Problem on the number of solutions to an inhomogeneous linear system

How many solutions are there to $$\begin{pmatrix}1&1&1\\1&1&0\\0&0&1\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix} = \begin{pmatrix}1\\2\\3\end{pmatrix}?$$

Solution
We observe that this equation is a linear system of the form $\bfA \bfx = \bfb$ with nonzero $\bfb$. Hence, it is an inhomogeneous linear system.
Recall that
Inhomogeneous linear equations

If $\text{ det } \mathbf{A} \neq 0$ then the inhomogeneous linear system $\mathbf{A} \bfx = \bfb$ has exactly one solution, $x=\mathbf{A}^{-1} \bfb$. If $\text{ det } \mathbf{A} = 0$ then $\mathbf{A} \bfx = \bfb$ has either no solutions or infinitely many solutions.
Identifying $\bfA$ as the matrix in the problem and computing its determinant, we have
Determinant of a 3x3 matrix

$$ \text{det } \begin{pmatrix}1&1&1\\1&1&0\\0&0&1\end{pmatrix} = 0$$
Hence, there are either no solutions or there are infinitely many of them.
We try to find a single solution to $$\begin{pmatrix}1&1&1\\1&1&0\\0&0&1\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix} = \begin{pmatrix}1\\2\\3\end{pmatrix}$$ by writing out each of the three equations that it corresponds to:
\begin{align}
x + y + z &= 1\\
x + y \phantom{+ z }&= 2\\
z &=3
\end{align}
We plug the value of $z$ into the first equation to see $$ \begin{align}x+y &=-2\\x+y&=\phantom{-}2 \end{align}$$
These equations are a contradiction. Thus, there are no solutions to the linear system.

Problem on the number of solutions to an inhomogeneous linear system

Solution

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