# To find: The linear transformation (T_2T_1)(v) for an arbitrary vector v in V. The vectors {v_1,v_2} is vasis for the vector space V. Given: The linea

To find:
The linear transformation $\left({T}_{2}{T}_{1}\right)\left(v\right)$ for an arbitrary vector v in V.
The vectors $\left\{{v}_{1},{v}_{2}\right\}$ is vasis for the vector space V.
Given:
The linear transformation with satisfying equations ${T}_{1}\left({v}_{1}\right)=3{v}_{1}+{v}_{2},$
and ${T}_{2}\left({v}_{2}\right)=-{v}_{1}+6{v}_{2}$ are given as
and
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Calculation:
To solve for ${T}_{2}\left({T}_{1}\left(v\right)\right)$ find ${T}_{1}\left(v\right)$ use properties of linear transformation.
${T}_{1}\left(v\right)={T}_{1}\left(a{v}_{1}+b{v}_{2}\right)$
${T}_{1}\left(v\right)=a{T}_{1}\left({v}_{1}\right)+b{T}_{1}\left({v}_{2}\right)$
Substitute $3{v}_{1}+{v}_{2}$ for ${T}_{1}\left({v}_{1}\right)$ and 0 for ${T}_{1}\left({v}_{2}\right)$ in the above expression.
${T}_{1}\left(v\right)=a\left(3{v}_{1}+{v}_{2}\right)+b\left(0\right)$
$=a\left(3{v}_{1}+{v}_{2}\right)$
Substitute $a\left(3{v}_{1}+{v}_{2}\right)$ for ${T}_{1}\left(v\right)$ to find ${T}_{2}\left({T}_{1}\left(v\right)\right)$ and it is written as,
${T}_{2}\left({T}_{1}\left(v\right)\right)={T}_{2}\left(a\left(3{v}_{1}+{v}_{2}\right)\right)$
${T}_{2}\left({T}_{1}\left(v\right)\right)=3a{T}_{2}\left({v}_{1}\right)+a{T}_{2}\left({v}_{2}\right)$
Substitute $-5{v}_{2}$ for ${T}_{2}\left({v}_{1}\right)$ and $-{v}_{1}+6{v}_{2}$ for ${T}_{2}\left({v}_{2}\right)$ in the above expression.
${T}_{2}\left({T}_{1}\left(v\right)\right)=3a\left(-5{v}_{2}\right)+a\left(-{v}_{1}+6{v}_{2}\right)$
$-15a{v}_{2}-a{v}_{1}+6a{v}_{2}$

Therefore, the linear transformation $\left({T}_{2}{T}_{1}\right)\left(v\right)$ is