$ \begin{vmatrix} x + 1 & x - 1 \\ x^2 + x + 1 & x^2 - x + 1 \end{vmatrix} $ is equal to:
$ \begin{vmatrix} x + 1 & x - 1 \\ x^2 + x + 1 & x^2 - x + 1 \end{vmatrix} $ is equal to:
Show Hint
- \( 2x^3 \)
- \( 2 \)
- \( 0 \)
- \( 2x^3 - 2 \)
The Correct Option is B
Solution and Explanation
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:
\[ \boxed{2} \]
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