JEE 2022 — Mathematics PYQ

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Question

JEE 2022 — Mathematics PYQ

JEE | Mathematics | 2022

Let $A = \begin{pmatrix} 2 & -1 \\ 0 & 2 \end{pmatrix}$ . If $B = I - {}^5C_1 (\text{adj } A) + {}^5C_2 (\text{adj } A)^2 - \dots - {}^5C_5 (\text{adj } A)^5$ , then the sum of all elements of the matrix $B$ is:

Choose the correct answer:

Explanation

Solution

1. Find the Adjoint of A:

For a $2 \times 2$ matrix $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$ , the adjoint is $\text{adj } A = \begin{pmatrix} d & -b \\ -c & a \end{pmatrix}$ .

\text{adj } A = \begin{pmatrix} 2 & 1 \\ 0 & 2 \end{pmatrix}

2. Simplify the expression for B:

The expression for $B$ follows the binomial expansion $(I - X)^n$ :

(I - X)^5 = I - {}^5C_1 X + {}^5C_2 X^2 - \dots - {}^5C_5 X^5

Let $X = \text{adj } A$ . Then:

B = (I - \text{adj } A)^5

3. Calculate $(I - \text{adj } A)$ :

I - \text{adj } A = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} - \begin{pmatrix} 2 & 1 \\ 0 & 2 \end{pmatrix} = \begin{pmatrix} -1 & -1 \\ 0 & -1 \end{pmatrix}

4. Find $B = \begin{pmatrix} -1 & -1 \\ 0 & -1 \end{pmatrix}^5$ :

Let $M = \begin{pmatrix} -1 & -1 \\ 0 & -1 \end{pmatrix} = -(I + N)$ , where $N = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$ .

Note that $N^2 = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}$ , so $N^k = 0$ for $k \ge 2$ .

Using binomial expansion:

M^5 = (-1)^5 (I + N)^5 = -1 (I + 5N + 10N^2 + \dots) = -(I + 5N)

B = -\left[ \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} + \begin{pmatrix} 0 & 5 \\ 0 & 0 \end{pmatrix} \right] = -\begin{pmatrix} 1 & 5 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} -1 & -5 \\ 0 & -1 \end{pmatrix}

5. Sum of elements:

Sum $= (-1) + (-5) + 0 + (-1) = -7$ .

Correct Option: (C) $-7$