Question

The determinant \(\begin{vmatrix} x+1 & x-1 \\ x^2+x+1 & x^2-x+1 \end{vmatrix}\) is equal to:

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

Hint:

When solving for determinants of \(2 \times 2\) matrices, always expand as \((a_{11} \cdot a_{22}) - (a_{12} \cdot a_{21})\) and simplify carefully.

Answer 1

The Correct Option is B

The given determinant is: \[ \begin{vmatrix} x+1 & x-1 \\ x^2+x+1 & x^2-x+1 \end{vmatrix}. \]

 Using the formula for the determinant of a \(2 \times 2\) matrix: \[ \text{Determinant} = \text{(Diagonal 1 product)} - \text{(Diagonal 2 product)}. \] 

We calculate: \[ \text{Diagonal 1 product} = (x+1)(x^2-x+1), \] \[ \text{Diagonal 2 product} = (x-1)(x^2+x+1). \] 

Expanding each term: \[ (x+1)(x^2-x+1) = x^3 - x^2 + x + x^2 - x + 1 = x^3 + x + 1, \] \[ (x-1)(x^2+x+1) = x^3 + x^2 + x - x^2 - x - 1 = x^3 - 1. \] 

Subtracting the two products: \[ \text{Determinant} = (x^3 + x + 1) - (x^3 - 1) = x^3 + x + 1 - x^3 + 1 = 2. \] 

Hence, the value of the determinant is \(2\), and the correct answer is (B).