JEE Mathematics 2023 Question with Solution

Given: $A = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 4 & -1 \\ 0 & 12 & -3 \end{pmatrix}$

Find: The sum of the diagonal elements of $(A+I)^{11}$.

From the solution working,

$$A^2 = A$$

Hence,

$$A^3 = A^4 = \cdots = A$$

So for every positive integer $k \ge 1$,

$$A^k = A$$

Now apply the binomial expansion:

$$(A+I)^{11} = \sum_{k=0}^{11} \binom{11}{k} A^k I^{11-k}$$

Using $A^k = A$ for $k \ge 1$, this becomes

$$(A+I)^{11} = \left(\binom{11}{0}+\binom{11}{1}+\cdots+\binom{11}{10}\right)A + I$$

Therefore,

$$(A+I)^{11} = (2^{11}-1)A + I = 2047A + I$$

Now the sum of diagonal elements is the trace:

$$\text{trace}(2047A + I) = 2047\,\text{trace}(A) + \text{trace}(I)$$

The diagonal entries of $A$ are $1, 4, -3$, so

$$\text{trace}(A) = 1+4-3 = 2$$

Also, for a $3 \times 3$ identity matrix,

$$\text{trace}(I) = 3$$

Hence,

$$\text{trace}((A+I)^{11}) = 2047 \times 2 + 3 = 4097$$

Therefore, the sum of the diagonal elements is $4097$. The numerical value matches option C, but the solution explicitly labels the correct option as D, so the source contains an option-label discrepancy.

Given: $A = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 4 & -1 \\ 0 & 12 & -3 \end{pmatrix}$

Find: The sum of diagonal elements of $(A+I)^{11}$.

First compute the key property shown in the solution:

$$A^2 = A$$

So the matrix is idempotent. Hence all higher powers satisfy

$$A^n = A \quad \text{for all } n \ge 1$$

Now expand:

$$(A+I)^{11} = \binom{11}{0}I + \binom{11}{1}A + \binom{11}{2}A^2 + \cdots + \binom{11}{11}A^{11}$$

Replacing every power $A^r$ with $A$ for $r \ge 1$,

$$(A+I)^{11} = I + \left[\binom{11}{1}+\binom{11}{2}+\cdots+\binom{11}{11}\right]A$$

Using

$$\sum_{k=0}^{11} \binom{11}{k} = 2^{11}$$

we get

$$\binom{11}{1}+\binom{11}{2}+\cdots+\binom{11}{11} = 2^{11}-1 = 2047$$

Thus,

$$(A+I)^{11} = I + 2047A$$

Now take trace:

$$\text{trace}(A) = 1+4-3 = 2$$

Therefore,

$$\text{trace}(I+2047A) = 3 + 2047 \times 2 = 4097$$

Therefore, the required sum is $4097$.