Given an $m \times n$ matrix $A$ and $\mathbf{b} \in \mathbb{R}^{m}$, prove that the vector $\mathbf{x} \in \mathbb{R}^{n}$ is the solution of the least-squares problem for $A \mathbf{x} \approx \mathbf{b}$ if and only if $A^{T}(A \mathbf{x}-\mathbf{b})=\mathbf{0}$. Let

$$A=\left[\begin{array}{cc} 1 & 2 \\ -3 & 1 \\ 1 & 3 \\ 4 & 1 \end{array}\right], \quad \mathbf{b}=\left[\begin{array}{c} 3 \\ 0 \\ -1 \\ 2 \end{array}\right]$$

Determine the solution of the least-squares problem for $A \mathbf{x} \approx \mathbf{b}$.