JEE Main 2025 Apr 2 Shift 1, Mathematics Q5: Algebra Of Matrices

Finding the entries of A: The condition $\det (A)=0$ gives $\alpha \beta +6=0$, i.e. $\alpha \beta =-6$. Combined with $\alpha +\beta =1$ and $\alpha >0$, the values are $\alpha =3$, $\beta =-2$.

$$A=\left[\begin{array}{c}3-1\\ 6-2\end{array}\right]$$

Key idempotent-like property: Compute $A^2$:

$$A^2=\left[\begin{array}{c}3-1\\ 6-2\end{array}\right]\left[\begin{array}{c}3-1\\ 6-2\end{array}\right]=\left[\begin{array}{c}9-6-3+2\\ 18-12-6+4\end{array}\right]=\left[\begin{array}{c}3-1\\ 6-2\end{array}\right]=A$$

So $A^2=A$, which means $A^k=A$ for every $k\ge 1$.

Expanding $(I+A{)}^8$: Using the binomial theorem (matrices $I$ and $A$ commute):

$$(I+A{)}^8=\sum _{k=0}^8(\genfrac{}{}{0ex}{}{8}{k})A^k=I+A\sum _{k=1}^8(\genfrac{}{}{0ex}{}{8}{k})=I+(2^8-1)A=I+255A$$

$$=\left[\begin{array}{c}10\\ 01\end{array}\right]+255\left[\begin{array}{c}3-1\\ 6-2\end{array}\right]=\left[\begin{array}{c}766-255\\ 1530-509\end{array}\right]$$

$\therefore$ Option (4) is correct.