Determinant of a $n\times n$ matrix in terms of $n-1\times n-1$ matrix Suppose we know  the determinant of matrix $A=(a_{ij})_{i,j=1}^{n-1}$. Can we express determinant of matrix $A'=(a_{ij})_{i,j=1}^{n}$ in terms of the determinant of matrix $A$?
We see that the matrix $A'$ differs from matrix $A$ only in the extra right-most column and bottom row. As a starting point, we can think of case $n=3$.

No, e.g., consider:
$$\begin{array}{ccc}
A = \left(\matrix{0 & 0 \\ 0 & 1}\right) & \quad\quad & B = \left(\matrix{1 & 0 \\ 0 & 0 }\right) \\
A' = \left(\matrix{0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 1}\right) & \quad\quad & B' = \left(\matrix{1 & 0 & 1 \\ 0 & 0 & 0 \\ 1 & 0 & 1}\right)
\end{array}$$
$A$ and $B$ have the same determinant (namely $0$) and $A'$ and $B'$ are obtained from them by adding the same third row and column, but $|A'| = -1 \neq 0 = |B'|$.
