If \(f(x) = \begin{vmatrix}x & 1 & 0\\ 0 & x & 1\\ 1 & 0 & x\end{vmatrix}\), then \(f(x)\) is equal to
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(x^{3} + 1\)
Along the first row the third entry is 0, so the expansion has only two terms.
\(\displaystyle f(x) = x\begin{vmatrix}x & 1\\ 0 & x\end{vmatrix} - 1\begin{vmatrix}0 & 1\\ 1 & x\end{vmatrix} + 0\)
\(f(x) = x(x^{2} - 0) - 1(0 - 1)\)
\(f(x) = x^{3} + 1\)
Check: at x = 1 the rows are (1, 1, 0), (0, 1, 1), (1, 0, 1), and \(1(1 - 0) - 1(0 - 1) + 0 = 2 = 1^{3} + 1\).