Exercise 3.4.7

See Exercise 1.6.8. Note that if we put those vectors as row vectors of $M$ just like what we’ve done in Exercise 1.6.8, we cannot interchange any two rows. However, we can also construct a matrix the $i$-th column $u_i$. And we can do the row operation including interchanging any two rows. The set of columns containing one pivot¹ forms an independent set.

So the set $\{u_1,u_2,u_5\}$ is a basis for ${ℝ}^3$.