Determinant of a $n\times n$ matrix in terms of $n-1\times n-1$ matrix

Question

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 $n=3$.

Answer 1

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 $|A'| = -1 \neq 0 = |B'|$.

Answer 2

In general no. You need to know the values of the entries of the extra row and column. Note that if it were possible, one could iterate the process and express the determinant of any matrix in terms of the value of its corner entry (say first row, first column).

Edit. As Rob pointed out in the comments, you need to know the internals of $A$ too.

Answer 3

No-ish, in that there exists a formula for computing the determinant perturbatively $$\det\left(\mathbf{A} + \mathbf{uv}^\textsf{T}\right) = \det\left(\mathbf{A}\right) + \mathbf{v}^\textsf{T}\mathrm{adj}\left(\mathbf{A}\right)\mathbf{u}$$ but the perturbing term $\mathbf{v}^\textsf{T}\mathrm{adj}\left(\mathbf{A}\right)\mathbf{u}$ is not solely a function of the determinant $\det\left(\mathbf{A}\right)$.

Answer 4

What connects the determinant of a matrix and its leading principal minor is Schur complement. In general, if a leading principal submatrix $A$ of a partitioned matrix $M$ is nonsingular, we have $$ \det\underbrace{\pmatrix{A&B\\ C&D}}_M=\det(A)\det(D-CA^{-1}B), $$ where $S=D-CA^{-1}B$ is called the Schur complement of $A$ in the partitioned matrix $M$.