Solution
= Solution
Write the <linear system> as $Ar=d$ with
$$
A=\begin{pmatrix}1&2&3\\3&4&5\\1&3&4\end{pmatrix},\qquad
r=\begin{pmatrix}x\\y\\z\end{pmatrix},\qquad
d=\begin{pmatrix}1\\2\\3\end{pmatrix}.
$$
Expansion along the first row gives $\det A=(16-15)-2(12-5)+3(9-4)=2$. Since \b[$\det A=2\ne0$], $A$ is an <invertible matrix>, and $r=A^{-1}d$ is the unique solution for this right side.