Solution to Linear Algebra Hoffman & Kunze Chapter 1.2 Exercise 1.2.4
Solution: These systems are not equivalent.
Call the two equations in the first system $E_1$ and $E_2$ and the equations in the second system $E’_1$ and $E’_2$. Then if $E’_2=aE_1+bE_2$ since $E_2$ does not have $x_1$ we must have $a=1/3$. But then to get the coefficient of $x_4$ we’d need $$7x_4=\dfrac13x_4+5bx_4.$$ That forces $b=\dfrac43$. But if $a=\dfrac13$ and $b=\dfrac43$ then the coefficient of $x_3$ would have to be $-2i\dfrac43$ which does not equal $1$. Therefore the systems cannot be equivalent.
