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 $|2adj(3Aadj(2A))|={2}^{\alpha }\cdot {3}^{\beta }\cdot {5}^{\gamma },\alpha ,\beta ,\gamma \in N$, then $\alpha +\beta +\gamma $ is equal to

Accepted Answer

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

Recognize that:

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

Apply properties of determinants:

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

Further simplify using $|adj(B)|=|B{|}^{n-1}$ 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