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Solution to Linear Algebra Hoffman & Kunze Chapter 1.2 Exercise 1.2.3

Solution: Yes the two systems are equivalent. We show this by writing each equation of the first system in terms of the second, and conversely.
\begin{alignat*}{1}
x_1-x_3 &= {\frac{-3}4}(-x_1+x_2+4x_3) + {\frac14}(x_1+3x_3+8x_3)\\
x_2+3x_3&={\frac14}(-x_1+x_2+4x_3) + {\frac14}(x_1+3x_3+8x_3)
\end{alignat*}
and
\begin{alignat*}{1}
-x_1+x_2+4x_3 &= -(x_1-x_3) + (x_2+3x_3)\\
x_1+3x_2+8x_3&= (x_1-x_3) + 3(x_2+3x_3)\\
{\frac12}x_1+x_2+{\frac52}x_3&= {\frac12}(x_1-x_3) + (x_2+3x_3)
\end{alignat*}