Good day to all!
I am calculating the determinant of the following $(K + 1) \times (K + 1)$ matrix? $$ {\bf A} = \left(\begin{matrix} a_{11} & a_{12} & 0 & \dots & 0 & 0 \\ a_{21} & a_{22} & a_{23} & \dots & 0 & 0 \\ \dots & \dots & \dots & \dots & \dots & \dots \\ a_{K1} & a_{K 2} & a_{K 3} & \dots & a_{K K} & c_{K K} \\ b_1 & b_2 & b_3 & \dots & b_K & b_{K + 1} \end{matrix}\right). $$
I have attempted to decompose it by the first row: $$ \det{\bf A} = \left(-1\right)^{1 + 1} a_{1 1} \cdot \det{\bf A}_{1, 1} + \left(-1\right)^{2 + 1} a_{1 2} \cdot \det{\bf A}_{2, 1}, $$ where ${\bf A}_{1, 1}$ is obtained from ${\bf A}$ by excluding its first column and first row: $$ {\bf A}_{1, 1} = \left(\begin{matrix} a_{22} & a_{23} & 0 & \dots & 0 & 0 \\ a_{32} & a_{33} & a_{34} & \dots & 0 & 0 \\ \dots & \dots & \dots & \dots & \dots & \dots \\ a_{K2} & a_{K 3} & a_{K 4} & \dots & a_{K K} & c_{K K} \\ b_2 & b_3 & b_4 & \dots & b_K & b_{K + 1} \end{matrix}\right). $$ ${\bf A}_{2, 1}$ is defined similarly.
${\bf A}_{1, 1}$ can be further decomposed, but at each step, I will have 2 non-zero elements at first row.
Is there a short way to calculate $\det{\bf A}$?