Let $A=\begin{bmatrix}\alpha & -1 \\ 6 & \beta\end{bmatrix}$, $\alpha>0$, such that $\det(A)=0$ and $\alpha+\beta=1$. If $I$ denotes $2 \times 2$ identity matrix, then the matrix $(I+A)^{8}$ is:

(D) $\begin{bmatrix}766 & -255 \\ 1530 & -509\end{bmatrix}$

(A) $\begin{bmatrix}4 & -1 \\ 6 & -1\end{bmatrix}$

(B) $\begin{bmatrix}257 & -64 \\ 514 & -127\end{bmatrix}$

(C) $\begin{bmatrix}1025 & -511 \\ 2024 & -1024\end{bmatrix}$

JEE Main · 2025 · 02 Apr 2025 · Shift 1 · Matrices

Let

A = [α 6 - 1 β]

α > 0

, such that

det (A) = 0

and

α + β = 1

. If

I

denotes

2 \times 2

identity matrix, then the matrix

(I + A)^{8}

is:

Answer: (D)

Given

det (A) = 0

α β + 6 = 0

, so

α β = - 6

.

With

α + β = 1

and

α β = - 6

, solving gives

α = 3, β = - 2

.

So

A = [36 - 1 - 2]

.

Computing

A^{2} = [36 - 1 - 2] [36 - 1 - 2] = [36 - 1 - 2] = A

.

Thus

A

is idempotent:

A^{2} = A

, so

A^{n} = A

for all

n \geq 1

.

Using binomial expansion:

(I + A)^{8} = I + (1 8) A + (2 8) A^{2} + \dots + (8 8) A^{8} = I + A (2^{8} - 1) = I + 255 A

= [1001] + 255 [36 - 1 - 2] = [7661530 - 255 - 509]

JEE Main · 2025 · 02 Apr 2025 · Shift 1 · Single correct option