← 2014 Paper 1
UPSC 2014 Maths Optional Paper 1 Q1b — Step-by-Step Solution
10 marks · Section A
Rank of a matrix · Linear Algebra · asked 7× in 13 yrs · Read the full method →
Question
Using elementary row or column operations, find the rank of the matrix
00311011−310−2−1120.
Technique
Standard row reduction to echelon form.
Solution
Strategy. Reduce to row echelon form via elementary row operations; rank = number of non-zero rows.
Step 1 — Swap to get a leading 1 in row 1
R1↔R4:
10301011−210−3012−1
Step 2 — Eliminate below pivot in column 1
R3→R3−3R1:
100010−21−216−3012−1
Step 3 — Swap rows to find leading entry in column 2
R2↔R4:
100011−20−2−3610−121
Step 4 — Eliminate below pivot in column 2
R3→R3+2R2:
10001100−2−3010−101
Step 5 — Swap zero row to bottom
R3↔R4:
10001100−2−3100−110
Row echelon form with three non-zero rows.
Answer
rank(A)=3.