If $$A = \begin{bmatrix}2 & n \\4 & 1 \end{bmatrix}$$ such that $$A^3 = 27\begin{bmatrix}4 & q \\p & r \end{bmatrix}$$, then $$p + q + r$$ equals ___________

$$A = \begin{bmatrix}2 & n \\4 & 1 \end{bmatrix}$$

=> $$A^2 = \begin{bmatrix}2 & n \\4 & 1 \end{bmatrix}\begin{bmatrix}2 & n \\4 & 1 \end{bmatrix}$$

=> $$A^2=\begin{bmatrix}4+4n & 3n \\12 & 4n+1 \end{bmatrix}$$

And, $$A^3 = \begin{bmatrix}4+4n & 3n \\12 & 4n+1 \end{bmatrix}\begin{bmatrix}2 & n \\4 & 1 \end{bmatrix}$$

=> $$A^3=\begin{bmatrix}8+20n & 4n^2+7n \\16n+28 & 16n+1 \end{bmatrix}\rightarrow1$$

And, it is given that $$A^3 = 27\begin{bmatrix}4 & q \\p & r \end{bmatrix}$$ 

=> $$A^3 = \begin{bmatrix}108 & 27q \\27p & 27r \end{bmatrix}\rightarrow2$$

Comparing each cell value in eq. 1 and eq. 2 - 

=> $$8+20n=108$$ => $$n=5$$

=> $$4n^2+7n=27q$$ => $$135=27q$$ => $$q=5$$

=> $$16n+28=27p$$ => $$108=27p$$ => $$p=4$$

=> $$16n+1=27r$$ => $$81=27r$$ => $$r=3$$

Thus, the value of $$p+q+r=4+5+3=12$$