1. **State the problem:** We are given a matrix equation $Ax = b$ where $A$ is a $3 \times 3$ matrix and $b$ is a $3 \times 1$ vector. We know: - $\begin{pmatrix}1 \\ 1 \\ 2\end{pmatrix}$ is a solution to $Ax = b$. - $\begin{pmatrix}1 \\ -1 \\ 1\end{pmatrix}$ is a solution to the homogeneous system $Ax = 0$. We want to find at least 3 more solutions to $Ax = b$ using the given information. 2. **Recall the key property:** If $u$ is a solution to $Ax = 0$ (homogeneous system), and $v$ is a solution to $Ax = b$, then for any scalar $s \in \mathbb{R}$, $$A(v + su) = Av + sAu = b + s0 = b.$$ This means $v + su$ is also a solution to $Ax = b$. 3. **Apply the property:** Given: - $v = \begin{pmatrix}1 \\ 1 \\ 2\end{pmatrix}$ solves $Ax = b$. - $u = \begin{pmatrix}1 \\ -1 \\ 1\end{pmatrix}$ solves $Ax = 0$. For any $s$, $$x = v + su = \begin{pmatrix}1 \\ 1 \\ 2\end{pmatrix} + s \begin{pmatrix}1 \\ -1 \\ 1\end{pmatrix} = \begin{pmatrix}1 + s \\ 1 - s \\ 2 + s\end{pmatrix}.$$ 4. **Find 3 specific solutions by choosing different values of $s$:** - For $s=1$: $x = \begin{pmatrix}2 \\ 0 \\ 3\end{pmatrix}$ - For $s=2$: $x = \begin{pmatrix}3 \\ -1 \\ 4\end{pmatrix}$ - For $s=-1$: $x = \begin{pmatrix}0 \\ 2 \\ 1\end{pmatrix}$ These are three distinct solutions to $Ax = b$. 5. **How to get infinitely many solutions to $Ax=0$ from a nontrivial solution:** If $u$ is a nontrivial solution to $Ax=0$, then for any scalar $s$, $su$ is also a solution because $$A(su) = sAu = s0 = 0.$$ Since $s$ can be any real number, there are infinitely many solutions along the line spanned by $u$. 6. **Construct infinitely many solutions to $Ax=b$ using $v$ and $u$:** For any $s \in \mathbb{R}$, $$x = v + su$$ is a solution to $Ax = b$, giving infinitely many solutions. **Final answer:** Three more solutions to $Ax = b$ are: $$\begin{pmatrix}2 \\ 0 \\ 3\end{pmatrix}, \quad \begin{pmatrix}3 \\ -1 \\ 4\end{pmatrix}, \quad \begin{pmatrix}0 \\ 2 \\ 1\end{pmatrix}.$$