Find the values of 'a' in a $4\times 4$ matrix(A) when the determinant is less than 2012

Question

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?

Answer 1

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.

Answer 2

$A \ =\begin{pmatrix} 7 & 1 & 3 & -2\\ -2 & 1 & -12 & -1 \\ 1 & 16 & -4 & a \\ 2 & 4 & 2 & 2 \\ \end{pmatrix}\cong \begin{pmatrix} 13 & 1 & 5 & -5\\ -5 & 1 & -25 & -3 \\ -14 & 16 & -24 & 2a-16 \\ 0 & 4 & 0 & 0 \\ \end{pmatrix}$

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?

Answer 3

You can simplify this matrix and make the determinant less tedious by performing row operations and introducing some zeros:

1. $R_4 \rightarrow R_1 + R_4 $

2. $ R_1 \rightarrow R_1 -2R_2$

The above two operations gives the following matrix:

$$ \begin{pmatrix} 11 & -1 & 27 & 0 \\ -2 & 1 & -12 & -1\\ 1 & 16 & -4 & a\\ 9 & 5 & 5 & 0 \end{pmatrix}$$

3. 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 $a = 2$, $\det A = 2012$