TeachingLinear Algebra – Systems of linear equations – Solution sets
A system of linear equations can now be written as a matrix equation \(A\mathbf{x}=\mathbf{b}\). Such a system is called homogeneous if \(\mathbf{b}=\mathbf{0}\). Otherwise, the system is called nonhomogeneous. A homogeneous system is always consistent, since: \(\mathbf{x}=\mathbf{0}\) is a solution (the trivial solution).
Example: If \(A=\begin{pmatrix}1&-1&0\\-1&2&1\\0&1&1\end{pmatrix}\), then we have:
\[A\mathbf{x}=\mathbf{0}:\quad\left(\left.\begin{matrix}1&-1&0\\-1&2&1\\0&1&1\end{matrix}\;\right|\;\begin{matrix}0\\0\\0\end{matrix}\right) \sim\left(\left.\begin{matrix}1&-1&0\\0&1&1\\0&1&1\end{matrix}\;\right|\;\begin{matrix}0\\0\\0\end{matrix}\right) \sim\left(\left.\begin{matrix}1&0&1\\0&1&1\\0&0&0\end{matrix}\;\right|\;\begin{matrix}0\\0\\0\end{matrix}\right).\]Note that the zeros right to the vertical bar never change and therefore could have been omitted. The solution is:
\[\left\{\begin{array}{l}x_1=-x_3\\x_2=-x_3\\x_3\;\text{is vrij}\end{array}\right.\quad\Longrightarrow\quad\mathbf{x}=\begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix} =\begin{pmatrix}-x_3\\-x_3\\x_3\end{pmatrix}=x_3\begin{pmatrix}-1\\-1\\1\end{pmatrix}.\]This is a line in \(\mathbb{R}^3\) through the origin.
Theorem: A homogeneous system of linear equations only has a nontrivial solution if there is at least one free variable.
Example: The homogeneous equation \(x_1-x_2+x_3=0\) has the solution
\[\left\{\begin{array}{l}x_1=x_2-x_3\\x_2\;\text{is vrij}\\x_3\;\text{is vrij}\end{array}\right.\quad\Longrightarrow\quad\mathbf{x} =\begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix}=\begin{pmatrix}x_2-x_3\\x_2\\x_3\end{pmatrix}=x_2\begin{pmatrix}-1\\1\\0\end{pmatrix} +x_3\begin{pmatrix}1\\0\\1\end{pmatrix}.\]The solution set \(\{\mathbf{x}=\begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix}\,:\,x_1-x_2+x_3=0\}=\text{Span}\left\{\begin{pmatrix}1\\1\\0\end{pmatrix}, \begin{pmatrix}-1\\0\\1\end{pmatrix}\right\}\) is a plane in \(\mathbb{R}^3\) through the origin.
Example: If \(A=\begin{pmatrix}1&-1&0\\-1&2&1\\0&1&1\end{pmatrix}\) and \(\mathbf{b}=\begin{pmatrix}2\\-1\\1\end{pmatrix}\), then we have:
\[A\mathbf{x}=\mathbf{b}:\quad\left(\left.\begin{matrix}1&-1&0\\-1&2&1\\0&1&1\end{matrix}\;\right|\;\begin{matrix}2\\-1\\1\end{matrix}\right) \sim\left(\left.\begin{matrix}1&-1&0\\0&1&1\\0&1&1\end{matrix}\;\right|\;\begin{matrix}2\\1\\1\end{matrix}\right) \sim\left(\left.\begin{matrix}1&0&1\\0&1&1\\0&0&0\end{matrix}\;\right|\;\begin{matrix}3\\1\\0\end{matrix}\right).\]The solution is:
\[\left\{\begin{array}{l}x_1=3-x_3\\x_2=1-x_3\\x_3\;\text{is vrij}\end{array}\right.\quad\Longrightarrow\quad\mathbf{x}=\begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix} =\begin{pmatrix}3-x_3\\1-x_3\\x_3\end{pmatrix}=\begin{pmatrix}3\\1\\0\end{pmatrix}+x_3\begin{pmatrix}-1\\-1\\1\end{pmatrix}.\]This is a line in \(\mathbb{R}^3\) not through the origin. This line is parallel to the line obtained in the first example.
Example: The nonhomogeneous equation \(x_1-x_2+x_3=1\) has the solution
\[\left\{\begin{array}{l}x_1=1+x_2-x_3\\x_2\;\text{is vrij}\\x_3\;\text{is vrij}\end{array}\right.\quad\Longrightarrow\quad\mathbf{x} =\begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix}=\begin{pmatrix}1+x_2-x_3\\x_2\\x_3\end{pmatrix}=\begin{pmatrix}1\\0\\0\end{pmatrix} +x_2\begin{pmatrix}1\\1\\0\end{pmatrix}+x_3\begin{pmatrix}-1\\0\\1\end{pmatrix}.\]This is a plane in \(\mathbb{R}^3\) not through the origin. This plane is parallel to the plane obtained in the second example.
Last modified on March 22, 2021
