1. **State the problem:** Find the least squares solution $\hat{x}$ to the system $A x = b$ where $$ A = \begin{pmatrix} 5 & 6 & 4 & 4 \\ 1 & 6 & 4 & 2 \\ 1 & 2 & 4 & 2 \\ 5 & 1 & 5 & 5 \\ 1 & 1 & 3 & 1 \\ 6 & 5 & 2 & 6 \end{pmatrix}, \quad b = \begin{pmatrix} -1 \\ -1 \\ 2 \\ 3 \\ -2 \\ -2 \end{pmatrix} $$ 2. **Formula used:** The least squares solution satisfies the normal equations $$ A^T A \hat{x} = A^T b $$ where $A^T$ is the transpose of $A$. 3. **Calculate $A^T A$:** $$ A^T = \begin{pmatrix} 5 & 1 & 1 & 5 & 1 & 6 \\ 6 & 6 & 2 & 1 & 1 & 5 \\ 4 & 4 & 4 & 5 & 3 & 2 \\ 4 & 2 & 2 & 5 & 1 & 6 \end{pmatrix} $$ Multiply $A^T$ by $A$: $$ A^T A = \begin{pmatrix} 5 & 1 & 1 & 5 & 1 & 6 \\ 6 & 6 & 2 & 1 & 1 & 5 \\ 4 & 4 & 4 & 5 & 3 & 2 \\ 4 & 2 & 2 & 5 & 1 & 6 \end{pmatrix} \times \begin{pmatrix} 5 & 6 & 4 & 4 \\ 1 & 6 & 4 & 2 \\ 1 & 2 & 4 & 2 \\ 5 & 1 & 5 & 5 \\ 1 & 1 & 3 & 1 \\ 6 & 5 & 2 & 6 \end{pmatrix} $$ Calculating each element (rounded to 3 decimals): $$ A^T A = \begin{pmatrix} 86 & 70 & 62 & 80 \\ 70 & 86 & 62 & 70 \\ 62 & 62 & 70 & 62 \\ 80 & 70 & 62 & 90 \end{pmatrix} $$ 4. **Calculate $A^T b$:** $$ A^T b = \begin{pmatrix} 5 & 1 & 1 & 5 & 1 & 6 \\ 6 & 6 & 2 & 1 & 1 & 5 \\ 4 & 4 & 4 & 5 & 3 & 2 \\ 4 & 2 & 2 & 5 & 1 & 6 \end{pmatrix} \times \begin{pmatrix} -1 \\ -1 \\ 2 \\ 3 \\ -2 \\ -2 \end{pmatrix} = \begin{pmatrix} -1 \\ -3 \\ 3 \\ -3 \end{pmatrix} $$ 5. **Solve the system $A^T A \hat{x} = A^T b$:** $$ \begin{pmatrix} 86 & 70 & 62 & 80 \\ 70 & 86 & 62 & 70 \\ 62 & 62 & 70 & 62 \\ 80 & 70 & 62 & 90 \end{pmatrix} \hat{x} = \begin{pmatrix} -1 \\ -3 \\ 3 \\ -3 \end{pmatrix} $$ Using numerical methods (e.g., Gaussian elimination or matrix inverse), the solution rounded to 3 decimals is: $$ \hat{x} = \begin{pmatrix} -0.500 \\ -0.500 \\ 0.500 \\ 0.000 \end{pmatrix} $$ 6. **Interpretation:** The vector $\hat{x}$ minimizes the squared error $\|A x - b\|^2$ and is the best approximate solution to the system. **Final answer:** $$ \hat{x} = \begin{pmatrix} -0.500 \\ -0.500 \\ 0.500 \\ 0.000 \end{pmatrix} $$