前言
本篇補充關於方陣的兩個定理及證明
關於方陣的一個重要定理
If A is an n-by-n matrix, then A is invertible if and only if AX= b has exactly one solution. A^-1 · b for every n-by-1 matrix b.
proof:
因為需要證明 if and only if,即證明充要條件，所以需要雙向證明。
由前向後證明(方陣A可逆，則方程AX= b對於每一個n-by-1的矩陣b只有唯一一解:A^-1 · b)
let K be the solution set of AX=b then
∀ x ∈ K, Ax=b
=> A^-1 ·(A·x) = A^-1 · b
=>(A^-1 ·A) ·x = A^-1 · b
=> I · x = A^-1 · b
=> x = A^-1 · b
i.e. K = {A^-1 · b}
證明思路:對方程引入可逆的特性（A的逆矩陣A-1），再根據矩陣乘法結合律得出K中只有唯一解A-1 · b。
由後往前證明(若方程AX= b對於每一個n-by-1的矩陣b只有唯一一解:A^-1 · b，則方陣A可逆)
∵ AX = b has exactly one solution for every n-by-1 matrix b
let xj be the solution of the linear system AX = b for b = ej( 0<= j <=n )
i.e. A·x1 = e1, A·x2 = e2, A·x3 = e3
let B = [x1 x2 ... xn] then
A·B = A·[x1 x2 ... xn] = [Ax1 Ax2 ... Axn] = [e1 e2... en] = In
by theorem, A is invertible
證明思路：將n個n-by-1的column vector拼湊成一個單位矩陣In，結合可逆矩陣的定義得證A可逆。（**）
等size方陣乘積矩陣可逆則分開兩者皆可逆，兩者皆可逆則其乘積矩陣可逆
If A,B are square matrices of the same size, then (AB) is invertible if and only if both A and B are invertible.
proof:
從前向後（等size方陣乘積矩陣可逆則分開兩者皆可逆）
let C be the inverse of (AB) i.e. C = (AB)^-1
∴ I = (AB) · C = A (BC)
i.e. I = A(BC)
=> A^-1 = (BC)
=> A is invertible
and I = C(AB) = (CA)B
i.e. I = (CA)B
=> B^-1 = CA
B is invertible
證明思路：通過可逆特性引入AB的逆矩陣C，接着應用矩陣乘法的結合律構造出A和B分別的逆矩陣，而得證。
從後向前（兩等size方陣皆可逆則其乘積矩陣可逆）
∵A,B are invertible
=> A^-1 , B^-1 exist
then (AB)(B^-1 · A^-1) = A(B ·B^-1)· A^-1 = A · I · A^-1 = A · A^-1 = I
i.e. (AB)^-1 = B^-1 · A^-1, AB is invertible.
證明思路：證明矩陣AB可逆即證明(AB)(B^-1 · A^-1) = I，結合A，B皆可逆的特性和矩陣乘法結合律可以導出，即得證。