The system \(A \vec{x} = \vec{b}\) does not have solutions. The normal equations are: \[A^T A = \begin{bmatrix} 24 & 6 \\ 6 & 14 \end{bmatrix}\] \[A^T \vec{b} = \begin{bmatrix} 0 \\ 4 \end{bmatrix}\] The reduced row echelon form (RREF) of \(A^TA \vec{x} = A^T\vec{b}\) is \[\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \end{bmatrix}\]

Problem:

Decide whether the linear system of equations \(A \vec{x} = \vec{b}\) has a solution. 

Then find a least squares solution. 

\[A = \begin{bmatrix} 
-4 & 1   \\ 
-2 & -2   \\ 
-2 & -3
\end{bmatrix}, \qquad 
\vec{b} = \begin{bmatrix} 
 1  \\ 
 -3  \\ 
 1
\end{bmatrix}\]

\begin{solution}
Solution:
The system \(A \vec{x} = \vec{b}\) does not have solutions. The normal equations are:
\[A^T A = \begin{bmatrix} 
24 & 6   \\ 
6 & 14
\end{bmatrix}\]
\[A^T \vec{b} = \begin{bmatrix} 
 0  \\ 
 4
\end{bmatrix}\]

The reduced row echelon form (RREF) of \(A^TA \vec{x} = A^T\vec{b}\) is
\[\begin{bmatrix} 
1 & 0 & 0   \\ 
0 & 1 & 0
\end{bmatrix}\] 

\end{solution}