Solution to Linear Algebra Hoffman & Kunze Chapter 1.3 Exercise 1.3.2
Solution: We have$$\rightarrow\left[\begin{array}{ccc}1&-3&0\\2&1&1\\3&-1&2\end{array}\right]\rightarrow\left[\begin{array}{ccc}1&-3&0\\0&7&1\\0&8&2\end{array}\right]\rightarrow\left[\begin{array}{ccc}1&-3&0\\0&1&1/7\\0&8&2\end{array}\right]$$ $$\rightarrow\left[\begin{array}{ccc}1&0&3/7\\0&1&1/7\\0&0&6/7\end{array}\right]\rightarrow\left[\begin{array}{ccc}1&0&3/7\\0&1&1/7\\0&0&1\end{array}\right]\rightarrow\left[\begin{array}{ccc}1&0&0\\0&1&1/7\\0&0&1\end{array}\right].$$Thus $A$ is row-equivalent to the identity matrix. It follows that the only solution to the system is $(0,0,0)$.
