Answer
Image of $(2,4)$ is $\left[ {\begin{array}{*{20}{c}}
{\sqrt 3 - 2}\\
{ - 2}\\
4
\end{array}} \right]$
The preimage of $w$ is $(2,0)$
Work Step by Step
We are given
$T\left[ {\begin{array}{*{20}{c}} {{v_1}}\\ {{v_2}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {\frac{{\sqrt 3 }}{2}{v_1} - \frac{1}{2}{v_2}}\\ {{v_1} - {v_2}}\\ { {v_2}} \end{array}} \right]$
a) Image of $v=(2,4)$ is
$\begin{array}{*{20}{l}}
{T\left[ {\begin{array}{*{20}{c}}
{{v_1}}\\
{{v_2}}
\end{array}} \right]}\\
{ = T\left[ {\begin{array}{*{20}{c}}
2\\
4
\end{array}} \right]}\\
{ = \left[ {\begin{array}{*{20}{c}}
{\frac{{\sqrt 3 }}{2}(2) - \frac{1}{2}(4)}\\
{2 - 4}\\
4
\end{array}} \right]}\\
{ = \left[ {\begin{array}{*{20}{c}}
{\sqrt 3 - 2}\\
{ - 2}\\
4
\end{array}} \right]}\\
{}
\end{array}$
b) Let the preimage of $w=(\sqrt{3},2,0)$ be $(a,b)$.
$\begin{array}{l}
\Rightarrow T\left[ {\begin{array}{*{20}{c}}
a\\
b
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
{ \sqrt 3 }\\
{ 2}\\
{ 0}
\end{array}} \right]\\
\Rightarrow \left[ {\begin{array}{*{20}{c}}
{\frac{{\sqrt 3 }}{2}a - \frac{{1}}{2}b}\\
{a - b}\\
{ b}
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
{ \sqrt 3 }\\
{ 2}\\
{ 0}
\end{array}} \right]
\end{array}$
Comparing each component of both vectors, we have
$\begin{array}{*{20}{l}}
{\frac{{\sqrt 3 }}{2}a - \frac{1}{2}b = \sqrt 3 ,\quad (1)}\\
{a - b = 2,\quad (2)}\\
{b = 0,\quad (3)}
\end{array}$
From equation $(3)$, we have $b=0$. Putting this in equation $(2)$, we get
$a-0=2\Rightarrow a=2$
Therefore, the preimage of $w$ is $(2,0)$
