「統計解析のための線形代数」復習筆記 26

練習 解下列連立一次方程式

\[\begin{align} \left\{ \begin{array}{ll} a_1+2a_2+a_3 = 2\\ 2a_1+a_2+a_3 = 3\\ a_1+a_2+2a_3 = 3 \end{array} \right. \end{align}\]

解

此連立方程組的擴大係數矩陣爲： \[(X \underline{y})=\left( \begin{array}{c} 1 & 2 & 1 & 2\\ 2 & 1 & 1 & 3\\ 1 & 1 & 2 & 3 \end{array} \right)\] 下面開始行變形： \[\left(\begin{array}{c} 1& 2& 1 & \vdots & 2\\ 2& 1& 1 & \vdots & 3\\ 1& 1& 2 & \vdots & 3\\ \end{array}\right) \begin{align} \left\{ \begin{array}{rr} (1)\\ (2)\\ (3) \end{array} \right. \end{align}\]

\[ \left(\begin{array}{r} 1& 2& 1 & \vdots & 2\\ 0& -3& -1 & \vdots & -1\\ 0& -1& 1 & \vdots & -1\\ \end{array}\right) \begin{align} \left\{ \begin{array}{l} (1)\\ (2)=(2)-2\times(1)\\ (3)=(3)-(1) \end{array} \right. \end{align}\]

\[\left(\begin{array}{r} 1& 0& 3 & \vdots & 4\\ 0& -4& 0 & \vdots & 0\\ 0& 1& -1 & \vdots & -1\\ \end{array}\right) \begin{align} \left\{ \begin{array}{l} (1)=(1)+2\times(3)\\ (2)=(2)+(3)\\ (3)=-1\times(3) \end{array} \right. \end{align}\]

\[\left(\begin{array}{r} 1& 0& 3 & \vdots & 4\\ 0& 1& 0 & \vdots & 0\\ 0& 1& -1 & \vdots & -1\\ \end{array}\right) \begin{align} \left\{ \begin{array}{l} (1)\\ (2)=(2)\div(-4)\\ (3) \end{array} \right. \end{align}\]

\[\left(\begin{array}{r} 1& 0& 3 & \vdots & 4\\ 0& 1& 0 & \vdots & 0\\ 0& 0& -1 & \vdots & -1 \end{array}\right) \begin{align} \left\{ \begin{array}{l} (1)\\ (2)\\ (3)=(3)-(2) \end{array} \right. \end{align}\]

\[\left(\begin{array}{r} 1& 0& 0 & \vdots & 1\\ 0& 1& 0 & \vdots & 0\\ 0& 0& 1 & \vdots & 1 \end{array}\right) \begin{align} \left\{ \begin{array}{l} (1)=(1)+3\times(3)\\ (2)\\ (3) \end{array} \right. \end{align}\]

至此，點線的左側矩陣變成了單位矩陣以後轉換結束。所求的 \(\underline{a}\) 便是點線右側的向量。

\[\therefore \underline{a}=\left( \begin{array}{c} a_1 \\ a_2 \\ a_3 \\ \end{array} \right)=\left( \begin{array}{c} 1 \\ 0 \\ 1 \\ \end{array} \right)\]