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Based on how a system must work I noticed the following simplification of matrix inversion for a given pattern of zeros in rows and columns and I would like to know if there is general proof that says this is always true. Let me show an example of the simplification.

A =
[  a1,  a2,   0,   0,  a5]
[  a6,  a7,   0,   0, a10]
[   0,   0, a13, a14,   0]
[   0,   0, a17, a18,   0]
[ a21, a22,   0,   0, a25]

This matrix has zeros in third and fourth row and columns except for the elements within the cross of those zeros. This matrix can represent a system of which variables 3 and 4 are independent of the others. Therefore when one inverts B to solve such a system one must preserve that variables 3 and 4 are still independent. Therefore it follows that the zeros that exist in B will preserve in B^-1 in the same position. Moreover, one can perform the matrix inversion by splitting in two matrices such as:

A1 =
[  a1,  a2,  a5]
[  a6,  a7, a10]
[ a21, a22, a25]

A2 =
[ a13, a14]
[ a17, a18]

And then just inverting them independently:

A1^-1 = 
[  b1,  b2,  b5]
[  b6,  b7, b10]
[ b21, b22, b25]

A2^-1 = 
[ b13, b14]
[ b17, b18]

The original matrix inversion is formed by combining them:

A^-1 =
[  b1,  b2,   0,   0,  b5]
[  b6,  b7,   0,   0, b10]
[   0,   0, b13, b14,   0]
[   0,   0, b17, b18,   0]
[ b21, b22,   0,   0, b25]

My point being is that this property holds for any other combinations where the zeros make this happen (for example a 10x10 matrix with 4 other variables being independent of which 2 are together and 2 alone, etc.)

Is there a known proof somewhere for this property of matrix inversion?

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What you have is essentially a Block Diagonal matrix in disguise.

You can reorder the rows to bring the fifth row up two rows, and reorder the columns in the same way (fifth column two steps to the left). The result is a similar matrix (it has merely been transformed by a permutation matrix) but it is now in block diagonal form. For these it is very straightforward to prove that its inverse is the block diagonal matrix consisting of the inverses of the blocks. You can then undo the permutation transformation.

The only thing left to prove is that the inverse of a transform is the transform of the inverse, i.e. $({P^{-1}AP})^{-1} = P^{-1}A^{-1}P$, but that is almost trivial.

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