Let the determinant of a square matrix A of order $m$ be $m-n$, where $m$ and $n$

satisfy $4 m+n=22$ and $17 m+4 n=93$.

If $\operatorname{det}(n \operatorname{adj}(\operatorname{adj}(m A)))=3^{a} 5^{b} 6^{c}$ then $a+b+c$ is equal to :

Given that $|A|=m-n$, and let's solve the system of linear equations to find the values of $m$ and $n$ : <br/><br/>$4m + n = 22$ ...... (1) <br/><br/>$17m + 4n = 93$ ....... (2) <br/><br/>We can multiply equation (1) by 4 to make the coefficients of $n$ in both equations equal: <br/><br/>$16m + 4n = 88$ ......... (3) <br/><br/>Now, subtract equation (3) from equation (2): <br/><br/>$(17m + 4n) - (16m + 4n) = 93 - 88$ <br/><br/>$m = 5$ <br/><br/>Now, we can find the value of $n$ by substituting $m$ into equation (1): <br/><br/>$4(5) + n = 22$ <br/><br/>$20 + n = 22$ <br/><br/>$n = 2$ <br/><br/>Since the order of matrix A is $m$, the order is 5. Now, let's find the determinant of $n \operatorname{adj}(\operatorname{adj}(mA))$: <br/><br/>We know that $\operatorname{det}(A) = m-n = 3$. <br/><br/> $$ \begin{aligned} & \therefore \operatorname{det}(n \operatorname{adj}(\operatorname{adj}(m A))) \\\\ & =|2 \operatorname{adj}(\operatorname{adj}(5 A))| \\\\ & =2^5|5 A|^{16} \\\\ & =2^5 5^{80}|A|^{16}=2^5 \cdot 3^{16} \cdot 5^{80} \\\\ & =3^{11} 5^{80} 6^5 \end{aligned} $$ <br/><br/>So, $a+b+c=96$