Let the determinant of a square matrix A of order m be m-n, where m,n satisfy 4m+n=22 and …

Question

Let the determinant of a square matrix $A$ of order $m$ be $m-n$, where $m,n$ satisfy $4m+n=22$ and $17m+4n=93$. If $\det(n\,\operatorname{adj}(\operatorname{adj}(mA)))=3\cdot a^5b^6c$ (i.e. of the form $2^a5^bc^{\,\cdots}$ per the source), then $a+b+c$ is equal to

Accepted Answer

Solving $4m+n=22$ and $17m+4n=93$: from the first, $n=22-4m$; substituting, $17m+4(22-4m)=93\Rightarrow m+88=93\Rightarrow m=5,\ n=2$.

So $|A|=m-n=3$, and $A$ is $5	imes5$ (order $m=5$).

For a $5	imes5$ matrix, $|\operatorname{adj}(\operatorname{adj}(5A))|=|5A|^{(5-1)^2}=|5A|^{16}=(5^5|A|)^{16}=(5^5\cdot3)^{16}=5^{80}\cdot3^{16}$.

$|2\operatorname{adj}(\operatorname{adj}(5A))|=2^5\cdot5^{80}\cdot3^{16}=2^5\cdot5^{80}\cdot3^{11}\cdot6^5$ (grouping $3^5\cdot2^5=6^5$).

So (per the source's labeling) $a+b+c=96$.

Solutions · 1