Archetype R is the linear transformation \begin{equation*} \ltdefn{T}{\complex{5}}{\complex{5}},\quad \lt{T}{\colvector{x_1\\x_2\\x_3\\x_4\\x_5}}= \colvector{-65 x_1 + 128 x_2 + 10 x_3 - 262 x_4 + 40 x_5\\ 36 x_1 - 73 x_2 - x_3 + 151 x_4 - 16 x_5\\ -44 x_1 + 88 x_2 + 5 x_3 - 180 x_4 + 24 x_5\\ 34 x_1 - 68 x_2 - 3 x_3 + 140 x_4 - 18 x_5\\ 12 x_1 - 24 x_2 - x_3 + 49 x_4 - 5 x_5} \end{equation*}
To establish that $R$ is injective we must begin with the assumption that $\lt{T}{\vect{x}}=\lt{T}{\vect{y}}$ and somehow arrive at the conclusion that $\vect{x}=\vect{y}$. Here we go, \begin{align*} \colvector{0\\0\\0\\0\\0} &=\lt{T}{\vect{x}}-\lt{T}{\vect{y}}\\ &=\lt{T}{\colvector{x_1\\x_2\\x_3\\x_4\\x_5}}-\lt{T}{\colvector{y_1\\y_2\\y_3\\y_4\\y_5}}\\ &= \colvector{-65 x_1 + 128 x_2 + 10 x_3 - 262 x_4 + 40 x_5\\ 36 x_1 - 73 x_2 - x_3 + 151 x_4 - 16 x_5\\ -44 x_1 + 88 x_2 + 5 x_3 - 180 x_4 + 24 x_5\\ 34 x_1 - 68 x_2 - 3 x_3 + 140 x_4 - 18 x_5\\ 12 x_1 - 24 x_2 - x_3 + 49 x_4 - 5 x_5}\\ &\quad\quad- \colvector{-65 y_1 + 128 y_2 + 10 y_3 - 262 y_4 + 40 y_5\\ 36 y_1 - 73 y_2 - y_3 + 151 y_4 - 16 y_5\\ -44 y_1 + 88 y_2 + 5 y_3 - 180 y_4 + 24 y_5\\ 34 y_1 - 68 y_2 - 3 y_3 + 140 y_4 - 18 y_5\\ 12 y_1 - 24 y_2 - y_3 + 49 y_4 - 5 y_5}\\ &= \colvector{-65 (x_1-y_1) + 128 (x_2-y_2) + 10 (x_3-y_3) - 262 (x_4-y_4) + 40 (x_5-y_5)\\ 36 (x_1-y_1) - 73 (x_2-y_2) - (x_3-y_3) + 151 (x_4-y_4) - 16 (x_5-y_5)\\ -44 (x_1-y_1) + 88 (x_2-y_2) + 5 (x_3-y_3) - 180 (x_4-y_4) + 24 (x_5-y_5)\\ 34 (x_1-y_1) - 68 (x_2-y_2) - 3 (x_3-y_3) + 140 (x_4-y_4) - 18 (x_5-y_5)\\ 12 (x_1-y_1) - 24 (x_2-y_2) - (x_3-y_3) + 49 (x_4-y_4) - 5 (x_5-y_5)}\\ &= \begin{bmatrix} -65&128&10&-262&40\\ 36&-73&-1&151&-16\\ -44&88&5&-180&24\\ 34&-68&-3&140&-18\\ 12&-24&-1&49&-5 \end{bmatrix} \colvector{x_1-y_1\\x_2-y_2\\x_3-y_3\\x_4-y_4\\x_5-y_5} \end{align*}
Now we recognize that we have a homogeneous system of 5 equations in 5 variables (the terms $x_i-y_i$ are the variables), so we row-reduce the coefficient matrix to \begin{equation*} \begin{bmatrix} \leading{1}&0&0&0&0\\ 0&\leading{1}&0&0&0\\ 0&0&\leading{1}&0&0\\ 0&0&0&\leading{1}&0\\ 0&0&0&0&\leading{1} \end{bmatrix} \end{equation*}
So the only solution is the trivial solution \begin{align*} x_1-y_1&=0&x_2-y_2&=0&x_3-y_3&=0&x_4-y_4&=0&x_5-y_5&=0 \end{align*} and we conclude that indeed $\vect{x}=\vect{y}$. By Definition ILT, $T$ is injective.