1. **Problem statement:** Confirm by multiplication that $x$ is an eigenvector of $A$ and find the eigenvalue for the first matrix and vector. 2. **Given:** $$A = \begin{bmatrix}1 & 2 \\ 3 & 2\end{bmatrix}, \quad x = \begin{bmatrix}1 \\ -1\end{bmatrix}$$ 3. **Recall:** If $x$ is an eigenvector of $A$, then $$A x = \lambda x$$ for some scalar $\lambda$ called the eigenvalue. 4. **Multiply $A$ by $x$:** $$A x = \begin{bmatrix}1 & 2 \\ 3 & 2\end{bmatrix} \begin{bmatrix}1 \\ -1\end{bmatrix} = \begin{bmatrix}1 \cdot 1 + 2 \cdot (-1) \\ 3 \cdot 1 + 2 \cdot (-1)\end{bmatrix} = \begin{bmatrix}1 - 2 \\ 3 - 2\end{bmatrix} = \begin{bmatrix}-1 \\ 1\end{bmatrix}$$ 5. **Check if $A x$ is a scalar multiple of $x$:** We want to find $\lambda$ such that $$\begin{bmatrix}-1 \\ 1\end{bmatrix} = \lambda \begin{bmatrix}1 \\ -1\end{bmatrix}$$ Comparing components: $$-1 = \lambda \cdot 1 \implies \lambda = -1$$ $$1 = \lambda \cdot (-1) \implies \lambda = -1$$ Both components give $\lambda = -1$, so $x$ is an eigenvector with eigenvalue $-1$. **Final answer:** $x$ is an eigenvector of $A$ with eigenvalue $\boxed{-1}$.