The system \(A \vec{x} = \vec{b}\) does not have solutions. The normal equations are: \[A^T A = \begin{bmatrix} 26 & 14 \\ 14 & 13 \end{bmatrix}\] \[A^T \vec{b} = \begin{bmatrix} -7 \\ -12 \end{bmatrix}\] The reduced row echelon form (RREF) of \(A^TA \vec{x} = A^T\vec{b}\) is \[\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & -1 \end{bmatrix}\]

Problem:

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

Then find a least squares solution. 

\[A = \begin{bmatrix} 
-4 & -3   \\ 
1 & 2   \\ 
3 & 0
\end{bmatrix}, \qquad 
\vec{b} = \begin{bmatrix} 
 4  \\ 
 0  \\ 
 3
\end{bmatrix}\]

\begin{solution}
Solution:
The system \(A \vec{x} = \vec{b}\) does not have solutions. The normal equations are:
\[A^T A = \begin{bmatrix} 
26 & 14   \\ 
14 & 13
\end{bmatrix}\]
\[A^T \vec{b} = \begin{bmatrix} 
 -7  \\ 
 -12
\end{bmatrix}\]

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

\end{solution}