The inverse of matrix N = [[2, 3], [1, 4]] is

$$\det(N) = \begin{vmatrix} 2 & 3 \\ 1 & 4 \end{vmatrix} = 2 \cdot 4 - 3 \cdot 1 = 8 - 3 = 5$$

Since $\det(N) = 5 \neq 0$, the inverse exists.

$$\begin{pmatrix} a & b \\ c & d \end{pmatrix}^{-1} = \frac{1}{ad - bc} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix}$$

Substitute $a=2$, $b=3$, $c=1$, $d=4$, $\det(N)=5$:

$$N^{-1} = \frac{1}{5} \begin{pmatrix} 4 & -3 \\ -1 & 2 \end{pmatrix}$$

First row, first column: $2 \cdot \frac{4}{5} + 3 \cdot \frac{-1}{5} = \frac{8}{5} - \frac{3}{5} = \frac{5}{5} = 1$

First row, second column: $2 \cdot \frac{-3}{5} + 3 \cdot \frac{2}{5} = \frac{-6}{5} + \frac{6}{5} = \frac{0}{5} = 0$

Second row, first column: $1 \cdot \frac{4}{5} + 4 \cdot \frac{-1}{5} = \frac{4}{5} - \frac{4}{5} = \frac{0}{5} = 0$

Second row, second column: $1 \cdot \frac{-3}{5} + 4 \cdot \frac{2}{5} = \frac{-3}{5} + \frac{8}{5} = \frac{5}{5} = 1$

$$N \cdot N^{-1} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = I$$

$\boxed{\dfrac{1}{5}\begin{pmatrix}4&-3\\-1&2\end{pmatrix}}$