匿名使用者
匿名使用者 發問時間: 科學數學 · 1 0 年前

線性代數證明可逆矩陣(需要什麼技巧?)

有四個矩陣,分別為A: n x n, B: n x k, C: k x k, D: k x n,

其中n,k是正整數,假設(A + BCD)是可逆的,

證明: (A + BCD)-1 = A-1 - A-1B(C-1 + DA-1B)-1DA-1

註:A-1是A的反矩陣

我是直接拿(A + BCD)(A-1 - A-1B(C-1 + DA-1B)-1DA-1) 下去求I,

可是乘不出來,請問需要什麼技巧嗎?

已更新項目:

另外我想問:

假設:X: n x n, Y: n x n, Z: n x n, W: n x n,

那麼 XYZ - XWZ = X(Y - W)Z 是對的嗎?

其中 X(Y - W)Z 的 Y - W 在什麼情況下要變成 Y + W 還是根本不會有此情況?

2 個已更新項目:

I為單位矩陣,

I - XZ - XYZ = I - X(I - Y)Z 是對的嗎?

3 個已更新項目:

To 淡淡:

Thank for your respondence.

It seems that parts of my concepts are correct.

But I just can't get identity matrix I by direct product.

Could you show that (A + BCD)(A -1 - A -1 B(C -1 + DA -1 B) -1 DA -1 ) = ... = I in great detail?

4 個已更新項目:

這題目困擾我一段時間了,

做到一個地方都會差個,+,- 號,

難不成書本題目有印錯?

5 個已更新項目:

唉~~看這裡比較清楚我的問題:

http://xs.to/xs.php?h=xs134&d=08012&f=faq007674.jp...

6 個已更新項目:

難道這題目不能直接乘?

1 個解答

評分
  • 1 0 年前
    最佳解答

    First, (A + BCD)-1 = A-1 - A-1B(C-1 + DA-1B)-1DA-1

    is proved by direct product. Notice that : the multiplication is not commutative. I mean AxB not equals BxA . However, you still can product them tern by tern. Just computation problems.

    Second, Matrix multiplication satisfies "distribution laws" so, A(B+C) = AB + AC is always true. so, XYZ - XWZ = X(Y - W)Z is always true.

    Third, I - XZ - XYZ = I - X(I - Y)Z is always true.

    I think you can check any linear algebra book, the matrix multiplication satisfies "distribution law", and I think you need a good reference book for linear algebra. Ask your friends in mathematics department, you will know which books are good.

    you can check NTU math website.

    http://www.math.ntu.edu.tw/home_c.htm

    2008-12-31 05:02:29 補充:

    Come on man. You make a stupid and obviously wrong -A -B = - (A+B) not -(A-B) . The fifth row should be I - B (I + CDA^(-1)B) (C.....)

    ~

    So, your computation are totally right except here. ^_^

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