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The problem:

A block diagonal matrix is a square matrix where nonzero element occurs in blocks along the diagonal. an example of a 4x4 block diagonal matrix with two 2x2 blocks is

$$ A_{} = \begin{pmatrix} {1} & {2} & {0} & {0} \\ {3} & {4} & {0} & {0} \\ {0} & {0} & {5} & {6} \\ {0} & {0} & {7} & {8} \\ \end{pmatrix} $$

Calculate the determinant of the matrix by hand using cofactor expansion along the first row

I'am confusing with all the zeros in the matrix, and using cofactor expansion along the first row? Could someone explain how to solve this kind of problem?

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Zeros are a good thing, as they mean there is no contribution from the cofactor there.

$$ \det A = 1 \cdot (-1)^{1 + 1} \det S_{11} + 2 \cdot (-1)^{1+2} \det S_{12} + 0 \cdot \dotsb + 0 \cdot \dotsb $$ with $$ S_{11} = \begin{pmatrix} \times & \times & \times & \times \\ \times & 4 & 0 & 0 \\ \times & 0 & 5 & 6 \\ \times & 0 & 7 & 8 \end{pmatrix} = \begin{pmatrix} 4 & 0 & 0 \\ 0 & 5 & 6 \\ 0 & 7 & 8 \end{pmatrix} \\ S_{12} = \begin{pmatrix} \times & \times & \times & \times \\ 3 & \times & 0 & 0 \\ 0 & \times & 5 & 6 \\ 0 & \times & 7 & 8 \end{pmatrix} = \begin{pmatrix} 3 & 0 & 0 \\ 0 & 5 & 6 \\ 0 & 7 & 8 \end{pmatrix} $$ where $S_{ij}$ is the matrix $A$ with row $i$ and column $j$ removed.

The determinants of $S_{11}$ and $S_{12}$ are then calculated again by expansion along the first row, e.g. $$ \det S_{11} = 4 \cdot (-1)^{1+1} \det \begin{pmatrix} \times & \times & \times \\ \times & 5 & 6 \\ \times & 7 & 8 \end{pmatrix} = 4 \det \begin{pmatrix} 5 & 6 \\ 7 & 8 \end{pmatrix} \\ \det S_{12} = 3 \cdot (-1)^{1+1} \det \begin{pmatrix} \times & \times & \times \\ \times & 5 & 6 \\ \times & 7 & 8 \end{pmatrix} = 3 \det \begin{pmatrix} 5 & 6 \\ 7 & 8 \end{pmatrix} $$ until one hits a $2\times2$ matrix where one knows the direct formula.

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  • $\begingroup$ Hmm I'am a bit confused where S11 and S12 comes from? $\endgroup$ – AdiT Jun 2 '16 at 21:30
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    $\begingroup$ I hope this version is easier to understand. $\endgroup$ – mvw Jun 2 '16 at 21:36

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