Chủ đề này có 4 trả lời

[vo van duc]

Cho ma trận $A=\begin{pmatrix} a\cos ^{2}t+b\sin^{2}t & (b-a)\sin t\cos t\\ (b-a)\sin t\cos t & a\sin ^{2}t+b\cos ^{2}t \end{pmatrix}$

Tính $A^{2012}$

[vo van duc]

Mới giải. Mọi người kiểm tra giúp tôi với nha!
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$A=\begin{pmatrix} a\cos ^{2}t+b\sin^{2}t & (b-a)\sin t\cos t\\ (b-a)\sin t\cos t & a\sin ^{2}t+b\cos ^{2}t \end{pmatrix}$

$=\begin{pmatrix} a(1-\sin ^{2}t)+b\sin ^{2}t & (b-a)\sin t\cos t\\ (b-a)\sin t\cos t & a(1-\cos ^{2}t)+b\cos ^{2}t \end{pmatrix}$

$=\begin{pmatrix} (b-a)\sin ^{2}t+a & (b-a)\sin t\cos t\\ (b-a)\sin t\cos t & (b-a)\cos ^{2}t+a \end{pmatrix}$

$=\begin{pmatrix} \frac{b-a}{2}(1-\cos 2t)+a & (b-a)\sin t\cos t\\ (b-a)\sin t\cos t & \frac{b-a}{2}(1+\cos 2t)+a \end{pmatrix}$

$=\begin{pmatrix} \frac{b-a}{2}(-\cos 2t)+\frac{a+b}{2} & \frac{b-a}{2}\sin 2t\\ \frac{b-a}{2}\sin 2t & \frac{b-a}{2}\cos 2t+\frac{a+b}{2} \end{pmatrix}$

$=\frac{b-a}{2}\begin{pmatrix} -\cos 2t & \sin 2t\\ \sin 2t & \cos 2t \end{pmatrix}+\frac{a+b}{2}\begin{pmatrix} 1 & 0\\ 0 & 1 \end{pmatrix}$

$=\frac{b-a}{2}B+\frac{a+b}{2}I$

Với $B=\begin{pmatrix} -\cos 2t & \sin 2t\\ \sin 2t & \cos 2t \end{pmatrix}$

$B^{2}=I$ $B^{3}=B$

$B^{2k}=I$, $B^{2k+1}=B$ $\forall k\in \mathbb{N}$

Suy ra:

$A^{n}=\left ( \frac{b-a}{2}.B+\frac{a+b}{2}.I \right )^{n}$

$=\sum_{k=0}^{n}C_{n}^{k}\left ( \frac{b-a}{2} \right )^{k}B^{k}\left ( \frac{a+b}{2} \right )^{n-k}$

$=\sum_{k=0}^{n}C_{n}^{k}\frac{(b-a)^{k}.(a+b)^{n-k}}{2^{n}}.B^{k}$

$=\sum_{k=0}^{[\frac{n}{2}]}C_{n}^{2k}\frac{(b-a)^{2k}.(a+b)^{n-2k}}{2^{n}}.B^{2k}+\sum_{k=0}^{[\frac{n}{2}]}C_{n}^{2k+1}\frac{(b-a)^{2k+1}.(a+b)^{n-2k-1}}{2^{n}}.B^{2k+1}$

$=\left ( \sum_{k=0}^{[\frac{n}{2}]}C_{n}^{2k}\frac{(b-a)^{2k}.(a+b)^{n-2k}}{2^{n}} \right ).I+\left ( \sum_{k=0}^{[\frac{n}{2}]}C_{n}^{2k+1}\frac{(b-a)^{2k+1}.(a+b)^{n-2k-1}}{2^{n}} \right ).B$

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Rút gọn được thì đẹp nhỉ!

Bài viết đã được chỉnh sửa nội dung bởi vo van duc: 12-01-2013 - 21:40

[phudinhgioihan]

$A^{n}=\left ( \frac{b-a}{2}.B+\frac{a+b}{2}.I \right )^{n}$

$=\sum_{k=0}^{n}C_{n}^{k}\left ( \frac{b-a}{2} \right )^{k}B^{k}\left ( \frac{a+b}{2} \right )^{n-k}$

$=\sum_{k=0}^{n}C_{n}^{k}\frac{(b-a)^{k}.(a+b)^{n-k}}{2^{n}}.B^{k}$

$=\sum_{k=0}^{[\frac{n}{2}]}C_{n}^{2k}\frac{(b-a)^{2k}.(a+b)^{n-2k}}{2^{n}}.B^{2k}+\sum_{k=0}^{[\frac{n}{2}]}C_{n}^{2k+1}\frac{(b-a)^{2k+1}.(a+b)^{n-2k-1}}{2^{n}}.B^{2k+1}$

$=\left ( \sum_{k=0}^{[\frac{n}{2}]}C_{n}^{2k}\frac{(b-a)^{2k}.(a+b)^{n-2k}}{2^{n}} \right ).I+\left ( \sum_{k=0}^{[\frac{n}{2}]}C_{n}^{2k+1}\frac{(b-a)^{2k+1}.(a+b)^{n-2k-1}}{2^{n}} \right ).B$

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Rút gọn được thì đẹp nhỉ!

Thế thì cho đẹp nào

$$2^nb^n=(b-a+b+a)^n=\sum_{k=0}^n C_n^k (b-a)^k(b+a)^{n-k}$$
$$=\sum_{k=0}^{[\frac{n}{2}]} C_{n}^{2k}(b-a)^{2k}(b+a)^{n-2k}+\sum_{k=0}^{[\frac{n}{2}]} C_n^{2k+1}(b-a)^{2k+1}(b+a)^{n-2k-1} $$


$$2^na^n=(b+a-(b-a))^n=\sum_{k=0}^n C_n^k (-1)^k(b-a)^k(b+a)^{n-k}$$
$$=\sum_{k=0}^{[\frac{n}{2}]}C_n^{2k}(b-a)^{2k}(b+a)^{n-2k}-\sum_{k=0}^{[\frac{n}{2}]}C_n^{2k+1}(b-a)^{2k+1}(b+a)^{n-2k-1}$$


Suy ra

$$\sum_{k=0}^{[\frac{n}{2}]}C_n^{2k}(b-a)^{2k}(b+a)^{n-2k}=2^{n-1}(a^n+b^n)$$
$$\sum_{k=0}^{[\frac{n}{2}]}C_n^{2k+1}(b-a)^{2k+1}(b+a)^{n-2k-1}=2^{n-1}(b^n-a^n)$$

Vậy $$A^n=2^{n-1}(a^n+b^n)I+2^{n-1}(b^n-a^n)B$$

$$=2^{n-1} \begin{pmatrix} a^n+b^n+(a^n-b^n)\cos 2t & (b^n-a^n)\sin 2t\\ \\ (b^n-a^n)\sin 2t&a^n+b^n+(b^n-a^n)\cos 2t \end{pmatrix}$$

[KaiPotter]

$=\frac{b-a}{2}B+\frac{a+b}{2}I$

Với $B=\begin{pmatrix} -\cos 2t & \sin 2t\\ \sin 2t & \cos 2t \end{pmatrix}$

$B^{2}=I$ $B^{3}=B$

$B^{2k}=I$, $B^{2k+1}=B$ $\forall k\in \mathbb{N}$

 

Em không hiểu lắm về đoạn này, lúc khai triển ra đâu có được $B^{2}=I$ $B^{3}=B$  đâu ạ??

[vo van duc]

Em không hiểu lắm về đoạn này, lúc khai triển ra đâu có được $B^{2}=I$ $B^{3}=B$  đâu ạ??

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