Consider the system \(Ax = b\) defined over the real numbers where \(A=\begin{bmatrix} 1 & 2 & 0 & -1 & 0 & 3 \\ 0 & 0 & 1 & -1 & 0 & 4 \\ 0 & 0 & 0 & 0 & 1 & 5 \end{bmatrix}\) and \(b = \begin{bmatrix} 6\\7\\8\end{bmatrix}\). Note that \(x\) is \(\begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \\ x_5 \\ x_6\end{bmatrix}\) here.

The augmented matrix representing the system is \(\left[\begin{array}{cccccc|c} 1 & 2 & 0 & -1 & 0 & 3 & 6 \\ 0 & 0 & 1 & -1 & 0 & 4 & 7 \\ 0 & 0 & 0 & 0 & 1 & 5 & 8 \end{array}\right],\) which is already in reduced row-echelon form with columns 1, 3, and 5 as pivot columns. Hence, the free variables are \(x_2\), \(x_4\) and \(x_6\).

Setting \(x_2 = s\), \(x_4 = t\), and \(x_6 = u\), we obtain the solution \(\begin{bmatrix} x_1\\x_2\\x_3\\x_4\\x_5\\x_6\end{bmatrix} = \begin{bmatrix} 6-2s+t-3u\\ s\\ 7+t- 4u\\ t\\ 8-5u\\ u\end{bmatrix}= \begin{bmatrix} 6\\0\\ 7\\ 0\\ 8\\ 0\end{bmatrix}+ s~\begin{bmatrix} -2\\1\\ 0\\ 0\\ 0\\ 0\end{bmatrix}+ t~\begin{bmatrix} 1\\0\\ 1\\ 1\\ 0\\ 0\end{bmatrix}+ u~\begin{bmatrix} -3\\0\\ -4\\ 0\\ -5\\ 1\end{bmatrix}\) for any choice of real numbers \(s\), \(t\), and \(u\).

Notice that each of
\(\begin{bmatrix} -2\\1\\ 0\\ 0\\ 0\\ 0\end{bmatrix}\),
\(\begin{bmatrix} 1\\0\\ 1\\ 1\\ 0\\ 0\end{bmatrix}\), and
\(\begin{bmatrix} -3\\0\\ -4\\ 0\\ -5\\ 1\end{bmatrix}\) is a solution
to the **homogeneous system** \(Ax = 0\).
(A homogeneous system of linear equations has all the right-hand side values
equal 0.)

In general, knowing the solutions of \(Ax=0\) tells us a lot about the solutions to \(Ax = b\). Suppose that both \(x^*\) and \(x'\) are solutions to \(Ax=b\); i.e. \(Ax^*=b\) and \(Ax'=b\). Then \(A(x^* - x') = b - b = 0\). (Why?) Hence, if \(\tilde{x} = x^* - x'\), then \(\tilde{x}\) is a solution to \(Ax = 0\).

As a result, once we have one solution to \(Ax = b\), say \(x^*\), then every other solution differs from \(x^*\) by a solution to \(Ax = 0\). In other words, the set of solutions to \(Ax = b\) is given by \(\{ x^* + d : Ad = 0\}\).

In short, knowing one solution to \(Ax=b\) and all the solutions to \(Ax = 0\) gives us all the solutions to \(Ax = b\).

Finally, note that \(Ax=0\) always has the tuple of 0's as a solution.
This solution is called the **trivial solution**. Any other solution
to \(Ax = 0\) is called a **nontrivial solution**.

Let \(A = \begin{bmatrix} 1 & 2 & -1 \\ 0 & -1 & 3\\ 1 & 1 & 2\end{bmatrix}\). Does the homogenous system \(Ax = 0\) have a nontrivial solution?

Let \(A \in \mathbb{F}^{m\times n}\) where \(\mathbb{F}\) denotes a field. Let \(N(A)\) denote the set \(\{d \in \mathbb{R} : Ad = 0\}\). Let \(x^*, x' \in \mathbb{F}^n\).

Show that \(A(x^* - x') = Ax^* - Ax'\).

Let \(z = x^* + x'\). Show that if \(x^*, x' \in N(A)\), then \(z \in N(A)\).

Let \(\alpha \in \mathbb{F}\). Let \(u = \alpha x^*\). Show that if \(x^* \in N(A)\), then \(u \in N(A)\).