Show that the following mappings are not linear:
(a) $F: \mathbf{R}^{2} \rightarrow \mathbf{R}^{2}$ defined by $F(x, y)=(x y, x)$
(b) $F: \mathbf{R}^{2} \rightarrow \mathbf{R}^{3}$ defined by $F(x, y)=(x+3,2 y, x+y)$
(c) $F: \mathbf{R}^{3} \rightarrow \mathbf{R}^{2}$ defined by $F(x, y, z)=(|x|, y+z)$
(a) Let $v=(1,2)$ and $w=(3,4) ;$ then $v+w=(4,6) .$ Also,
\[
F(v)=(1(2), 1)=(2,1) \quad \text { and } \quad F(w)=(3(4), 3)=(12,3)
\]
Hence,
\[
F(v+w)=(4(6), 4)=(24,6) \neq F(v)+F(w)
\]
(b) Because $F(0,0)=(3,0,0) \neq(0,0,0), F$ cannot be linear.
(c) Let $v=(1,2,3)$ and $k=-3 .$ Then $k v=(-3,-6,-9) .$ We have
\[
F(v)=(1,5) \text { and } k F(v)=-3(1,5)=(-3,-15)
\]
Thus,
\[
F(k v)=F(-3,-6,-9)=(3,-15) \neq k F(v)
\]
Accordingly, $F$ is not linear.