将\(A\)列分块，设\(A=(\alpha_1,\alpha_2,\alpha_3)\)，其中
\begin{equation*}
\alpha_1=(1,1,-1)^T,\alpha_2=(1,0,0)^T,\alpha_3=(0,1,0)^T.
\end{equation*}
先正交化，令
\begin{equation*}
\begin{array}{l}
\beta_1=\alpha_1=(1,1,-1)^T,\\
\beta_2=\alpha_2-\frac{\alpha_2\cdot\beta_1}{\beta_1\cdot\beta_1}\beta_1=\alpha_2-\frac{1}{3}\beta_1=(\frac{2}{3},-\frac{1}{3},\frac{1}{3})^T,\\
\beta_3=\alpha_3-\frac{\alpha_3\cdot\beta_1}{\beta_1\cdot\beta_1}\beta_1-\frac{\alpha_3\cdot\beta_2}{\beta_2\cdot\beta_2}\beta_2=\alpha_3-\frac{1}{3}\beta_1+\frac{1}{2}\beta_2=(0,\frac{1}{2},\frac{1}{2})^T,
\end{array}
\end{equation*}
再单位化，得
\begin{equation*}
\begin{array}{c}\gamma_1=\frac{1}{\|\beta_1\|}\beta_1=\frac{\sqrt{3}}{3}\beta_1=(\frac{\sqrt{3}}{3},\frac{\sqrt{3}}{3},-\frac{\sqrt{3}}{3})^T,\\
\gamma_2=\frac{1}{\|\beta_2\|}\beta_2=\frac{\sqrt{6}}{2}\beta_2=(\frac{\sqrt{6}}{3},-\frac{\sqrt{6}}{6},\frac{\sqrt{6}}{6})^T,\\
\gamma_3=\frac{1}{\|\beta_3\|}\beta_3=\sqrt{2}\beta_3=(0,\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2})^T,
\end{array}
\end{equation*}
则\(\gamma_1,\gamma_2,\gamma_3\)是\(\mathbb{R}^3\)的一个标准正交基。令
\begin{equation*}
Q=(\gamma_1,\gamma_2,\gamma_3)=\begin{pmatrix}
\frac{\sqrt{3}}{3}&\frac{\sqrt{6}}{3}&0\\
\frac{\sqrt{3}}{3}&-\frac{\sqrt{6}}{6}&\frac{\sqrt{2}}{2}\\
-\frac{\sqrt{3}}{3}&\frac{\sqrt{6}}{6}&\frac{\sqrt{2}}{2}
\end{pmatrix},
\end{equation*}
则\(Q^TQ=E_3\)。注意到
\begin{equation*}
\begin{array}{l}
\alpha_1=\beta_1=\sqrt{3}\gamma_1,\\
\alpha_2=\frac{1}{3}\beta_1+\beta_2=\frac{\sqrt{3}}{3}\gamma_1+\frac{\sqrt{6}}{3}\gamma_2,\\
\alpha_3=\frac{1}{3}\beta_1-\frac{1}{2}\beta_2+\beta_3=\frac{\sqrt{3}}{3}\gamma_1-\frac{\sqrt{6}}{6}\gamma_2+\frac{\sqrt{2}}{2}\gamma_3,
\end{array}
\end{equation*}
所以\(A=QR\)，其中
\begin{equation*}
R=\begin{pmatrix}
\sqrt{3}&\frac{\sqrt{3}}{3}&\frac{\sqrt{3}}{3}\\
0&\frac{\sqrt{6}}{3}&-\frac{\sqrt{6}}{6}\\
0&0&\frac{\sqrt{2}}{2}
\end{pmatrix}.
\end{equation*}
