Let A be a matrix of order 3 3 and |A|=5.

Question

Let $A$ be a matrix of order $3 	imes 3$ and $|A|=5$. If $|2 \operatorname{adj}(3 A \operatorname{adj}(2 A))|=2^\alpha \cdot 3^\beta \cdot 5^\gamma, \alpha, \beta, \gamma \in N$, then $\alpha+\beta+\gamma$ is equal to

Accepted Answer

To find the expression $|2 \operatorname{adj}(3 A \operatorname{adj}(2 A))|$, we break it down as follows:

Recognize that:

$    |2 \operatorname{adj}(3 A \operatorname{adj}(2 A))| = 2^3 |3A (\operatorname{adj}(2A))|^2    $

Apply properties of determinants:

$    = 2^3 (3^3)^2 |A|^2 \left|\operatorname{adj}(2A)ight|^2    $

Further simplify using $|\operatorname{adj}(B)| = |B|^{n-1}$ for a $3 	imes 3$ matrix:

$    = 2^3 \cdot 3^6 \cdot 5^2 \cdot (|2A|^2)^2    $

Simplify $|2A|$:

$    = 2^3 \cdot 3^6 \cdot 5^2 \cdot (2^3)^4 \cdot |A|^4    $

Continue to simplify:

$    = 2^3 \cdot 3^6 \cdot 5^2 \cdot (2^3)^4 \cdot 5^4    $

Expand and combine powers:

$    = 2^{15} \cdot 3^6 \cdot 5^6    $

Therefore, $\alpha = 15$, $\beta = 6$, and $\gamma = 6$. So, $\alpha + \beta + \gamma = 15 + 6 + 6 = 27$.

Solutions · 1