Find the determinant of the following 4x4 matrix: \[A = \begin{pmatrix} 3 & 2 & 1 & 4 \\ 0 & 1 & 0 & 2 \\ 1 & 0 & 2 & 1 \\ 0 & 1 & 0 & 1 \end{pmatrix}\]

Step 1 ：Firstly, let's expand the determinant along the first row. Therefore, we have: \[\text{det}(A) = 3\text{det}(A_1) - 2\text{det}(A_2) + \text{det}(A_3) + 4\text{det}(A_4)\] where \(A_1\), \(A_2\), \(A_3\), and \(A_4\) are the 3x3 matrices that result after removing the first row and the respective columns from the original matrix.

Step 2 ：We obtain: \[A_1 = \begin{pmatrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 1 & 0 & 1 \end{pmatrix}, A_2 = \begin{pmatrix} 0 & 0 & 2 \\ 1 & 2 & 1 \\ 0 & 0 & 1 \end{pmatrix}, A_3 = \begin{pmatrix} 0 & 1 & 2 \\ 1 & 0 & 1 \\ 0 & 1 & 1 \end{pmatrix}, A_4 = \begin{pmatrix} 0 & 1 & 0 \\ 1 & 0 & 2 \\ 0 & 1 & 0 \end{pmatrix}\]

Step 3 ：Next, we calculate the determinants for each of these matrices. The determinant of a 3x3 matrix can be computed by using the formula: \[\text{det}\begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix} = aei + bfg + cdh - ceg - bdi - afh\]

Step 4 ：Therefore, we have: \[\text{det}(A_1) = 1(2*1 - 0*1) - 0(0*1 - 2*1) + 2(1*0 - 0*1) = 2\]

Step 5 ：\[\text{det}(A_2) = 0(2*1 - 0*1) - 0(0*1 - 2*1) + 2(1*0 - 0*1) = 0\]

Step 6 ：\[\text{det}(A_3) = 0(0*1 - 2*1) - 1(1*1 - 1*1) + 2(1*1 - 0*1) = 1\]

Step 7 ：\[\text{det}(A_4) = 0(0*0 - 2*0) - 1(1*0 - 2*0) + 0(1*1 - 0*1) = 0\]

Step 8 ：Finally, substitute these values into the equation from step 1: \[\text{det}(A) = 3*2 - 2*0 + 1 + 4*0 = 6 + 1 = 7\]