The value of \(\begin{vmatrix}x^2 - x + 1&x - 1\\x + 1&x + 1\end{vmatrix}\) is equal to
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(x^3 - x^2 + 2\)
Expanding the 2 × 2 determinant:
\( \displaystyle \Delta = (x^2 - x + 1)(x + 1) - (x - 1)(x + 1) \)
By the sum of cubes, \((x + 1)(x^2 - x + 1) = x^3 + 1\), and \((x - 1)(x + 1) = x^2 - 1\):
\(\Delta = x^3 + 1 - (x^2 - 1)\)
\(\Delta = x^3 - x^2 + 2\)
Shortcut: Put \(x = 0\): \(\Delta = (1)(1) - (-1)(1) = 2\). Only Options (2) and (3) give 2. Put \(x = 1\): \(\Delta = (1)(2) - (0)(2) = 2\); Option (3) gives 2 and Option (2) gives 4.
NCERT (2026-27 edition):
Determinant
