asked in 教育與參考考試 · 1 decade ago



give reasons or counterexamples


Let A be a n*n matrix. If the column of A are linearly independent,then

Ax=b has exactly one solution for each vector b

(2)If the square matrices A and B are invertible,then the product AB is


1 Answer

  • Anonymous
    1 decade ago
    Favorite Answer

    (1) True.

    There is a theorem says that

    "The column vectors of an n-by-n matrix A are linearly independent iff Ax=b has exactly one solution for every column vector b."

    You can prove it in this way:

    Suppose A = [c_1 | c_2 | ... | c_n]. Since c_i's are linearly independent, the equation

    c_1 x_1 + c_2 x_2 + ... + c_n x_n = 0 (*)

    has only the trivial solution x_1 = x_2 = ... = x_n.

    But the equation can be rewritten as Ax=0, where x = transpose of [x_1, x_2, ... . x_n]. By above we know Ax=0 has only the zero solution x = transpose of [0, 0, ... , 0]. Thus A is invertible. And so Ax=b has exactly one solution for any b. QED

    (2) True.

    Since A and B are invertible, we let the inverses be A' and B', respectively. Then we can see


    Hence AB is invertible; besieds, the inverse is B'A'. QED

    2008-03-22 15:54:14 補充:

    這些都是很基本的定理, 任何一本初等線性代數的書都應該可以找到.

    Source(s): myself
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