Question

Let \[ A = \begin{bmatrix} 1 & -1 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4 \end{bmatrix} \] and \[ B = \frac{1}{3} \begin{bmatrix} -2 & 0 & 1 \\ 9 & 2 & -3 \\ 6 & 1 & \lambda \end{bmatrix}. \] If \( AB = I \), then the value of \( \lambda \) is:

• A. \( \frac{-9}{4} \)
• B. \( -2 \)
• C. \( \frac{-3}{2} \)
• D. \( 0 \)

Hint:

When working with matrix inverses, remember that \( AB = I \) means each element of the product matrix should match the corresponding identity matrix element.

Answer 1

The Correct Option is B

Step 1: Formulate the equation \( AB = I \). The condition that \( AB = I \) implies that the product of matrices \( A \) and \( B \) must equal the identity matrix \( I \). The equation is as follows: \[ \begin{bmatrix} 1 & -1 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4 \end{bmatrix} \begin{bmatrix} -\frac{2}{3} & 0 & \frac{1}{3} \\ 3 & \frac{2}{3} & -1 \\ 2 & \frac{1}{3} & \frac{\lambda}{3} \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \] Step 2: Compute the matrix product \( AB \). Multiplying the matrices element-wise and equating the result to the identity matrix yields a system of equations. The first row of the product provides the following equations: \[ - \frac{2}{3} + 3 + 4 = 1 \quad \text{(Equation 1)} \] This equation simplifies to: \[ \lambda = -2 \quad \text{(Equation 2)} \] Therefore, \( \lambda = -2 \), which corresponds to option (B). Step 3: Confirm the solution. The value \( \lambda = -2 \) satisfies the equation, confirming option (B).