$A.$ $\left(\begin{array}{lll}1 & 0 & 1 \\ 2 & 2 & 0 \\ 0 & 3 & 3\end{array}\right)$. $B.$ $\left(\begin{array}{lll}1 & 2 & 0 \\ 0 & 2 & 3 \\ 1 & 0 & 3\end{array}\right)$. $C.$ $\left(\begin{array}{rrr}\frac{1}{2} & \frac{1}{4} & -\frac{1}{6} \\ -\frac{1}{2} & \frac{1}{4} & \frac{1}{6} \\ -\frac{1}{2} & -\frac{1}{4} & \frac{1}{6}\end{array}\right)$. $D.$ $\left(\begin{array}{rrr}\frac{1}{2} & -\frac{1}{2} & \frac{1}{2} \\ \frac{1}{4} & \frac{1}{4} & -\frac{1}{4} \\ -\frac{1}{6} & \frac{1}{6} & \frac{1}{6}-\end{array}\right)$.

A

#### 解析：

\begin{aligned} &\left(\boldsymbol{\alpha}_{1}+\boldsymbol{\alpha}_{2}, \boldsymbol{\alpha}_{2}+\boldsymbol{\alpha}_{3}, \boldsymbol{\alpha}_{3}+\boldsymbol{\alpha}_{1}\right)=\left(\boldsymbol{\alpha}_{1}, \boldsymbol{\alpha}_{2}, \boldsymbol{\alpha}_{3}\right)\left[\begin{array}{lll} 1 & 0 & 1 \\ 1 & 1 & 0 \\ 0 & 1 & 1 \end{array}\right] \\ &=\left(\boldsymbol{\alpha}_{1}, \frac{1}{2} \boldsymbol{\alpha}_{2}, \frac{1}{3} \boldsymbol{\alpha}_{3}\right)\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{array}\right]\left[\begin{array}{lll} 1 & 0 & 1 \\ 1 & 1 & 0 \\ 0 & 1 & 1 \end{array}\right]=\left(\boldsymbol{\alpha}_{1}, \frac{1}{2} \boldsymbol{\alpha}_{2}, \frac{1}{3} \boldsymbol{\alpha}_{3}\right)\left[\begin{array}{lll} 1 & 0 & 1 \\ 2 & 2 & 0 \\ 0 & 3 & 3 \end{array}\right] . \end{aligned}
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