Find the values of 'a' in a $4\times 4$ matrix(A) when the determinant is less than 2012 The matrix is $A \ =\begin{pmatrix}
                    7 & 1 & 3 & -2\\
                   -2 & 1 & -12 & -1 \\
                    1 & 16 & -4 & a \\
                    2 & 4 & 2 & 2 \\
                   \end{pmatrix}$
Where $\det(A)\lt  2012$
I have tried reducing the first columns to 0's to reduce it to a $3\times3$ determinant. When I do that it gets messy fast. Is there a special method to solve this problem?
 A: $A \ =\begin{pmatrix}
                    7 & 1 & 3 & -2\\
                   -2 & 1 & -12 & -1 \\
                    1 & 16 & -4 & a \\
                    2 & 4 & 2 & 2 \\
                   \end{pmatrix}\cong \begin{pmatrix}
                    13 & 1 & 5 & -5\\
                   -5 & 1 & -25 & -3 \\
                    -14 & 16 & -24 & 2a-16 \\
                  0 & 4 & 0 & 0 \\
                   \end{pmatrix}$  
S0, you would get $\det(A)=-4\det\begin{pmatrix}
                    13 & 5 & -5\\
                   -5 & -25 & -3 \\
                    -14  & -24 & 2a-16 \\
                  \end{pmatrix}$
Can you finish this and fill the gaps?
A: Not sure if there is any special trick, but you can kill off the $(1,4)$-th and $(4,4)$-th elements of $A$ easily using the second row. This leaves you two $3\times3$ determinants to compute, if you Laplace-expand $\det A$ along the last column.
A: You can simplify this matrix and make the determinant less tedious by performing row operations and introducing some zeros:  


*

*$R_4 \rightarrow R_1 + R_4  $

*$ R_1 \rightarrow R_1 -2R_2$
The above two operations gives the following matrix:
$$ \begin{pmatrix} 
11 & -1 & 27 & 0 \\
-2 & 1 & -12 & -1\\
1 & 16 & -4 & a\\
9 & 5 & 5 & 0
\end{pmatrix}$$


*Expand along column 4:


$$ -1 \det\begin{bmatrix}
11 & -1 & 27\\
1 & 16 & -4\\
9 & 5 & 5
\end{bmatrix} - a \det\begin{bmatrix}
11 & -1 & 27\\
-2 & 1 & -12\\
9 & 5 & 5
\end{bmatrix} $$
After computing, you get the following:
$$2612-300a < 2012$$
$$  a > 2$$
when $a = 2$, $\det A = 2012$
