Proof of rank of linear algebraic matrix and adjoint matrix At (3), I really don't understand why when R (a) & lt; n-1, all the N-1 order subformulas of a are 0? Why (3) step, R (a)

Theorem: R (a) = R A has non-zero r-order subformulas, and all R + 1-order subformulas are 0
If a has a subexpression of order n-1 which is not equal to 0, then the rank of a is at least n-1

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