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# 數學問題？

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• 2 decades ago

Chapter 1 高斯消去法(Haussiam Elimination)

Matrix mode : Ax=b

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

下列條件互為等價：

A-1 exists

det(A) 0

The row vector of A are l.I ( Linear Indep. )

The column vector of A are l.I

The RREF of A is the Identity matrix

A is full rank

Ax=0 x=0 ( There is no nozero x such that Ax=0 )

<PS>

The "pivots" are the first nonzero entries in these rows.

The number of pivots is the "rank".

三角矩陣

下三角矩陣：

Def : lij=0 whenever i < j

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

Ux=y , LUx=b Ly=b

先解 y

再解 Ux = y 的 x

Forward Substitution

aij , bi are given.

演算法：

For i=1 : n

if lii=0 exit

for j=1 ...i-1

end

<PS>

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