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# 線性代數考古問題

TRUE or FALSE

give reasons or counterexamples

(1)

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

invertible

### 1 Answer

- Anonymous1 decade agoFavorite 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

(B'A')(AB)=B'(A'A)B=B'B=I.

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