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UPSC Maths 2013 Paper 1 Q2b-ii — Solution 8 marks · Section A
Question
Find the rank of the matrix
A = [ 1 2 3 4 5 2 3 5 8 12 3 5 8 12 17 5 8 12 17 23 8 12 17 23 30 ] . A=\begin{bmatrix}1 & 2 & 3 & 4 & 5\\ 2 & 3 & 5 & 8 & 12\\ 3 & 5 & 8 & 12 & 17\\ 5 & 8 & 12 & 17 & 23\\ 8 & 12 & 17 & 23 & 30\end{bmatrix}. A = 1 2 3 5 8 2 3 5 8 12 3 5 8 12 17 4 8 12 17 23 5 12 17 23 30 . Technique
Spot the obvious row dependency by inspection (R 3 = R 1 + R 2 R_3=R_1+R_2 R 3 = R 1 + R 2
Solution
Strategy. Row reduce; identify dependent rows.
Step 1 — Spot the obvious relation
By inspection, R 3 = R 1 + R 2 R_3=R_1+R_2 R 3 = R 1 + R 2
( 1 + 2 , 2 + 3 , 3 + 5 , 4 + 8 , 5 + 12 ) = ( 3 , 5 , 8 , 12 , 17 ) = R 3 ✓ . (1+2,\;2+3,\;3+5,\;4+8,\;5+12)=(3,5,8,12,17)=R_3\;\checkmark. ( 1 + 2 , 2 + 3 , 3 + 5 , 4 + 8 , 5 + 12 ) = ( 3 , 5 , 8 , 12 , 17 ) = R 3 ✓ . So R 3 − R 1 − R 2 = 0 R_3-R_1-R_2=0 R 3 − R 1 − R 2 = 0
Step 2 — Row-reduce the remaining four rows { R 1 , R 2 , R 4 , R 5 } \{R_1,R_2,R_4,R_5\} { R 1 , R 2 , R 4 , R 5 }
After zeroing R 3 R_3 R 3
[ 1 2 3 4 5 2 3 5 8 12 5 8 12 17 23 8 12 17 23 30 ] → R 2 − 2 R 1 , R 4 − 5 R 1 , R 5 − 8 R 1 [ 1 2 3 4 5 0 − 1 − 1 0 2 0 − 2 − 3 − 3 − 2 0 − 4 − 7 − 9 − 10 ] . \begin{bmatrix}1 & 2 & 3 & 4 & 5\\ 2 & 3 & 5 & 8 & 12\\ 5 & 8 & 12 & 17 & 23\\ 8 & 12 & 17 & 23 & 30\end{bmatrix}\xrightarrow{R_2-2R_1,\;R_4-5R_1,\;R_5-8R_1}\begin{bmatrix}1 & 2 & 3 & 4 & 5\\ 0 & -1 & -1 & 0 & 2\\ 0 & -2 & -3 & -3 & -2\\ 0 & -4 & -7 & -9 & -10\end{bmatrix}. 1 2 5 8 2 3 8 12 3 5 12 17 4 8 17 23 5 12 23 30 R 2 − 2 R 1 , R 4 − 5 R 1 , R 5 − 8 R 1 1 0 0 0 2 − 1 − 2 − 4 3 − 1 − 3 − 7 4 0 − 3 − 9 5 2 − 2 − 10 . Continue: R 4 − 2 R 2 R_4-2R_2 R 4 − 2 R 2 R 5 − 4 R 2 R_5-4R_2 R 5 − 4 R 2
[ 1 2 3 4 5 0 − 1 − 1 0 2 0 0 − 1 − 3 − 6 0 0 − 3 − 9 − 18 ] . \begin{bmatrix}1 & 2 & 3 & 4 & 5\\ 0 & -1 & -1 & 0 & 2\\ 0 & 0 & -1 & -3 & -6\\ 0 & 0 & -3 & -9 & -18\end{bmatrix}. 1 0 0 0 2 − 1 0 0 3 − 1 − 1 − 3 4 0 − 3 − 9 5 2 − 6 − 18 . Finally, R 5 − 3 R 4 R_5-3R_4 R 5 − 3 R 4
[ 1 2 3 4 5 0 − 1 − 1 0 2 0 0 − 1 − 3 − 6 0 0 0 0 0 ] . \begin{bmatrix}1 & 2 & 3 & 4 & 5\\ 0 & -1 & -1 & 0 & 2\\ 0 & 0 & -1 & -3 & -6\\ 0 & 0 & 0 & 0 & 0\end{bmatrix}. 1 0 0 0 2 − 1 0 0 3 − 1 − 1 0 4 0 − 3 0 5 2 − 6 0 . Three nonzero rows, one zero row — so among { R 1 , R 2 , R 4 , R 5 } \{R_1,R_2,R_4,R_5\} { R 1 , R 2 , R 4 , R 5 }
R 5 = 3 R 4 − 2 R 2 − 3 R 1 . R_5=3R_4-2R_2-3R_1. R 5 = 3 R 4 − 2 R 2 − 3 R 1 . (Quick verification: 3 R 4 − 2 R 2 − 3 R 1 = ( 15 , 24 , 36 , 51 , 69 ) − ( 4 , 6 , 10 , 16 , 24 ) − ( 3 , 6 , 9 , 12 , 15 ) = ( 8 , 12 , 17 , 23 , 30 ) = R 5 3R_4-2R_2-3R_1=(15,24,36,51,69)-(4,6,10,16,24)-(3,6,9,12,15)=(8,12,17,23,30)=R_5 3 R 4 − 2 R 2 − 3 R 1 = ( 15 , 24 , 36 , 51 , 69 ) − ( 4 , 6 , 10 , 16 , 24 ) − ( 3 , 6 , 9 , 12 , 15 ) = ( 8 , 12 , 17 , 23 , 30 ) = R 5
Step 3 — Conclude
Two independent linear dependencies among 5 rows:
R 3 = R 1 + R 2 R_3=R_1+R_2 R 3 = R 1 + R 2 R 5 = 3 R 4 − 2 R 2 − 3 R 1 R_5=3R_4-2R_2-3R_1 R 5 = 3 R 4 − 2 R 2 − 3 R 1
leave 3 independent rows (R 1 , R 2 , R 4 R_1,R_2,R_4 R 1 , R 2 , R 4
Answer
rank ( A ) = 3. \boxed{\;\operatorname{rank}(A)=3.\;} rank ( A ) = 3.