Before you can read the parent note on cofactor expansion, you must own every symbol it throws at you. Below, each idea is built on the previous one — read top to bottom, skip nothing.
Picture it as a spreadsheet: number the rows top-to-bottom 1,2,…,n and the columns left-to-right 1,2,…,n. Figure s01 labels these row and column indices on our example matrix.
Why the topic needs it: the determinant is only defined for square matrices, and the general formulas run over 1,2,…,n — so you must know what n counts before any sum makes sense.
Read it like a street address: a23 means "go to row 2, then column 3". In the matrix above, a23=5 and a31=1.
Why the topic needs it: every formula in the parent note multiplies specific entries aij; you must be able to pick the exact number out of the grid instantly.
We have not yet said how to compute detA for a big matrix — that is the whole point of the parent note. Here we are only fixing what the symbol means so later formulas like detA=… read as "the number detA equals …". The upcoming sections (§4, §5) give the actual recipes for the small cases.
Why the topic needs it: the parent note opens straight into "detA=∑…" — you cannot read that line unless det already means something to you.
Why the topic needs it: the expansion formula detA=∑j=1naijCij is nothing but "walk along a line and add up the pieces". Without ∑ you can't read it.
Everything the parent note does recurses down until the leftover matrices are tiny. So we must nail the smallest cases where det is defined outright, no recipe needed.
Why the topic needs it: cofactor expansion turns a 3×3 into 2×2 pieces, and a 4×4 into 3×3 into 2×2 into 1×1. These two tiny cases are the floor of the recursion — the recipe stops here because these are defined directly.
Why the topic needs it: the minor is the smaller determinant each step of the expansion produces. It is how the recursion physically happens — a 3×3 minor-hunt gives 2×2 leftovers, which give 1×1 leftovers.
Why does raising −1 to a power give this? Because −1 multiplied by itself an even number of times returns to +1, and an odd number of times lands on −1. So the parity (even-or-odd-ness) of i+j is all that matters.
Why the topic needs it: without this sign the smaller determinants would add up to the wrong number. The sign encodes a deeper fact (permutation parity, from the Leibniz formula for determinants) — but for computing, it is simply the checkerboard.
First fix one particular row, call its number i (any single value from 1 to n that you choose). Then let j sweep across that row. Now every piece of the parent formula is defined:
detA=§4 add upj=1∑n§2 entryaij§7 sign(−1)i+j§6 minorMij(row i fixed)
Read it aloud: "Pick one row i. Walk along that row; for each entry, multiply by its checkerboard sign and its minor; add them all."