※ 引述《rr8r8r8r8tw (amp)》之銘言：
: 幾題證明實在不知如何下手
: http://i754.photobucket.com/albums/xx185/rpd11140/DSC_0001.jpg
Sol:
A = [a b] b = [e] v = [x]
[c d] [f] [y]
-1
若 det(A) ≠ 0,則 v = A b
0
-1
若有另一解 v' => Av' = b => v' = A b => v = v' => 解是唯一的
0
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若解是唯一的, 則存在一矩陣K,使得 KA = U 且 Uv = Kb(U為上三角矩陣)
因為 det(U) ≠ 0 => det(K)det(A) ≠ 0 => det(K) ≠ 0 且
det(A) ≠ 0
(這個方向我不確定可不可以這樣寫,因為書籍引理的方向可能不同)
: http://i754.photobucket.com/albums/xx185/rpd11140/DSC_0002.jpg
39.
Sol:
|a11 a12 a13| |b11 a12 a13|
|a21 a22 a23| + |b21 a22 a33|
|a31 a32 a33| |b31 a32 a33|
(降階)
= a11 |a22 a23| - a21 |a12 a13| + a31 |a12 a13|
|a32 a33| |a32 a33| |a22 a23|
+ b11 |a22 a23| -b21 |a12 a13| + b31 |a12 a13|
|a32 a33| |a32 a33| |a22 a23|
= (a11 + b11) |a22 a23| -(a21 + b21)|a12 a13| + (a31 + b31) |a12 a13|
|a32 a33| |a32 a33| |a22 a23|
(合併)
=|a11 + b11 a12 a13|
|a21 + b21 a22 a23|
|a31 + b31 a32 a33|
40.
Sol:
|1+a 1 1 | | 1+a 1 1 |
| 1 1+b 1 | = | 1+0 1+b 1 |
| 1 1 1+c| | 1+0 1 1+c|
| 1 1 1 | | a 1 1 |
= | 1 1+b 1 | + | 0 1+b 1 | (利用39題性質)
| 1 1 1+c| | 0 1 1+c|
| 1 1 1 | | a 1 1 |
= | 0 b 0 | + | 0 1+b 1 | (行列式列運算)
| 0 0 c | | 0 1 1+c|
= bc + a[(1+b)(1+c)-1]
= bc + a(bc+b+c)
= bc + abc + ab + ac
= abc(1 + 1/a + 1/b + 1/c)
其實直接展開也行啦
: http://i754.photobucket.com/albums/xx185/rpd11140/DSC_0003.jpg
Sol:
| cos x 0 sin x |
| sin 0 -cos x |
| sin x - cos x 1 sin x + cos x|
= -| cos x sin x |
| sin x -cos x |
2 2 2 2
= -(- cos x - sin x) = cos x + sin x = 1 ≠ 0
=> x 無解
: http://i754.photobucket.com/albums/xx185/rpd11140/DSC_0005.jpg
39.
Sol:
因為 det(AB) = det(A)det(B) = det(I) = 1
=> det(A) ≠ 0 且 det(B) ≠ 0 (當然也可用簡單反證法論證)
40.
Sol:
因為 AB 是 singular => det(AB) = 0
又 det(AB) = det(A)det(B) = 0
=> det(A) = 0 or det(B) = 0
=> A is singular or B is singular.
: http://i754.photobucket.com/albums/xx185/rpd11140/DSC_0006.jpg
62. 2 2
Sol: A = A => det(A ) = det(A)
2
=> [det(A)] = det(A)
=> det(A) = 0 or det(A) = 1
63.
Sol:
因為 S is singular => det(S) = 0 => det(SB) = det(S)det(B) = 0
所以 SB is singular for all n x n matrix B.
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