Let the determinant of a square matrix  A  of order m  be $m-n$ ,where m and n satisfy $4m+n=22$ and if $17m+4n=93$ , $\det (n.adj(adj\left(mA\right)\left)\right)=3^a{5}^b{6}^{c}thena+b-c=$

$4m+n=22 17m+4n=93$

Solving above two equations, we get m=5 and n = 2
$\therefore$ A is a square matrix of order 5 and $\left|A\right|$   = 5-2=3

Now, we know that adj  $\left(KA\right)$ $=K^{n-1}$  adjA, where A is a matrix of order n
$adj\left(mA\right)=adj\left(5A\right)=5^{5-1}\left(adjA\right)=5^4\left(adjA\right)$

Again, adj  $\left(5^{4}adjA\right)$  $={\left(5^4\right)}^{4}adjA\left(adjA\right)=5^{16}|A{|}^{5-2}.A=5^{16}{3}^{3}A$

Now, det $\left(nadj\left(adjmA\right)\right)=\det ({2.5}^{16}{.3}^3.A)$

$={\left({2.5}^{16}{.3}^3\right)}^5\det A=2^5{.5}^{80}{.3}^{15}.3=2^5{.5}^{80}{.3}^{16}=6^5{.5}^{80}{.3}^{11}$

$\therefore a=11,b=80andc=5$  $\Rightarrow a+b-c=80+11-5=86$