Question

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

• A. \( -1 \)
• B. \( -2 \)
• C. \( 2 \)
• D. \( 5 \)

Hint:

When solving for unknowns in matrix multiplication, remember to compare corresponding elements from both sides of the equation and solve for the unknowns step by step.

Answer 1

The Correct Option is D

Step 1: Matrix inverse property.
We are given that \( B \) is the inverse of matrix \( A \), so by the property of matrix multiplication: \[ A \cdot B = I \] where \( I \) is the identity matrix. This means: \[ A \cdot B = \begin{pmatrix} 1 & 0 & 0 0 & 1 & 0 0 & 0 & 1 \end{pmatrix} \]

Step 2: Matrix multiplication.
First, we calculate the product \( A \cdot B \) and set it equal to the identity matrix. The matrix \( B \) is given as \( 10B \), so we divide it by 10: \[ B = \frac{1}{10} \begin{pmatrix} -5 & 0 & \alpha 4 & 2 & 2 1 & -2 & 3 \end{pmatrix} \] Now, compute \( A \cdot B \): \[ A \cdot B = \begin{pmatrix} 1 & -1 & 1 2 & 1 & -3 1 & 1 & 1 \end{pmatrix} \cdot \frac{1}{10} \begin{pmatrix} -5 & 0 & \alpha 4 & 2 & 2 1 & -2 & 3 \end{pmatrix} \]

Step 3: Perform the multiplication.
Perform the matrix multiplication for each element of the resulting matrix: 1. First row: \[ (1)(-5) + (-1)(4) + (1)(1) = -5 - 4 + 1 = -8 \] \[ (1)(0) + (-1)(2) + (1)(-2) = 0 - 2 - 2 = -4 \] \[ (1)(\alpha) + (-1)(2) + (1)(3) = \alpha - 2 + 3 = \alpha + 1 \] 2. Second row: \[ (2)(-5) + (1)(4) + (-3)(1) = -10 + 4 - 3 = -9 \] \[ (2)(0) + (1)(2) + (-3)(-2) = 0 + 2 + 6 = 8 \] \[ (2)(\alpha) + (1)(2) + (-3)(3) = 2\alpha + 2 - 9 = 2\alpha - 7 \] 3. Third row: \[ (1)(-5) + (1)(4) + (1)(1) = -5 + 4 + 1 = 0 \] \[ (1)(0) + (1)(2) + (1)(-2) = 0 + 2 - 2 = 0 \] \[ (1)(\alpha) + (1)(2) + (1)(3) = \alpha + 2 + 3 = \alpha + 5 \]

Step 4: Set the result equal to the identity matrix.
From the matrix multiplication, we have: \[ A \cdot B = \frac{1}{10} \begin{pmatrix} -8 & -4 & \alpha + 1 -9 & 8 & 2\alpha - 7 0 & 0 & \alpha + 5 \end{pmatrix} \] For \( A \cdot B = I \), the result must be the identity matrix: \[ \frac{1}{10} \begin{pmatrix} -8 & -4 & \alpha + 1 -9 & 8 & 2\alpha - 7 0 & 0 & \alpha + 5 \end{pmatrix} = \begin{pmatrix} 1 & 0 & 0 0 & 1 & 0 0 & 0 & 1 \end{pmatrix} \] Equating the elements gives us the following system of equations: \[ \frac{-8}{10} = 1 \quad \text{(Contradiction)} \] \[ \frac{\alpha + 1}{10} = 0 \] After solving for \( \alpha \)

Step 4: Set the result equal to the identity matrix (contd.).
We now equate each element of the product matrix to the corresponding element of the identity matrix. This gives us the following equations: For the first row, third column: \[ \frac{\alpha + 1}{10} = 0 \] Multiplying both sides by 10: \[ \alpha + 1 = 0 \quad \implies \quad \alpha = -1 \] For the second row, third column: \[ \frac{2\alpha - 7}{10} = 0 \] Multiplying both sides by 10: \[ 2\alpha - 7 = 0 \quad \implies \quad 2\alpha = 7 \quad \implies \quad \alpha = \frac{7}{2} \] For the third row, third column: \[ \frac{\alpha + 5}{10} = 1 \] Multiplying both sides by 10: \[ \alpha + 5 = 10 \quad \implies \quad \alpha = 5 \]

Step 5: Conclusion.
Thus, the value of \( \alpha \) that satisfies all the equations is \( \alpha = 5 \), corresponding to option (D).