Let A be a matrix of order \(3\times 3\) and \(|A|=5\). If \(|2adj(3\mathrm{A}adj(2\mathrm{A}))|={2}^{\alpha }\cdot {3}^…

First, use property (3) for \(\text{adj}(2A)\): Since \(n=3\), \(\text{adj}(2A)={2}^{3−1}\text{adj}(A)=4\text{ adj}(A)\)

Now substitute back:

\(3A⋅\text{adj}(2A)=3A⋅4\text{ adj}(A)=12⋅(A⋅\text{adj}(A))\)

Using property (4): \(A⋅\text{adj}(A)=\mathrm{∣}A\mathrm{∣}I=5I\)

\(\text{  }⟹\text{  }3A⋅\text{adj}(2A)=12\times 5I=60I\)

For scalar identity matrix \(kI\), \(\text{adj}(kI)={k}^{n−1}I\) (for \(n=3\), this is \({k}^{2}I\)):

\(\text{adj}(60I)={60}^{2}I\)

First, \(2⋅\text{adj}(60I)=2\times {60}^{2}I\) Using property (1), determinant of \(mI\) (3x3) is \({m}^{3}\):

\(∣2\times {60}^{2}I∣={(2\times {60}^{2})}^{3}\)

Write \(60={2}^{2}\times 3\times 5\), so:

\({(2\times ({2}^{2}\times 3\times 5{)}^{2})}^{3}={2}^{3}\times ({2}^{2}\times 3\times 5{)}^{6}\) \(={2}^{3}\times {2}^{12}\times {3}^{6}\times {5}^{6}=\)

Comparing with \({2}^{\alpha }⋅{3}^{\beta }⋅{5}^{\gamma }\):

\(\alpha =15\), \(\beta =6\), \(\gamma =6\)

\(\alpha +\beta +\gamma =15+6+6=27\)