Let A be a square matrix of order 3 such that $\det(A)=-2$ and $\det(3\text{adj}(-6\text{adj}(3A)))=2^{m+n}\cdot 3^{mn},$ $m>n$. Then $4m + 2n$ is equal to:
Solution:
Using the properties of determinants and adjoint matrices:
1) $|kA| = k^n|A|$
2) $\text{adj}(kA) = k^{n-1}\text{adj}(A)$
3) $|\text{adj}(\text{adj}(A))| = |A|^{(n-1)^2}$
$$
\begin{aligned}
\text{Expression} &= \det(3 \text{adj}(-6 \text{adj}(3A))) \\
&= 3^3 |\text{adj}(-6 \text{adj}(3A))| \\
&= 3^3 \left| (-6)^{3-1} \text{adj}(\text{adj}(3A)) \right| \quad (\because n=3) \\
&= 3^3 \left| 36 \text{adj}(\text{adj}(3A)) \right| \\
&= 3^3 \cdot (36)^3 \left| \text{adj}(\text{adj}(3A)) \right| \\
&= 3^3 \cdot (6^2)^3 \cdot |3A|^{(3-1)^2} \\
&= 3^3 \cdot 6^6 \cdot |3A|^4 \\
&= 3^3 \cdot (2 \cdot 3)^6 \cdot (3^3 |A|)^4 \\
&= 3^3 \cdot 2^6 \cdot 3^6 \cdot 3^{12} \cdot |A|^4 \\
&= 2^6 \cdot 3^{(3+6+12)} \cdot (-2)^4 \\
&= 2^6 \cdot 3^{21} \cdot 2^4 \\
&= 2^{10} \cdot 3^{21}
\end{aligned}
$$
Comparing this with $2^{m+n} \cdot 3^{mn}$:
$$
\begin{aligned}
mn &= 21 \\
m+n &= 10
\end{aligned}
$$
Since $m > n$, the factors of 21 that sum to 10 are 7 and 3.
$$ \therefore m = 7, \quad n = 3 $$
Value required:
$$
\begin{aligned}
4m + 2n &= 4(7) + 2(3) \\
&= 28 + 6 \\
&= 34
\end{aligned}
$$
Ans. (34)