\[S(a+bx+c{ x }^{ 2 }+d{ x }^{ 3 })=\left[ \begin{matrix} -a & +4b & +c & +2d \\ a & +5b & -2c & +2d \end{matrix}\quad \quad \begin{matrix} 4a & -b & +6c & -d \\ a & & +2c & +5d \end{matrix} \right]\]
# Creating matrix into S
(S <- matrix(c(-1,4,1,1,4,-1,5,0,1,6,-2,2,2,-1,2,5), 4,4))
## [,1] [,2] [,3] [,4]
## [1,] -1 4 1 2
## [2,] 4 -1 6 -1
## [3,] 1 5 -2 2
## [4,] 1 0 2 5
# RREF
(answer <- rref(S))
## [,1] [,2] [,3] [,4]
## [1,] 1 0 0 0
## [2,] 0 1 0 0
## [3,] 0 0 1 0
## [4,] 0 0 0 1
The identity matrix is nonsingular and have null space therefore it is injective. Uās dimension is 4 and kernel is 0 therefore it is also surjective. Lastly, transformation is injective and surjective therefore it is also invertible as well as per ILTIS theorem.