How to determine is a matrix is invertible? $$      M=
        \left[
        \begin{array}
        -1 & k & 3 & -2\\
        2 & 1 & 1 & k \\
        1 & k+1 & 4 & k-2\\
        2 & 1 & 1 & k+1\\
        \end{array}
        \right]
$$
What must $k$ be if the matrix $M$ is invertible ?
However what I reached after some trying is that M is not invertible for any values of $k$ because after some basic row operations I get a zero row and hence $\det(M)= 0$ and hence M is not invertible.
Is there any value for M to be invertible or not ?
 A: There must be an error in your calculation of determinant.  what you should get is $$det(M)=6-2k$$
and hence $M$ is invertible for all $k\neq 3$.
A: Here is an ad-hoc reasoning that can be done in your head, rather than compute the determinant:
$$M= \begin{bmatrix}
        1 & k & 3 & -2\\
        2 & 1 & 1 & k \\
        1 & k+1 & 4 & k-2\\
        2 & 1 & 1 & k+1 
\end{bmatrix}$$
The last column is the only column where entry 2 and 4 are different (and these entries are different no matter what $k$ is). So if we're trying to construct an arbitrary column as a linear combination of column, the coefficient of the last common is given by the difference between the second and fourth entry of our target.
Whether we reach our goal then depends on whether we can get the first three entries to be what we want by a linear combination of the first three column, that is, whether
$$\begin{bmatrix}
        1 & k & 3 \\
        2 & 1 & 1 \\
        1 & k+1 & 4 
\end{bmatrix}$$
is invertible.
Subtracting the third column from the second gives
$$\begin{bmatrix}
        1 & k-3 & 3 \\
        2 & 0 & 1 \\
        1 & k-3 & 4 
\end{bmatrix}$$
and now the third column is the only column where entry 1 and 3 are different. With exactly the same reasoning as before the $3\times 3$ matrix is invertible if
$$\begin{bmatrix}
        1 & k-3  \\
        2 & 0 
\end{bmatrix}$$
is -- and that is clearly exactly when $k\ne 3$.
