NotesMITLinearAlgebra

A* B

a11   a12    a13    a14               

a21   a22 .................

..................................      *   

.............................a44

A: m * n, B: n * p

Cij= SUM(Aik*Bkj), k -> [1,n];

matix* several column vector , the corresponding 

severalrow vector * matrix.

m*1* 1*p

A*B= sum ( columns of A * rows of B )

[A11,A12]    * [ B11, B12 ]    = [ A11*B11 + A12*B21,A11*B12+A12*B22 ]

[A21,A22]       [ B21, B22 ]      [ A21*B11+ A22*B22, A21*B12+A22*B22 ]

Ifthere exists A‘ of A . that A‘ * A = I ( we call it invertible,non-singular.) , the A‘ is the inverse of A.

andif A is a square matrix A‘*A = I = A*A‘;

[ 1 3 ; 2 6 ];

1.           the linear combination of columns of A. and all lies on (1,2) socant combine(1,0)

2.           there is a lema? say that Ax = 0 and x != 0. then A has noinverse.

Intuition , that says that some/all columns of Acould form zero vector. So that they are linear dependent ? means lies onthe same direction.

Sothere will get no A‘ in this case.

Howcan we get the inverse of A?

Gauss- Jordan ( solve 2 equations at once )

13

27* [ a ; b ] = [ 1 ; 0 ]

13

27 * [ c; d ] = [ 0 ; 1 ]

andhow can we solve it , use following ideas.

AI --> I A^-1

look more detail for 3.乘法和逆矩阵.mp4later.