A ( B C ) = ( A B ) C A(BC)=(AB)C A(BC)=(AB)C

A ( B + C ) = A B + A C ( A + B ) C = A C + B C A(B+C) = AB + AC \\ (A+B)C = AC + BC A(B+C)=AB+AC(A+B)C=AC+BC

设 A A A 为方阵，如果存在方阵 A − 1 A^{-1} A−1 使得 A A − 1 = A − 1 A = I AA^{-1} = A^{-1}A = I AA−1=A−1A=I则方阵 A A A 可逆， A − 1 A^{-1} A−1 为 A A A 的逆矩阵。方阵的逆若存在，则是唯一的，即一个方阵不可能有两个或以上的逆。如果方阵 A A A 和 B B B 均可逆，则其乘积 A B AB AB 也可逆 ( A B ) − 1 =   B − 1 A − 1 ( A B ) − 1 ( A B ) =   B − 1 A − 1 A B =   B − 1 ( A − 1 A ) B （ 结 合 律 ） =   B − 1 I B =   I \begin{aligned} (AB)^{-1} =& \ B^{-1}A^{-1} \\ \\ (AB)^{-1}(AB) =& \ B^{-1}A^{-1}AB \\ =& \ B^{-1}(A^{-1}A)B （结合律）\\ =& \ B^{-1}IB \\ =& \ I \end{aligned} (AB)−1=(AB)−1(AB)====​ B−1A−1 B−1A−1AB B−1(A−1A)B（结合律） B−1IB I​三个（或多个）方阵乘积的逆 ( A B C ) − 1 =   C − 1 B − 1 A − 1 ( A 1 A 2 … A n ) − 1 =   A n − 1 … A 2 − 1 A 1 − 1 \begin{aligned} (ABC)^{-1} =& \ C^{-1}B^{-1}A^{-1} \\ (A_1A_2 \dots A_n)^{-1} =& \ A_n^{-1} \dots A_2^{-1}A_1^{-1} \end{aligned} (ABC)−1=(A1​A2​…An​)−1=​ C−1B−1A−1 An−1​…A2−1​A1−1​​ 若方阵 A A A 可逆，则方阵 A A A 满足消去律，即 A B = A C    ⟹    B = C AB = AC \implies B=C AB=AC⟹B=C证明 A B = A C    ⟹    A − 1 A B = A − 1 A C    ⟹    I B = I C    ⟹    B = C AB = AC \implies A^{-1}AB = A^{-1}AC \implies IB=IC \implies B = C AB=AC⟹A−1AB=A−1AC⟹IB=IC⟹B=C

矩阵相加的转置 ( A + B ) T = A T + B T (A + B)^T = A^T + B^T (A+B)T=AT+BT

矩阵乘积的转置 ( A B ) T =   B T A T ( A B C ) T =   C T B T A T \begin{aligned} (AB)^T =& \ B^TA^T \\ (ABC)^T =& \ C^TB^TA^T \end{aligned} (AB)T=(ABC)T=​ BTAT CTBTAT​ 矩阵的逆和转置操作可以交换，即 ( A T ) − 1 = ( A − 1 ) T (A^{T})^{-1}=(A^{-1})^T (AT)−1=(A−1)T证明： A T ( A − 1 ) T = ( A − 1 A ) T = I A^T(A^{-1})^T = (A^{-1}A)^T = I AT(A−1)T=(A−1A)T=I所以 ( A − 1 ) T (A^{-1})^T (A−1)T 即是 A T A^T AT 的逆，即 ( A T ) − 1 = ( A − 1 ) T (A^{T})^{-1}=(A^{-1})^T (AT)−1=(A−1)T