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UPSC 2016 Maths Optional Paper 1 Q1b-i — Step-by-Step Solution 7 marks · Section A
Solution of system of linear equations · Linear Algebra · asked 7× in 13 yrs · Read the full method →
Question
Using elementary row operations, find the condition that the linear equations
x − 2 y + z = a 2 x + 7 y − 3 z = b 3 x + 5 y − 2 z = c \begin{aligned}x-2y+z&=a\\ 2x+7y-3z&=b\\ 3x+5y-2z&=c\end{aligned} x − 2 y + z 2 x + 7 y − 3 z 3 x + 5 y − 2 z = a = b = c
have a solution.
Technique
Row-reduce [ A ∣ b ] [A\,|\,\mathbf b] [ A ∣ b ] ; when the coefficient part develops a zero row, the matching right-hand entry must vanish for consistency (Rouché–Capelli).
Solution
( 1 − 2 1 a 2 7 − 3 b 3 5 − 2 c ) . \left(\begin{array}{ccc|c}1&-2&1&a\\2&7&-3&b\\3&5&-2&c\end{array}\right). 1 2 3 − 2 7 5 1 − 3 − 2 a b c .
Step 2 — Eliminate the first column
R 2 → R 2 − 2 R 1 R_2\to R_2-2R_1 R 2 → R 2 − 2 R 1 and R 3 → R 3 − 3 R 1 R_3\to R_3-3R_1 R 3 → R 3 − 3 R 1 :
( 1 − 2 1 a 0 11 − 5 b − 2 a 0 11 − 5 c − 3 a ) . \left(\begin{array}{ccc|c}1&-2&1&a\\0&11&-5&b-2a\\0&11&-5&c-3a\end{array}\right). 1 0 0 − 2 11 11 1 − 5 − 5 a b − 2 a c − 3 a .
Step 3 — Eliminate the second column from R 3 R_3 R 3
R 3 → R 3 − R 2 R_3\to R_3-R_2 R 3 → R 3 − R 2 :
( 1 − 2 1 a 0 11 − 5 b − 2 a 0 0 0 c − 3 a − ( b − 2 a ) ) = ( 1 − 2 1 a 0 11 − 5 b − 2 a 0 0 0 c − a − b ) . \left(\begin{array}{ccc|c}1&-2&1&a\\0&11&-5&b-2a\\0&0&0&\,c-3a-(b-2a)\,\end{array}\right)=\left(\begin{array}{ccc|c}1&-2&1&a\\0&11&-5&b-2a\\0&0&0&\,c-a-b\,\end{array}\right). 1 0 0 − 2 11 0 1 − 5 0 a b − 2 a c − 3 a − ( b − 2 a ) = 1 0 0 − 2 11 0 1 − 5 0 a b − 2 a c − a − b .
Step 4 — Read off the consistency condition
The coefficient matrix has rank 2 2 2 (third row of coefficients is all zero). The system is consistent iff the augmented matrix also has rank 2 2 2 , i.e. the last right-hand entry vanishes:
c − a − b = 0. c-a-b=0. c − a − b = 0.
Answer
The system has a solution ⟺ a + b = c . \boxed{\;\text{The system has a solution}\iff a+b=c.\;} The system has a solution ⟺ a + b = c .