Q.If $f(x) = \begin{vmatrix} (1 + x)^{17} & (1 + x)^{19} & (1 + x)^{23} \\ (1 + x)^{23} & (1 + x)^{29} & (1 + x)^{34} \\ (1 + x)^{41} & (1 + x)^{43} & (1 + x)^{47} \end{vmatrix} = A + Bx + Cx^2 + \ldots$, then $A =$ ________ .

Concept: Determinant Equality Equation — The constant term $A$ is $f(0)$, so evaluate the determinant at $x = 0$.

Step 1: Set $x = 0$. Then each entry becomes $1$ raised to the given power, which is $1$: …

We are asked for the constant term $A$ in the expansion of $f(x)$ as a polynomial in $x$. The determinant is a polynomial in $x$ because each entry is a binomial expansion in $x$. The constant term of any polynomial $P(x)$ is simply $P(0)$. So instead of expanding the whole determinant, we just plug $x=0$ into every entry.

1. Evaluate at $x=0$.

When $x=0$, each $(1+x)^n$ becomes $1^n = 1$. So the matrix becomes:

2. Recognize the structure.

All nine entries are $1$. This is a matrix where every row is identical — the first row is $(1,1,1)$, and so are the second and third rows.

3. Determinant of a matrix with two equal rows is zero. …

Method: Extracting a Specific Coefficient from a Determinant-Valued Polynomial

Use this method whenever a determinant whose entries are functions of $x$ is said to equal a polynomial $A + Bx + Cx^2 + \dots$, and you're asked for one particular coefficient (commonly the constant term $A$).

Steps

Step 1: Recognise that the determinant is a polynomial in $x$

Since each entry is a function of $x$ (here, a power of $(1+x)$), the determinant $f(x)$ expands into some polynomial $A + Bx + Cx^2 + \cdots$ — you don't need the whole expansion to get ONE coefficient.

Step 2: Recall that the constant term equals $f(0)$

For any polynomial $P(x) = A + Bx + Cx^2 + \cdots$, setting $x=0$ gives $P(0) = A$ (every other term vanishes because it has a positive power of $x$). So instead of expanding the full determinant symbolically, substitute $x = 0$ into every entry.

Step 3: Simplify the resulting numeric determinant …

Common Mistakes

Mistake 1: Trying to fully expand the symbolic determinant in $x$

Why it's wrong: expanding a $3\times3$ determinant whose entries are powers like $(1+x)^{17}, (1+x)^{19}, \ldots$ symbolically is extremely long and unnecessary when only the constant term is needed. Correct approach: substitute $x=0$ first — this immediately turns every entry into a plain number.

Mistake 2: Assuming different exponents mean different values at $x=0$ …