Determine the entry {\color{red}b} in
\begin{pmatrix}
A11 & A12 & A13 & A14 \\
A21 & A22 & A23 & A24 \\
A31 & A32 & {\color{red}b} & A34 \\
A41 & A42 & A43 & A44
\end{pmatrix},
such that the matrix has eigenvalues
\color{blue} \lambda_1 = L1, \color{blue} \lambda_2 = L2
\color{blue} \lambda_3 = L3 and \color{blue} \lambda_4 = L4.
\color{red} b
=
A33
We can use that the sum of the eigenvalues is the trace of the matrix.
Using this observation, we need to find {\color{red}b} satisfying
negParens(A11) +negParens(A22) +
{\color{red}b} + negParens(A44) =
{\color{blue}negParens(L1) + negParens(L2) +negParens(L3) + negParens(L4) } =
{\color{blue}L1 + L2 + L3 + L4}.
Solving for {\color{red}b} yields
{\color{red}b} = A33.