Define the linear transformation
\[T:C^3 -> C^2,\] \[T \left( \begin{bmatrix} x1 \\ x2 \\ x3 \end{bmatrix}\right) = \begin{bmatrix} 2x1-x2+5x3 \\ -4x1+2x2-10x3 \end{bmatrix} \]
For finding the linear transformation, following conditions must met:
First property:
\[T(X + Y) = \left( \begin{bmatrix} x1 \\ x2 \\ x3 \end{bmatrix} + \begin{bmatrix} y1 \\ y2 \\ y3 \end{bmatrix} \right) \]
\[ = T \left( \begin{bmatrix} x1 + y1 \\ x2 + y2\\ x3 + y3 \end{bmatrix}\right) \]
Now apply the above rule
\[ \begin{bmatrix} 2(x1 + y1) - (x2 + y2) + 5(x3 + y3) \\ -4(x1 + y1) + 2(x2 + y2) - 10(x3 + y3) \end{bmatrix}\]
\[ \begin{bmatrix} 2x1 + 2y1 - x2 - y2 + 5x3 + 5y3 \\ -4x1 - 4y1 + 2x2 + 2y2 - 10x3 - 10y3) \end{bmatrix}\]
\[ = \begin{bmatrix} 2x1 -x2 +5x3\\ -4x1 +2x2 -10x3 \end{bmatrix} + \begin{bmatrix} 2y1 -y2 +5y3\\ -4y1 +2y2 -10y3 \end{bmatrix} \]
\[T(X + Y) = \left( \begin{bmatrix} x1\\ x2 \end{bmatrix} + \begin{bmatrix} y1\\ y2 \end{bmatrix} \right) \]
And it proves that
\[T(X + Y) = T(x)+T(y) \]
Second property:
\[T(CX) = \left( C\begin{bmatrix} x1 \\ x2 \\ x3 \end{bmatrix} \right) \]
\[ = \left( \begin{bmatrix} Cx1 \\ Cx2 \\ Cx3 \end{bmatrix} \right) \]
Now apply the second property in the above rule
\[ = \begin{bmatrix} 2(Cx1) -(Cx2) + 5(Cx3)\\ -4(Cx1) +2(Cx2) - 10(Cx3) \end{bmatrix} \]
\[ = C \begin{bmatrix} 2x1 -x2 + 5x3\\ -4x1 +2x2 - 10x3 \end{bmatrix} \]
\[ = CT\left( \begin{bmatrix} x1 \\ x2 \\ x3 \end{bmatrix} \right) \]
Hence it is proved
\[ = CT( x) \]