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Let a and B be matrices of order n, ab = a + B. It is proved that: (1) A-E and B-E are invertible; (2) AB = ba
It is proved that: (1) because (A-E) (B-E) = ab - (a + b) + e = e, A-E and B-E are reversible. (2) from (1) we know that E = (a − E) (B − E) & nbsp; & nbsp; = (B − E) (a − E) & nbsp; & nbsp; = BA − (a + b) + e, so AB = a + B = ba
Let a and B be square matrices of order n and E be identity matrices. It is proved that if E-Ab is invertible, then e-Ba is also invertible, and the inverse of e-Ba is obtained
(E-Ab) a = a-aba = a (e-Ba) = > A = (E-Ab) ^ (- 1) a (e-Ba) e = e-Ba + Ba = e-Ba + B (E-Ab) ^ (- 1) a (e-Ba) = (E + B (E-Ab) ^ (- 1) a) (e-Ba) so e-Ba is reversible, and (e-Ba) ^ (- 1) = e + B (E-Ab) ^ (- 1) a
Determinant order of matrix ab | ab | = | BA |, right or wrong, why
Yes, | ab | = | a |, B | = | B |, a | = | BA |, matrix multiplication has no commutative law, while number multiplication has commutative law
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