The system \(A \vec{x} = \vec{b}\) does not have solutions. The normal equations are: \[A^T A = \begin{bmatrix} 29 & -14 \\ -14 & 18 \end{bmatrix}\] \[A^T \vec{b} = \begin{bmatrix} -20 \\ 7 \end{bmatrix}\] 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. 

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

\begin{solution}
Solution:
The system \(A \vec{x} = \vec{b}\) does not have solutions. The normal equations are:
\[A^T A = \begin{bmatrix} 
29 & -14   \\ 
-14 & 18
\end{bmatrix}\]
\[A^T \vec{b} = \begin{bmatrix} 
 -20  \\ 
 7
\end{bmatrix}\]

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}