**Solution to Linear Algebra Hoffman & Kunze Chapter 1.3 Exercise 1.3.5**

Solution: Call the first matrix $A$ and the second matrix $B$. The matrix $A$ is row-equivalent to $$A’=\left[\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right]$$ and the matrix $B$ is row-equivalent to $$B’=\left[\begin{array}{ccc}1&0&1/2\\0&1&3/2\\0&0&0\end{array}\right].$$

By Theorem 3 page 7 $AX=0$ and $A’X=0$ have the same solutions. Similarly $BX=0$ and $B’X=0$ have the same solutions. Now if $A$ and $B$ are row-equivalent then $A’$ and $B’$ are row equivalent. Thus if $A$ and $B$ are row equivalent then $A’X=0$ and $B’X=0$ must have the same solutions. But $B’X=0$ has infinitely many solutions and $A’X=0$ has only the trivial solution $(0,0,0)$. Thus $A$ and $B$ cannot be row-equivalent.