若用配方法将二次型`f(x_1,x_2,x_3)=2x_1^2+x_2^2+4x_3^2+2x_1x_2-2x_2x_3`化为规范形，则规范形和所用的变换矩阵为
A．`f(y_1,y_2,y_3)=y_1^2+y_2^2+y_3^2`，\begin{bmatrix} -1/\sqrt{2} & -1/\sqrt{2} &-1/\sqrt{2} \\ 0& \sqrt{2}& \sqrt{2} \\ 0 & 0 & -1/\sqrt{2} \end{bmatrix}
B．`f(y_1,y_2,y_3)=y_1^2+y_2^2+y_3^2`，\begin{bmatrix} 1/\sqrt{2} & -1/\sqrt{2} &-1/\sqrt{2} \\ 0& \sqrt{2}& \sqrt{2} \\ 0 & 0 &- 1/\sqrt{2} \end{bmatrix}
C．`f(y_1,y_2,y_3)=y_1^2+y_2^2+y_3^2`，\begin{bmatrix} 1/\sqrt{2} &1/\sqrt{2} &-1/\sqrt{2} \\ 0& \sqrt{2}& \sqrt{2} \\ 0 & 0 & 1/\sqrt{2} \end{bmatrix}
D．`f(y_1,y_2,y_3)=y_1^2+y_2^2+y_3^2`，\begin{bmatrix} 1/\sqrt{2} & -1/\sqrt{2} &-1/\sqrt{2} \\ 0& \sqrt{2}& \sqrt{2} \\ 0 & 0 & 1/\sqrt{2} \end{bmatrix}