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Please, help solve this question:

Given the partitioned matrix \begin{equation} P=\left( \begin{array} {c,c} A \quad B \\ C \quad 0 \end{array} \right) \end{equation} where A is a 2x2 block matrix, B is a 2x1 matrix, C is a 1x2 matrix and 0 is a 1x1 zero matrix, such that \begin{equation} A=\left( \begin{array} {c,c,c} A_{11} \quad A_{12}\quad B_{11} \\ A_{21} \quad A_{22} \quad B_{21}\\ C_{11} \quad \, C_{12} \quad \, 0 \end{array} \right) \end{equation} Additionally, $A_{11}, A_{12}, A_{21}$ and $A_{22}$ themselves are 2x2, 2x3, 3x2 and 3x3 matrices, respectively. So, we are dealing with matrices of matrices of matrices. Now, I know that $Det(P) = Det(A)Det(-C A^{-1} B)$ but I have difficulty in manipulating the block matrix multiplication, $C A^{-1} B$. Please, how do I solve it? Thank you.

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  • $\begingroup$ I'm not sure what you're trying to solve. Note however that $CA^{-1}B$ is a scalar, so you don't need the determinant there. $\endgroup$ – JPi Feb 26 '14 at 14:30
  • $\begingroup$ Are the matrices $C$, $C_{11}$ and $C_{12}$ related in any way? If this is the case, decomposing $A^{-1}$ just like $A$ and looking at $AA^{-1}=I=A^{-1}A$ (the last column/row, to be precise) should give some insight. $\endgroup$ – Roland Feb 26 '14 at 14:35
  • $\begingroup$ Don't get too confused because of the matrixception: It's still a matrix of numbers; you just give it some structure. $\endgroup$ – Roland Feb 26 '14 at 14:40
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Simply take determinants on both sides of the identity

$$ \left( \begin{array} {c,c} A \quad 0 \\ C \quad I \end{array} \right) \left( \begin{array} {cc} I & A^{-1}B \\ 0 & -CA^{-1}B \end{array} \right)= \left( \begin{array} {cc} A \quad B \\ C \quad 0 \end{array} \right) $$

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