设二次型 $f\left(x_{1}, x_{2}, x_{3}\right)=2\left(a_{1} x_{1}+a_{2} x_{2}+a_{3} x_{3}\right)^{2}+\left(b_{1} x_{1}+b_{2} x_{2}+b_{3} x_{3}\right)^{2}$, 记
$$
\boldsymbol{\alpha}=\left(\begin{array}{l}
a_{1} \\
a_{2} \\
a_{3}
\end{array}\right), \quad \boldsymbol{\beta}=\left(\begin{array}{l}
b_{1} \\
b_{2} \\
b_{3}
\end{array}\right) .
$$
( I ) 证明二次型 $f$ 对应的矩阵为 $2 \boldsymbol{\alpha} \boldsymbol{\alpha}^{\mathrm{T}}+\boldsymbol{\beta} \boldsymbol{\beta}^{\mathrm{T}}$;
(III)若 $\boldsymbol{\alpha}, \boldsymbol{\beta}$ 正交且均为单位向量,证明 $f$ 在正交变换下的标准形为 $2 y_{1}^{2}+y_{2}^{2}$.
$\text{A.}$
$\text{B.}$
$\text{C.}$
$\text{D.}$