Determinant and symmetric positive definite matrix

Question

Let $\mathbf{A}$ be an $N \times N$ real symmetric (or complex Hermitian) positive definite matrix, such that $\mathrm{det}(\mathbf{A})=1$.

Is it possible to recover the first top-left element $[\mathbf{A}]_{1,1}$ from all the other elements, i.e. $[\mathbf{A}]_{1,1} = f([\mathbf{A}]_{2,1},\ldots,[\mathbf{A}]_{N,N})?$ Which is the function $f$ that does the job?

Thanks!

Answer 1

Expand the determinant along the top row:

$$1 = \det(A) = A_{11} M_{11} - A_{12} M_{12} + \ldots + (-1)^{N+1} A_{1N} M_{1N} $$

where $M_{ij}$ is the determinant of the matrix obtained by removing row $i$ and column $j$. Note that $M_{11} > 0$ because the matrix is positive definite. Solve for $A_{11}$.

Answer 2

You can develop the determinant along the first line $$\det(A) = \sum_{i=1}^N a_{1i}C_{1j}$$ where $C_{ij}$ are the cofactors thus : $$a_{11} = \frac{1}{C_{11}}\left(\det A - \sum_{i=2}^N a_{1i}C_{1j}\right)$$