Question

$ \begin{vmatrix} x + 1 & x - 1 \\ x^2 + x + 1 & x^2 - x + 1 \end{vmatrix} $ is equal to:

• A. \( 2x^3 \)
• B. \( 2 \)
• C. \( 0 \)
• D. \( 2x^3 - 2 \)

Hint:

The determinant of a \( 2 \times 2 \) matrix is calculated as the product of the main diagonal elements minus the product of the off-diagonal elements.

Answer 1

The Correct Option is B

The determinant of a \( 2 \times 2 \) matrix is given by:

\[ \begin{vmatrix} a & b \\ c & d \end{vmatrix} = ad - bc. \]

For the given matrix:

\[ \begin{vmatrix} x + 1 & x - 1 \\ x^2 + x + 1 & x^2 - x + 1 \end{vmatrix}, \]

we have:

\[ a = x + 1, \quad b = x - 1, \quad c = x^2 + x + 1, \quad d = x^2 - x + 1. \]

Step 1: Calculate the determinant.

\[ \text{Determinant} = (x + 1)(x^2 - x + 1) - (x - 1)(x^2 + x + 1). \]

Step 2: Expand the terms.

\[ (x + 1)(x^2 - x + 1) = x^3 - x^2 + x + x^2 - x + 1 = x^3 + x + 1, \]

\[ (x - 1)(x^2 + x + 1) = x^3 + x^2 + x - x^2 - x - 1 = x^3 - 1. \]

Step 3: Simplify the determinant.

\[ \text{Determinant} = (x^3 + x + 1) - (x^3 - 1) = x^3 + x + 1 - x^3 + 1 = 2. \]

Final Answer:

\[ \boxed{2} \]