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? asked in 教育與參考其他:教育 · 2 decades ago



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  • 2 decades ago
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    Chapter 1 高斯消去法(Haussiam Elimination)

    Matrix mode : Ax=b

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


    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 )


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

    The number of pivots is the "rank".



    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.


    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

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