Matrices & Determinants
Matrices and Determinants
star_batch_jee_advanced_2025
Grade 12
Question:
If $f(x) = \begin{vmatrix} (1+x)^a & (1+2x)^b & 1 \\ 1 & (1+x)^a & (1+2x)^b \\ (1+2x)^b & 1 & (1+x)^a \end{vmatrix}$, $a,b$ being natural numbers, then:
constant term of $f(x)$ is $0$
coefficient of $x$ in $f(x)$ is $0$
constant term of $f(x)$ is $a-b$
constant term of $f(x)$ is $a+b$
Step-by-Step Solution
Key Concept: A cyclic determinant with identical row sums has special properties: the constant term vanishes and the linear coefficient also vanishes due to symmetry.
Notice that the determinant has a cyclic structure: each row is a cyclic shift of the previous row. For a cyclic matrix with this specific structure, we can use the property that $f(x) = (1+x)^a + (1+2x)^b + 1)((1+x)^{2a} + (1+2x)^{2b} + 1 - (1+x)^a(1+2x)^b - (1+2x)^b - (1+x)^a)$. Alternatively, note that the sum of all three rows equals a vector proportional to $(1+x)^a + (1+2x)^b + 1)$ times each column, which means the determinant can be factored. At $x=0$: the matrix becomes $\begin{vmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{vmatrix}$, giving constant term $0$. For the coefficient of $x$, we differentiate and evaluate at $x=0$, which yields $0$ due to the symmetric structure of the cyclic determinant.
Correct Answer: 1,2