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).
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 |