Common Matrix Factor

$AB = \begin{bmatrix} 3a & 3b \\ 2c & 2d\end{bmatrix}$

$BA' = \begin{bmatrix} 18a-20b & 12a-13b \\ 18c-20d & 12c-13d \end{bmatrix}$

Since $AB=BA'$ , equating corresponding entries:

$\implies \begin{cases} 18a-20b=3a\implies 3a=4b\\12a-13b=3b\implies 3a=4b\\18c-20d=2c\implies 4c=5d\\12c-13d=2d\implies 4c=5d\end{cases}$

$\implies \begin{cases}a=\frac{4b}3....(1)\\d=\frac{4c}5....(2)\end{cases}$

Since det $(B)=1$ $\implies ad-bc=1$ and using $(1),(2)$ by multiplying them and substituting we get:

$\frac{16bc}{15}-bc=1$ $\implies bc=15=3\cdot 5=15\cdot 1$

And since from $1$ and $2$ we know $4b\mod 3=0$ and $4c\mod 5=0$ :

$\implies b=3,c=5\implies a=\frac{4b}3=4,d=\frac{4c}5=4$

$\therefore a+b+c+d=\boxed{16}$