UPSC NDA I 2025 Mathematics Paper

Letωbe a non-real cube root of unity, soω3=1and1+ω+ω2=0.
Consider the determinant
∆(x)=|

x+1	ω	ω2
ω	x+ω2	1
ω2	1	x+ω

|=0.
Step 1 - Column operation: Replace the first column byC1−C2:
∆(x)=|

x2−1	2k+1	1
k−1	k+2	1
0	3	1

|.
Step 2 - Expansion along the third row:
∆(x)=−3|

k2−1	1
k2−1	1

| + |

k2−1	2k+1
k2−1	k+2

|
which simplifies to
∆(x)=x(x2−1)−x(ω+ω2)=x(x2−1)+x=x3
Step 3 - Equate to zero:
∆(x)=0⇒x3=0⇒x=0
∴The root of the equation isx=0.