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## Chapter 3 Exercise A

1. Solution: If $T$ is linear, then $(0,0)=T(0,0,0)=(b,0)$by 3.11, hence $b=0$. We also have $T(1,1,1)=T(1,1,0)+T(0,0,1),$it is equivalent to $(1+b,6+c)=(b-2,6)+(3+b,0)=(1+2b,6).$Thus $6+c=6$ implies $c=0$. Conversely, if $b=c=0$, $T$ is obviously linear. See…