Chapter 1 高斯消去法(Haussiam Elimination)

Matrix mode : Ax=b

If A-1 exist,then answer is x=A-1b .


  1. A-1 exists
  2. det(A) 0
  3. The row vector of A are l.I ( Linear Indep. )
  4. The column vector of A are l.I
  5. The RREF of A is the Identity matrix
  6. A is full rank
  7. Ax=0 x=0 ( There is no nozero x such that Ax=0 )


  1. The "pivots" are the first nonzero entries in these rows.
  2. The number of pivots is the "rank".


  1. 下三角矩陣:

    Def : lij=0  whenever   i < j

  2. 上三角矩陣 Def : uij=0  whenever   i > j

[Thm] 如果 G=(gij) 為三角矩陣, 則 det(G) = g11 . g22 ... gnn . [Cor] 設 G 為三角矩陣,若 G 為 nonsingular
gii 0  for  all  i=1 ...n.

[Cor] G 若為三角矩陣,Gx=b 有唯一解 G 的主對角線皆不為 0

A=LU L:lower U:upper

slove Ax=b is the same to LUx=b

  1. Ux=y , LUx=b Ly=b
  2. 先解 y
  3. 再解 Ux = yx

Forward Substitution

aij , bi  are  given.


For i=1 : n

if lii=0 exit
for j=1 ...i-1



上三角系統 Ux=y 解法也類似 Forward Substitution, 先解yn , 再解 yn-1 ... 此法稱為 "Backward Subsitution"。

Example :

Exercise : Prove det(A)=g11 g22