Homogeneous linear system: A system of linear equations is homogeneous if its matrix equation is \(A\overrightarrow{x}=\overrightarrow{0}\). Note that \(\overrightarrow{0}\) is always a solution called the trivial solution. Any nonzero solution is called a nontrivial solution.

Example.

1. \[\begin{eqnarray*} \begin{array}{rcrcrcr} x_1&+&x_2 &-&x_3&=&0\\ &&3x_2&-&2x_3&=&0 \end{array} \end{eqnarray*}\] The corresponding matrix equation \(A\overrightarrow{x}=\overrightarrow{0}\) has the solution set \[\left\lbrace s \left[\begin{array}{r} 1\\ 2\\ 3 \end{array} \right] \; |\; s\in \mathbb R \right\rbrace\] which is also denoted by \(\operatorname{Span}\left\lbrace \left[\begin{array}{r} 1\\ 2\\ 3 \end{array} \right]\right\rbrace\). This solution set corresponds to the points on the line in the 3-space \(\mathbb R^3\) passing through the point \((1,2,3)\) and the origin \((0,0,0)\). Recall that the vector \(\left[\begin{array}{r} 1\\ 2\\ 3 \end{array} \right]\) is the position vector of the point \((1,2,3)\) which is a directed line segment from the origin \((0,0,0)\) to the point \((1,2,3)\).


2. \[x_1-x_2-2x_3=0\] The corresponding matrix equation \(A\overrightarrow{x}=\overrightarrow{0}\) has the solution set \[\left\lbrace s \left[\begin{array}{r}1\\ 1\\ 0 \end{array} \right] +t \left[\begin{array}{r}2\\ 0\\ 1 \end{array} \right] \; |\; s,t\in \mathbb R \right\rbrace =\operatorname{Span}\left\lbrace \left[\begin{array}{r} 1\\ 1\\0 \end{array} \right],\;\left[\begin{array}{r} 2\\0\\1 \end{array} \right] \right\rbrace.\] This solution set corresponds to the points on the plane in the 3-space \(\mathbb R^3\) passing through the points \((1,1,0)\), \((2,0,1)\), and the origin \((0,0,0)\).

Remark. If \(A\overrightarrow{x}=\overrightarrow{0}\) has \(k\) free variables, then its solution set is the span of \(k\) vectors.

The solution set of \(A\overrightarrow{x}=\overrightarrow{0}\) is \(\operatorname{Span}\{\overrightarrow{v_1},\ldots,\overrightarrow{v_k}\}\) for some vectors \(\overrightarrow{v_1},\ldots,\overrightarrow{v_k}\).
The solution set of \(A\overrightarrow{x}=\overrightarrow{b}\) is \(\{\overrightarrow{p}+\overrightarrow{v}\:|\: A\overrightarrow{v}=\overrightarrow{0}\}\) where \(A\overrightarrow{p}=\overrightarrow{b}.\)
So a nonhomogenous solution is a sum of a particular solution and a homogeneous solution. To justify it, let \(\overrightarrow{y}\) be a solution of \(A\overrightarrow{x}=\overrightarrow{b}\), i.e., \(A\overrightarrow{y}=\overrightarrow{b}\). Then \[A(\overrightarrow{y}-\overrightarrow{p})=\overrightarrow{b}-\overrightarrow{b}=\overrightarrow{0}.\] Then \((\overrightarrow{y}-\overrightarrow{p})=\overrightarrow{v}\) where \(A\overrightarrow{v}=\overrightarrow{0}\). Thus \(\overrightarrow{y}=\overrightarrow{p}+\overrightarrow{v}\).
Geometrically we get the solution set of \(A\overrightarrow{x}=\overrightarrow{b}\) by shifting the solution set of \(A\overrightarrow{x}=\overrightarrow{0}\) to the point whose position vector is \(\overrightarrow{p}\) along the vector \(\overrightarrow{p}\).

Example. The nonhomogeneous system \(x_1-x_2-2x_3=-2\) has a particular solution \(\overrightarrow{p}=\left[\begin{array}{r} 1\\ 1\\ 1 \end{array} \right]\). The corresponding homogeneous system \(x_1-x_2-2x_3=0\) has the solution set \[\left\lbrace s \left[\begin{array}{r}1\\ 1\\ 0 \end{array} \right] +t \left[\begin{array}{r}2\\ 0\\ 1 \end{array} \right] \; |\; s,t\in \mathbb R \right\rbrace.\] Thus the solution set of the nonhomogeneous system \(x_1-x_2-2x_3=-2\) is \[\left\lbrace \left[\begin{array}{r} 1\\ 1\\ 1 \end{array} \right] +s \left[\begin{array}{r}1\\ 1\\ 0 \end{array} \right] +t \left[\begin{array}{r}2\\ 0\\ 1 \end{array} \right] \; |\; s,t\in \mathbb R \right\rbrace.\]

The solution set of \(A\overrightarrow{x}=\overrightarrow{b}\) is a translation of that of \(A\overrightarrow{x}=\overrightarrow{0}\)

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