<s>[INST] 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?
 [/INST] $A \ =\begin{pmatrix}
                    7 & 1 & 3 & -2\\
                   -2 & 1 & -12 & -1 \\
                    1 & 16 & -4 & a \\
                    2 & 4 & 2 & 2 \\
                   \end{pmatrix}\cong \begin{pmatrix}
                    13 & 1 & 5 & -5\\
                   -5 & 1 & -25 & -3 \\
                    -14 & 16 & -24 & 2a-16 \\
                  0 & 4 & 0 & 0 \\
                   \end{pmatrix}$  
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?
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