Answer
$T(v)=(3a+b)v_1+(2b-a)v_2$
Work Step by Step
$v$ is an arbitrary vector in $V$
Since $\{v_1,v_2\}$ is a basic for $V$, we obtain:
$$v=av_1+av_2$$
Hence,
$$T(v)=T(av_1+bv_2) \\
=aT(v_1)+bT(v_2)\\
=a(3v_1-v_2)+b(v_1+2v_2)\\
=3av_1-av_2+bv_1+2bv_2\\
=(3a+b)v_1+(2b-a)v_2$$
