← 2018 Paper 1
UPSC 2018 Maths Optional Paper 1 Q3a — Step-by-Step Solution 13 marks · Section A
Solution of system of linear equations · Linear Algebra · asked 7× in 13 yrs · Read the full method →
Question
For the system of linear equations
x + 3 y − 2 z = − 1 5 y + 3 z = − 8 x − 2 y − 5 z = 7 \begin{aligned}x+3y-2z&=-1\\ 5y+3z&=-8\\ x-2y-5z&=7\end{aligned} x + 3 y − 2 z 5 y + 3 z x − 2 y − 5 z = − 1 = − 8 = 7 determine which of the following statements are true and which are false:
(i) The system has no solution.
(ii) The system has a unique solution.
(iii) The system has infinitely many solutions.
Technique
Rouché–Capelli: compare rank A \operatorname{rank}A rank A rank [ A ∣ b ] \operatorname{rank}[A|\mathbf b] rank [ A ∣ b ] det A = 0 \det A=0 det A = 0 < n <n < n
Solution
Step 1 — Coefficient matrix and its determinant
A = ( 1 3 − 2 0 5 3 1 − 2 − 5 ) , b = ( − 1 − 8 7 ) . A=\begin{pmatrix}1&3&-2\\0&5&3\\1&-2&-5\end{pmatrix},\qquad \mathbf b=\begin{pmatrix}-1\\-8\\7\end{pmatrix}. A = 1 0 1 3 5 − 2 − 2 3 − 5 , b = − 1 − 8 7 . Expand det A \det A det A
det A = 1 ⋅ det ( 5 3 − 2 − 5 ) − 0 + 1 ⋅ det ( 3 − 2 5 3 ) = ( − 25 + 6 ) + ( 9 + 10 ) = − 19 + 19 = 0. \det A=1\cdot\det\!\begin{pmatrix}5&3\\-2&-5\end{pmatrix}-0+1\cdot\det\!\begin{pmatrix}3&-2\\5&3\end{pmatrix}=(−25+6)+(9+10)=-19+19=0. det A = 1 ⋅ det ( 5 − 2 3 − 5 ) − 0 + 1 ⋅ det ( 3 5 − 2 3 ) = ( − 25 + 6 ) + ( 9 + 10 ) = − 19 + 19 = 0. So det A = 0 \det A=0 det A = 0 not uniquely solvable — statement (ii) is already settled as false .
Step 2 — Row-reduce to compare ranks
R 3 → R 3 − R 1 R_3\to R_3-R_1 R 3 → R 3 − R 1 ( 1 , − 2 , − 5 ∣ 7 ) − ( 1 , 3 , − 2 ∣ − 1 ) = ( 0 , − 5 , − 3 ∣ 8 ) (1,-2,-5\,|\,7)-(1,3,-2\,|\,-1)=(0,-5,-3\,|\,8) ( 1 , − 2 , − 5 ∣ 7 ) − ( 1 , 3 , − 2 ∣ − 1 ) = ( 0 , − 5 , − 3 ∣ 8 )
( 1 3 − 2 − 1 0 5 3 − 8 0 − 5 − 3 8 ) → R 3 → R 3 + R 2 ( 1 3 − 2 − 1 0 5 3 − 8 0 0 0 0 ) . \left(\begin{array}{ccc|c}1&3&-2&-1\\0&5&3&-8\\0&-5&-3&8\end{array}\right)\xrightarrow{R_3\to R_3+R_2}\left(\begin{array}{ccc|c}1&3&-2&-1\\0&5&3&-8\\0&0&0&0\end{array}\right). 1 0 0 3 5 − 5 − 2 3 − 3 − 1 − 8 8 R 3 → R 3 + R 2 1 0 0 3 5 0 − 2 3 0 − 1 − 8 0 . The last row is 0 = 0 0=0 0 = 0 consistent , no contradiction.
Step 3 — Read off ranks
rank ( A ) = 2 \operatorname{rank}(A)=2 rank ( A ) = 2 rank ( [ A ∣ b ] ) = 2 \operatorname{rank}([A\,|\,\mathbf b])=2 rank ([ A ∣ b ]) = 2 3 3 3 3 − 2 = 1 3-2=1 3 − 2 = 1 infinitely many solutions.
Solving: from row 2, 5 y + 3 z = − 8 ⇒ y = − 8 − 3 z 5 5y+3z=-8\Rightarrow y=\frac{-8-3z}{5} 5 y + 3 z = − 8 ⇒ y = 5 − 8 − 3 z x = − 1 − 3 y + 2 z = 19 z + 19 5 x=-1-3y+2z=\frac{19z+19}{5} x = − 1 − 3 y + 2 z = 5 19 z + 19 z = t z=t z = t
( x , y , z ) = ( 19 ( t + 1 ) 5 , − 8 − 3 t 5 , t ) , t ∈ R . \big(x,y,z\big)=\left(\frac{19(t+1)}{5},\ \frac{-8-3t}{5},\ t\right),\quad t\in\mathbb R. ( x , y , z ) = ( 5 19 ( t + 1 ) , 5 − 8 − 3 t , t ) , t ∈ R . Step 4 — Verdict
Answer
(i) False , (ii) False , (iii) True. \boxed{\;\text{(i) False},\qquad\text{(ii) False},\qquad\text{(iii) True.}\;} (i) False , (ii) False , (iii) True.