Assume that $\mathbf{x}_{p}$ is a particular solution to the equation $A \mathbf{x}=\mathbf{b}$ with $\mathbf{x}, \mathbf{b} \in \mathbb{R}^{3}$ and a real $3 \times 3$ matrix $A$. Explain why the general solution to $A \mathbf{x}=\mathbf{b}$ is given by $\mathbf{x}=\mathbf{x}_{p}+\mathbf{h}$ where $\mathbf{h}$ is any vector such that $A \mathbf{h}=\mathbf{0}$.

Now assume that $A$ is a real symmetric $3 \times 3$ matrix with three different eigenvalues $\lambda_{1}, \lambda_{2}$ and $\lambda_{3}$. Show that eigenvectors of $A$ with respect to different eigenvalues are orthogonal. Let $\mathbf{x}_{k}$ be a normalised eigenvector of $A$ with respect to the eigenvalue $\lambda_{k}$, $k=1,2,3$. Show that the linear system

$$\left(A-\lambda_{k} I\right) \mathbf{x}=\mathbf{b}$$

where $I$ denotes the $3 \times 3$ unit matrix, is solvable if and only if $\mathbf{x}_{k} \cdot \mathbf{b}=0$. Show that the general solution is given by

$$\mathbf{x}=\sum_{i \neq k} \frac{\mathbf{b} . \mathbf{x}_{i}}{\lambda_{i}-\lambda_{k}} \mathbf{x}_{i}+\beta \mathbf{x}_{k}, \quad \beta \in \mathbb{R}$$

[Hint: consider the components of $\mathbf{x}$ and $\mathbf{b}$ with respect to a basis of eigenvectors of $A$.]

Consider the matrix $A$ and the vector $\mathbf{b}$

$$A=\left(\begin{array}{ccc} -\frac{1}{2} \sqrt{2}+\frac{1}{6} \sqrt{3} & \frac{1}{2} \sqrt{2}+\frac{1}{6} \sqrt{3} & -\frac{1}{3} \sqrt{3} \\ \frac{1}{2} \sqrt{2}+\frac{1}{6} \sqrt{3} & -\frac{1}{2} \sqrt{2}+\frac{1}{6} \sqrt{3} & -\frac{1}{3} \sqrt{3} \\ -\frac{1}{3} \sqrt{3} & -\frac{1}{3} \sqrt{3} & \frac{2}{3} \sqrt{3} \end{array}\right), \quad \mathbf{b}=\left(\begin{array}{c} \sqrt{2}+\sqrt{3} \\ -\sqrt{2}+\sqrt{3} \\ -2 \sqrt{3} \end{array}\right)$$

Verify that $\frac{1}{\sqrt{3}}(1,1,1)^{T}$ and $\frac{1}{\sqrt{2}}(1,-1,0)^{T}$ are eigenvectors of $A$. Show that $A \mathbf{x}=\mathbf{b}$ is solvable and find its general solution.