Let A be a $$3 \times 3$$ real matrix such that $$A\begin{pmatrix}1\\0\\1\end{pmatrix} = 2\begin{pmatrix}1\\0\\1\end{pmatrix}$$, $$A\begin{pmatrix}-1\\0\\1\end{pmatrix} = 4\begin{pmatrix}-1\\0\\1\end{pmatrix}$$, $$A\begin{pmatrix}0\\1\\0\end{pmatrix} = 2\begin{pmatrix}0\\1\\0\end{pmatrix}$$. Then, the system $$(A - 3I)\begin{pmatrix}x\\y\\z\end{pmatrix} = \begin{pmatrix}1\\2\\3\end{pmatrix}$$ has

The given conditions tell us that $$\begin{pmatrix}1\\0\\1\end{pmatrix}$$, $$\begin{pmatrix}-1\\0\\1\end{pmatrix}$$, $$\begin{pmatrix}0\\1\\0\end{pmatrix}$$ are eigenvectors of $$A$$ with eigenvalues 2, 4, 2 respectively.

So $$A - 3I$$ has eigenvalues $$2-3 = -1$$, $$4-3 = 1$$, $$2-3 = -1$$ corresponding to the same eigenvectors.

Since $$\det(A - 3I) = (-1)(1)(-1) = 1 \neq 0$$, the matrix $$A - 3I$$ is invertible.

Therefore the system $$(A - 3I)\begin{pmatrix}x\\y\\z\end{pmatrix} = \begin{pmatrix}1\\2\\3\end{pmatrix}$$ has a unique solution.

To verify, let us express $$\begin{pmatrix}1\\2\\3\end{pmatrix}$$ in terms of the eigenvectors. Let:

$$\begin{pmatrix}1\\2\\3\end{pmatrix} = a\begin{pmatrix}1\\0\\1\end{pmatrix} + b\begin{pmatrix}-1\\0\\1\end{pmatrix} + c\begin{pmatrix}0\\1\\0\end{pmatrix}$$

From the second component: $$c = 2$$.

From the first component: $$a - b = 1$$.

From the third component: $$a + b = 3$$.

Solving: $$a = 2, b = 1$$.

Then the solution is:

$$\begin{pmatrix}x\\y\\z\end{pmatrix} = \frac{2}{-1}\begin{pmatrix}1\\0\\1\end{pmatrix} + \frac{1}{1}\begin{pmatrix}-1\\0\\1\end{pmatrix} + \frac{2}{-1}\begin{pmatrix}0\\1\\0\end{pmatrix} = \begin{pmatrix}-2\\0\\-2\end{pmatrix} + \begin{pmatrix}-1\\0\\1\end{pmatrix} + \begin{pmatrix}0\\-2\\0\end{pmatrix} = \begin{pmatrix}-3\\-2\\-1\end{pmatrix}$$

The answer is Option A: unique solution.