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UPSC 2021 Maths Optional Paper 1 Q1c — Step-by-Step Solution

10 marks · Section A

Algebra of matrices · Linear Algebra · asked 5× in 13 yrs · Read the full method →

Question

$\Delta(x)=\begin{vmatrix}f(x+\alpha) & f(x+2\alpha) & f(x+3\alpha)\\ f(\alpha) & f(2\alpha) & f(3\alpha)\\ f'(\alpha) & f'(2\alpha) & f'(3\alpha)\end{vmatrix}$. Find $\lim_{x\to 0}\Delta(x)/x$.

Technique

Taylor expand row 1; observe row 1 = row 2 + $x\cdot$row 3 + $O(x^2)$; use multilinearity of determinant.

Solution

Strategy. At $x=0$: row 1 = $(f(\alpha),f(2\alpha),f(3\alpha))$ = row 2. So $\Delta(0)=0$ (two identical rows).

Use Taylor expansion of row 1 around $x=0$.

Step 1 — Expand row 1 to first order

$f(x+k\alpha)=f(k\alpha)+x\cdot f'(k\alpha)+O(x^2)$ for $k=1,2,3$.

So row 1 of $\Delta(x)$: $(f(\alpha)+xf'(\alpha),\;f(2\alpha)+xf'(2\alpha),\;f(3\alpha)+xf'(3\alpha))+O(x^2)$.

= row 2 + $x\cdot$row 3 + $O(x^2)$.

Step 2 — Linearity in row 1

$\Delta(x)=\det\begin{pmatrix}\text{row 2}+x\cdot\text{row 3}+O(x^2)\\ \text{row 2}\\ \text{row 3}\end{pmatrix}$.

Determinant is linear in row 1: $\Delta(x)=\det\begin{pmatrix}\text{row 2}\\ \text{row 2}\\ \text{row 3}\end{pmatrix}+x\det\begin{pmatrix}\text{row 3}\\ \text{row 2}\\ \text{row 3}\end{pmatrix}+O(x^2)$.

Both determinants have two identical rows (rows 2,2 and rows 3,3 respectively), so both are $0$.

Wait — that gives $\Delta(x)=O(x^2)$, so $\Delta(x)/x\to 0$.

Let me re-examine. First determinant: rows are (row 2, row 2, row 3) — two identical rows (1 and 2), so determinant = 0.

Second determinant: rows are (row 3, row 2, row 3) — two identical rows (1 and 3), so determinant = 0.

So both leading terms vanish, and $\Delta(x)=O(x^2)$.

Therefore $\Delta(x)/x\to 0$ as $x\to 0$.

Answer