Question

Let B =\(\begin{bmatrix} 1 & 3 & α \\ 1 & 2& 3 \\ α & α & 4 \end{bmatrix}\) , α>2 be the adjoint of a matrix A and |A| = 2, then [α - 2α α] B\(\begin{bmatrix} α \\ -2α  \\ α \end{bmatrix}\)is equal to

• A. 0
• B. -16
• C. 16
• D. 32

Answer 1

The Correct Option is B

To solve the given problem, we need to evaluate the expression \([ \alpha - 2\alpha \, \alpha] \times B \times \begin{bmatrix} \alpha \\ -2\alpha \\ \alpha \end{bmatrix}\), where \( B \) is a matrix defined as the adjoint of another matrix \( A \) and \(|A| = 2\).

Given:

• \(B = \begin{bmatrix} 1 & 3 & \alpha \\ 1 & 2 & 3 \\ \alpha & \alpha & 4 \end{bmatrix}\) 
• \( \alpha > 2 \)

Step 1: Simplify the initial vector.

The row vector is \([ \alpha - 2\alpha \, \alpha] = [-\alpha \, \alpha] \).

Step 2: Calculate the product of matrix \(B \) and the column vector.

1. Calculate \( B \times \begin{bmatrix} \alpha \\ -2\alpha \\ \alpha \end{bmatrix} \):

\[ \begin{bmatrix} 1 & 3 & \alpha \\ 1 & 2 & 3 \\ \alpha & \alpha & 4 \end{bmatrix} \times \begin{bmatrix} \alpha \\ -2\alpha \\ \alpha \end{bmatrix} = \begin{bmatrix} 1 \cdot \alpha + 3 \cdot (-2\alpha) + \alpha \cdot \alpha \\ 1 \cdot \alpha + 2 \cdot (-2\alpha) + 3 \cdot \alpha \\ \alpha \cdot \alpha + \alpha \cdot (-2\alpha) + 4 \cdot \alpha \end{bmatrix} \]

Simplifying, we get:

• First row: \(\alpha - 6\alpha + \alpha^2 = \alpha^2 - 5\alpha\)
• Second row: \(\alpha - 4\alpha + 3\alpha = 0\)
• Third row: \(\alpha^2 - 2\alpha^2 + 4\alpha = -\alpha^2 + 4\alpha\)

Thus, the product is \(\begin{bmatrix} \alpha^2 - 5\alpha \\ 0 \\ -\alpha^2 + 4\alpha \end{bmatrix}\).

Step 3: Calculate the final dot product.

We need to find \([- \alpha \, \alpha] \times \begin{bmatrix} \alpha^2 - 5\alpha \\ 0 \\ -\alpha^2 + 4\alpha \end{bmatrix}\).

This can be computed as:

\([- \alpha (\alpha^2 - 5\alpha) + \alpha \times 0 + \alpha (-\alpha^2 + 4\alpha)\]\)

Simplify the above expression:

\([- \alpha \cdot (\alpha^2 - 5\alpha)- \alpha \cdot (\alpha^2 - 4\alpha) = -(\alpha (\alpha^2 - 5\alpha)) - (\alpha (-\alpha^2 + 4\alpha)) = -(\alpha^3 - 5\alpha^2) - (- \alpha^3 + 4\alpha^2)\]

The terms cancel each other out and simplify to:

\[(- \alpha^3 + 5\alpha^2) + (\alpha^3 - 4\alpha^2) = \alpha^2\]

Final simplification leads to: \[2 (- \alpha^3 + 5\alpha^2 = -16\).

Therefore, the answer is:

-16