Question

Let \[ A = \begin{pmatrix} 1 & -1 & 1 \\ 2 & 1 & -3 \\ 1 & 1 & 1 \end{pmatrix}, \quad 10B = \begin{pmatrix} 4 & 2 & 2 \\ -5 & 0 & \alpha \\ 1 & -2 & 3 \end{pmatrix} \] If \( B \) is the inverse of \( A \), then the value of \( \alpha \) is:

• A. 4
• B. \( -4 \)
• C. 3
• D. 5

Hint:

When solving for unknowns in matrix products, focus on the relevant row and column to simplify the calculation. Always check the consistency of the result with the identity matrix.

Answer 1

The Correct Option is D

We are given that \( B \) is the inverse of \( A \), which means: \[ A \times B = I \] where \( I \) is the identity matrix: \[ I = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} \] We are also given that: \[ 10B = \begin{pmatrix} 4 & 2 & 2 \\ -5 & 0 & \alpha \\ 1 & -2 & 3 \end{pmatrix} \] Thus: \[ B = \frac{1}{10} \begin{pmatrix} 4 & 2 & 2 \\ -5 & 0 & \alpha \\ 1 & -2 & 3 \end{pmatrix} \] To find the value of \( \alpha \), we multiply \( A \) and \( B \) and set the result equal to the identity matrix. We focus on the second row and third column of the product \( A \times B \), since this will involve \( \alpha \). The second row of \( A \) is: \[ (2, 1, -3) \] The third column of \( B \) is: \[ \left( \frac{2}{10}, \frac{\alpha}{10}, \frac{3}{10} \right) = \left( \frac{1}{5}, \frac{\alpha}{10}, \frac{3}{10} \right) \] Now, calculate the dot product of these two vectors: \[ 2 \times \frac{1}{5} + 1 \times \frac{\alpha}{10} + (-3) \times \frac{3}{10} = 0 \] Simplifying the terms: \[ \frac{2}{5} + \frac{\alpha}{10} - \frac{9}{10} = 0 \] Multiplying through by 10 to eliminate fractions: \[ 4 + \alpha - 9 = 0 \] \[ \alpha - 5 = 0 \] \[ \alpha = 5 \] Thus, the value of \( \alpha \) is 5, making the correct answer Option D.