{pmatrix}\equiv(-1)\begin{pmatrix}
5 \\
5  \\
5  \\
5 \\
5+a 
\end{pmatrix}
$$
which is true for $a=-15$. Since now we have 5 dimensional eigenspace we are finished and conclude that for $a_1=0$ and $a_2=-15$ the matrix $C$ is not invertible.
For the second part just use Cramer's rule which will give you
$$
A^{-1}=\frac 1{a(3+a)}\begin{pmatrix} 1 & -a \\ a & 3a \end{pmatrix}
$$ 
which indeed is defined for all $a\in\mathbb{R}\backslash\{0,-3\}$
