## 每週問題 June 29, 2015

Let

$\displaystyle A=\begin{bmatrix} 1&0&0\\ 0&1&0\\ 3&1&2 \end{bmatrix}$.

Find all matrices $B$ such that $AB=BA$.

$B=[b_{ij}]$ 使得 $AB=BA$，代入數值可得包含9個變數 $b_{ij}$$1\le i,j\le 3$，的齊次方程，化簡即得所求。下面介紹另一個方法。令 $C$ 為一可逆矩陣。若 $AB=BA$

$\displaystyle (C^{-1}AC)(C^{-1}BC)=C^{-1}ABC=C^{-1}BAC=(C^{-1}BC)(C^{-1}AC)$

$(C^{-1}AC)(C^{-1}BC)=(C^{-1}BC)(C^{-1}AC)$

$\displaystyle AB=C(C^{-1}AC)(C^{-1}BC)C^{-1}=C(C^{-1}BC)(C^{-1}AC)C^{-1}=BA$

$\displaystyle S=\left[\!\!\begin{array}{rrc} 1&0&0\\ 0&1&0\\ -3&-1&1 \end{array}\!\!\right]$

$\displaystyle E=\begin{bmatrix} a&b&0\\ c&d&0\\ 0&0&e \end{bmatrix}$

\displaystyle\begin{aligned} B&=SES^{-1}\\ &=\left[\!\!\begin{array}{rrc} 1&0&0\\ 0&1&0\\ -3&-1&1 \end{array}\!\!\right] \begin{bmatrix} a&b&0\\ c&d&0\\ 0&0&e \end{bmatrix} \begin{bmatrix} 1&0&0\\ 0&1&0\\ 3&1&1 \end{bmatrix}\\ &=\begin{bmatrix} a&b&0\\ c&d&0\\ -3a-c+3e&-3b-d+e&e \end{bmatrix}.\end{aligned}

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