JEE Main 2025 Apr 3 Shift 1, Mathematics Q1: Algebra Of Matrices

For an $n\times n$ matrix $B$, use $|kB|=k^n|B|$ and $|\text{adj}B|=|B{|}^{n-1}$. Here $n=3$, so $|kB|=k^3|B|$ and $|\text{adj}B|=|B{|}^2$.

Peel the expression from the inside out:

$|\text{adj}(2A)|=|2A{|}^2=(2^3|A|{)}^2=2^6|A{|}^2.$

Inside the outer adj, let $M=3A\text{adj}(2A)$. Then

$|M|=3^3|A|\cdot |\text{adj}(2A)|=27\cdot |A|\cdot 2^6|A{|}^2=2^6\cdot 3^3\cdot |A{|}^3.$

$|\text{adj}(M)|=|M{|}^2=2^{12}\cdot 3^6\cdot |A{|}^6.$

Finally,

$|2\text{adj}(M)|=2^3\cdot |\text{adj}(M)|=2^{15}\cdot 3^6\cdot |A{|}^6.$

Substituting $|A|=5$: $|A{|}^6=5^6$, so the expression equals $2^{15}\cdot 3^6\cdot 5^6$.

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