设 $\boldsymbol{\alpha}_1=\left(\begin{array}{r}1 \\ -1 \\ 1\end{array}\right), \boldsymbol{\alpha}_2=\left(\begin{array}{l}1 \\ 2 \\ 3\end{array}\right), \boldsymbol{\alpha}_3=\left(\begin{array}{r}-2 \\ 1 \\ -4\end{array}\right)$ 和 $\boldsymbol{\beta}_1=\left(\begin{array}{l}1 \\ 1 \\ 1\end{array}\right), \boldsymbol{\beta}_2=\left(\begin{array}{c}1 \\ 0 \\ -2\end{array}\right), \boldsymbol{\beta}_3=\left(\begin{array}{l}0 \\ 1 \\ 2\end{array}\right)$ 为 $\mathbb{R}^3$ 的两组基, 则 $\boldsymbol{\beta}_1$, $\boldsymbol{\beta}_2, \boldsymbol{\beta}_3$ 到 $\boldsymbol{\alpha}_1, \boldsymbol{\alpha}_2, \boldsymbol{\alpha}_3$ 的过渡矩阵为
$\text{A.}$ $\left(\begin{array}{rrr}5 & -4 & -6 \\ 1 & 0 & 1 \\ 10 & 8 & 11\end{array}\right)$
$\text{B.}$ $\left(\begin{array}{rrr}5 & 1 & -10 \\ -4 & 0 & 8 \\ -6 & 1 & 11\end{array}\right)$.
$\text{C.}$ $\left(\begin{array}{ccc}\frac{1}{2} & \frac{3}{2} & \frac{1}{2} \\ \frac{15}{4} & \frac{7}{4} & \frac{9}{4} \\ -2 & 0 & -1\end{array}\right)$.
$\text{D.}$ $\left(\begin{array}{rrr}\frac{1}{2} & \frac{15}{4} & -2 \\ \frac{3}{2} & \frac{7}{4} & 0 \\ \frac{1}{2} & \frac{9}{4} & -1\end{array}\right)$.