Matrices – Powers of Matrices – JEE Main 2 Apr 2025 Shift 1

Solution:

Given $\det(A) = 0$:
$$ \alpha\beta – (-6) = 0 \Rightarrow \alpha\beta = -6 $$

Given $\alpha + \beta = 1$. The values of $\alpha$ and $\beta$ are roots of the quadratic equation:
$$ t^2 – (\alpha+\beta)t + \alpha\beta = 0 \Rightarrow t^2 – t – 6 = 0 $$
$$ (t-3)(t+2) = 0 \Rightarrow t = 3, -2 $$

Since $\alpha > \beta$, we have $\alpha = 3$ and $\beta = -2$.
$$ A = \begin{bmatrix} 3 & -1 \\ 6 & -2 \end{bmatrix} $$

Calculate $A^2$:
$$ A^2 = \begin{bmatrix} 3 & -1 \\ 6 & -2 \end{bmatrix} \begin{bmatrix} 3 & -1 \\ 6 & -2 \end{bmatrix} = \begin{bmatrix} 9-6 & -3+2 \\ 18-12 & -6+4 \end{bmatrix} = \begin{bmatrix} 3 & -1 \\ 6 & -2 \end{bmatrix} = A $$

Since $A^2 = A$, it follows that $A^n = A$ for all $n \in \mathbb{N}$.

Using the Binomial expansion for matrices (since $I$ and $A$ commute):
$$ (I+A)^8 = \binom{8}{0}I + \binom{8}{1}A + \binom{8}{2}A^2 + \dots + \binom{8}{8}A^8 $$

Substitute $A^k = A$:
$$ (I+A)^8 = I + A \left( \binom{8}{1} + \binom{8}{2} + \dots + \binom{8}{8} \right) $$

The sum of binomial coefficients is $\sum_{r=0}^n \binom{n}{r} = 2^n$. Thus, $\sum_{r=1}^8 \binom{8}{r} = 2^8 – 1 = 255$.
$$ (I+A)^8 = I + 255A $$

$$ = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} + 255 \begin{bmatrix} 3 & -1 \\ 6 & -2 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} + \begin{bmatrix} 765 & -255 \\ 1530 & -510 \end{bmatrix} $$

$$ = \begin{bmatrix} 766 & -255 \\ 1530 & -509 \end{bmatrix} $$

Ans. (4)