Step 1: Set up the equation \( AB = I \)
We know that \( AB = I \), so multiplying the matrices \( A \) and \( B \) should yield the identity matrix \( I \). The equation is: \[ \begin{bmatrix} 1 & -1 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4 \end{bmatrix} \begin{bmatrix} -\frac{2}{3} & 0 & \frac{1}{3} \\ 3 & \frac{2}{3} & -1 \\ 2 & \frac{1}{3} & \frac{\lambda}{3} \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \]
Step 2: Calculate the product of \( A \) and \( B \)
We multiply the matrices element by element and equate the result to the identity matrix. The equations formed from the first row of the resulting matrix are: \[ - \frac{2}{3} + 3 + 4 = 1 \quad \text{(First equation)} \] which simplifies to: \[ \lambda = -2 \quad \text{(Second equation)} \] Thus, \( \lambda = -2 \), which corresponds to option (B).
Step 3: Verify the options
The value \( \lambda = -2 \) satisfies the equation, matching option (B).
Show that \( R \) is an equivalence relation. Also, write the equivalence class \([2]\).
| List-I | List-II |
| (A) Absolute maximum value | (I) 3 |
| (B) Absolute minimum value | (II) 0 |
| (C) Point of maxima | (III) -5 |
| (D) Point of minima | (IV) 4 |
In number theory, it is often important to find factors of an integer \( N \). The number \( N \) has two trivial factors, namely 1 and \( N \). Any other factor, if it exists, is called a non-trivial factor of \( N \). Naresh has plotted a graph of some constraints (linear inequations) with points \( A(0, 50) \), \( B(20, 40) \), \( C(50, 100) \), \( D(0, 200) \), and \( E(100, 0) \). This graph is constructed using three non-trivial constraints and two trivial constraints. One of the non-trivial constraints is \( x + 2y \geq 100 \).

Based on the above information, answer the following questions: