1. **State the problem:** We have a matrix equation $A\mathbf{x} = \mathbf{b}$ where $A$ is a $3 \times 3$ matrix, $\mathbf{b}$ is a $3 \times 1$ vector. Given: - $\mathbf{x}_p = \begin{pmatrix}1 \\ 1 \\ 2\end{pmatrix}$ is a particular solution to $A\mathbf{x} = \mathbf{b}$. - $\mathbf{x}_h = \begin{pmatrix}1 \\ -1 \\ 1\end{pmatrix}$ is a solution to the homogeneous system $A\mathbf{x} = \mathbf{0}$. 2. **Recall the principle:** The general solution to $A\mathbf{x} = \mathbf{b}$ is given by: $$\mathbf{x} = \mathbf{x}_p + \mathbf{x}_h$$ where $\mathbf{x}_p$ is any particular solution and $\mathbf{x}_h$ is any solution to the homogeneous system. 3. **Find more solutions:** Since $\mathbf{x}_h$ is a solution to the homogeneous system, any scalar multiple of $\mathbf{x}_h$ is also a solution to $A\mathbf{x} = \mathbf{0}$. Therefore, more solutions to $A\mathbf{x} = \mathbf{b}$ are: $$\mathbf{x} = \mathbf{x}_p + t \mathbf{x}_h = \begin{pmatrix}1 \\ 1 \\ 2\end{pmatrix} + t \begin{pmatrix}1 \\ -1 \\ 1\end{pmatrix} = \begin{pmatrix}1 + t \\ 1 - t \\ 2 + t\end{pmatrix}$$ for any scalar $t$. 4. **Find at least 3 more solutions by choosing different values of $t$:** - For $t=1$: $$\mathbf{x} = \begin{pmatrix}1+1 \\ 1-1 \\ 2+1\end{pmatrix} = \begin{pmatrix}2 \\ 0 \\ 3\end{pmatrix}$$ - For $t=2$: $$\mathbf{x} = \begin{pmatrix}1+2 \\ 1-2 \\ 2+2\end{pmatrix} = \begin{pmatrix}3 \\ -1 \\ 4\end{pmatrix}$$ - For $t=-1$: $$\mathbf{x} = \begin{pmatrix}1-1 \\ 1+1 \\ 2-1\end{pmatrix} = \begin{pmatrix}0 \\ 2 \\ 1\end{pmatrix}$$ These are three distinct solutions to $A\mathbf{x} = \mathbf{b}$. **Final answer:** $$\begin{pmatrix}2 \\ 0 \\ 3\end{pmatrix}, \quad \begin{pmatrix}3 \\ -1 \\ 4\end{pmatrix}, \quad \begin{pmatrix}0 \\ 2 \\ 1\end{pmatrix}$$