If \( \begin{vmatrix}1 & x & x^2\\x^2 & 1 & x\\x & x^2 & 1\end{vmatrix}=(x^{\lambda}-1)^{\mu} \) then the value of λ and μ are given by
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( \lambda=3,\mu=2 \)
Expanding the determinant along the first row:
\( \displaystyle \begin{vmatrix}1 & x & x^2\\x^2 & 1 & x\\x & x^2 & 1\end{vmatrix}=1(1-x^3)-x(x^2-x^2)+x^2(x^4-x) \)
\( \displaystyle =1-x^3-0+x^6-x^3 \)
\( \displaystyle =x^6-2x^3+1 \)
\( \displaystyle =(x^3-1)^2 \)
Comparing with \( (x^{\lambda}-1)^{\mu} \), λ = 3 and μ = 2.
Shortcut: Put x = 2. The left side is \( 1(1-8)-2(4-4)+4(16-2)=-7+56=49 \). On the right, λ = 3, μ = 2 gives \( 7^2=49 \), while the other options give \( 3^2=9 \), \( 3^3=27 \) and \( 7^3=343 \).