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 surjective we must begin with a totally arbitrary element of the codomain, $\vect{v}$ and somehow find an input vector $\vect{u}$ such that $\lt{T}{\vect{u}}=\vect{v}$. We desire, \begin{align*} \lt{T}{\vect{u}}&=\vect{v}\\ \colvector{-65 u_1 + 128 u_2 + 10 u_3 - 262 u_4 + 40 u_5\\ 36 u_1 - 73 u_2 - u_3 + 151 u_4 - 16 u_5\\ -44 u_1 + 88 u_2 + 5 u_3 - 180 u_4 + 24 u_5\\ 34 u_1 - 68 u_2 - 3 u_3 + 140 u_4 - 18 u_5\\ 12 u_1 - 24 u_2 - u_3 + 49 u_4 - 5 u_5} &= \colvector{v_1\\v_2\\v_3\\v_4\\v_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{u_1\\u_2\\u_3\\u_4\\u_5} &= \colvector{v_1\\v_2\\v_3\\v_4\\v_5} \end{align*}
We recognize this equation as a system of equations in the variables $u_i$, but our vector of constants contains symbols. In general, we would have to row-reduce the augmented matrix by hand, due to the symbolic final column. However, in this particular example, the $5\times 5$ coefficient matrix is nonsingular and so has an inverse (Theorem NI, Definition MI). \begin{equation*} \inverse{ \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} } = \begin{bmatrix} -47 & 92 & 1 & -181 & -14 \\ 27 & -55 & \frac{7}{2} & \frac{221}{2} & 11\\ -32 & 64 & -1 & -126 & -12\\ 25 & -50 & \frac{3}{2} & \frac{199}{2} & 9 \\ 9 & -18 & \frac{1}{2} & \frac{71}{2} & 4 \end{bmatrix} \end{equation*} so we find that \begin{align*} \colvector{u_1\\u_2\\u_3\\u_4\\u_5} &= \begin{bmatrix} -47 & 92 & 1 & -181 & -14 \\ 27 & -55 & \frac{7}{2} & \frac{221}{2} & 11\\ -32 & 64 & -1 & -126 & -12\\ 25 & -50 & \frac{3}{2} & \frac{199}{2} & 9 \\ 9 & -18 & \frac{1}{2} & \frac{71}{2} & 4 \end{bmatrix} \colvector{v_1\\v_2\\v_3\\v_4\\v_5}\\ &= \colvector{-47 v_1 + 92 v_2 + v_3 - 181 v_4 - 14 v_5\\ 27 v_1 - 55 v_2 + \frac{7}{2} v_3 + \frac{221}{2} v_4 + 11 v_5\\ -32 v_1 + 64 v_2 - v_3 - 126 v_4 - 12 v_5\\ 25 v_1 - 50 v_2 + \frac{3}{2} v_3 + \frac{199}{2} v_4 + 9 v_5\\ 9 v_1 - 18 v_2 + \frac{1}{2} v_3 + \frac{71}{2} v_4 + 4 v_5} \end{align*}
This establishes that if we are given any output vector $\vect{v}$, we can use its components in this final expression to formulate a vector $\vect{u}$ such that $\lt{T}{\vect{u}}=\vect{v}$. So by Definition SLT we now know that $T$ is surjective. You might try to verify this condition in its full generality (i.e. evaluate $T$ with this final expression and see if you get $\vect{v}$ as the result), or test it more specifically for some numerical vector $\vect{v}$ (see Exercise SLT.C20).