matrix inverses

row reduction method to find A^-1

write matrix on one side and identity matrix on other, then put matrix into rref and new matrix on the right is the inverse

inverse method to solve linear system

Ax→ = b→ has a unique solution given by x→ = A^-1b→

if A and B are invertible then so is AB and (AB)^-1 =

B^-1A^-1

if A_1,A_2,…,A_k are all invertible then A₁A₂…A_k is invertible and

(A₁A₂…A_k)^-1 =

A_k^-1A_k-1^-1…A₂^-1A₁^-1

A is invertible with inverse B if

AB = I_n = BA

determinant of A (2 × 2)

ad - bc

if det(A) ≠ 0, then

A is invertible and A^-1 = 1/det(a) * [d -b]

                                                       [-c a]

if A is invertible then [A^k]^-1 =

[A^-1]^k

if A is invertible and c is a nonzero scalar, then (cA)^-1

(1/c)A^-1

[A^T]^-1 =

[A^-1]^T

invertible matrix theorem, where all of the statements are equivalent (all true or all false) and A is n x n

A is invertible, homogenous system Ax→ = 0→ has a unique solution x→ = 0→, A is row equivalent to I, A can be written as a product of elementary matrices, for every b→ ∈ Rⁿ, Ax→=b→ is consistent, rank (A) = n, there exists B n x n such that AB = I and C such that CA = I