Let $A$ be a matrix of order $3 \times 3$ and $|A|=5$. If $|2\text{adj}(3A\text{adj}(2A))|=2^{\alpha}\cdot3^{\beta}\cdot5^{\gamma}$, where $\alpha, \beta, \gamma \in \mathbb{N}$, then $\alpha+\beta+\gamma$ is equal to
- (1) 25
- (2) 26
- (3) 27
- (4) 28
Solution:
$$|kA| = k^n|A|$$
$$|\text{adj}(A)| = |A|^{n-1}$$
$$|2\text{adj}(3A\text{adj}(2A))| = 2^3|\text{adj}(3A\text{adj}(2A))|$$
$$= 2^3(|3A\text{adj}(2A)|)^{3-1}$$
$$= 2^3(|3A| \cdot |\text{adj}(2A)|)^2$$
$$= 2^3(3^3|A| \cdot (|2A|)^{3-1})^2$$
$$= 2^3(3^3|A| \cdot (|2A|)^2)^2$$
$$= 2^3(3^3|A| \cdot (2^3|A|)^2)^2$$
$$= 2^3(3^3|A| \cdot 2^6|A|^2)^2$$
$$= 2^3(2^6 \cdot 3^3|A|^3)^2$$
$$= 2^3 \cdot 2^{12} \cdot 3^6|A|^6$$
$$= 2^{15} \cdot 3^6|A|^6$$
$$= 2^{15} \cdot 3^6 \cdot 5^6$$
$$2^{\alpha} \cdot 3^{\beta} \cdot 5^{\gamma} = 2^{15} \cdot 3^6 \cdot 5^6$$
$$\alpha=15, \beta=6, \gamma=6$$
$$\alpha+\beta+\gamma = 15+6+6 = 27$$
Ans. (3)