Let $A=\left(\begin{array}{ccc}1 & 0 & 0 \\ 0 & 4 & -1 \\ 0 & 12 & -3\end{array}\right)$. Then the sum of the diagonal elements of the matrix $(A+I)^{11}$ is equal to:

JEE Main · 2023 · 31 Jan 2023 · Shift 1 · Matrices

Let

A = 1000412 0 - 1 - 3

. Then the sum of the diagonal elements of the matrix

(A + I)^{11}

is equal to:

Answer: (C)

The lower block

(412 - 1 - 3)

squares to itself, and the leading entry

1

is unchanged, so

A^{2} = A

and hence

A^{n} = A

for every

n \geq 1

. Expanding by the binomial theorem,

(A + I)^{11} = I + (\sum_{k = 1}^{11} (k 11)) A = I + (2^{11} - 1) A = I + 2047 A

. The trace of

A

1 + 4 - 3 = 2

, so the sum of the diagonal entries of

(A + I)^{11}

3 + 2047 \times 2 = 4097

JEE Main · 2023 · 31 Jan 2023 · Shift 1 · Single correct option