← 2014 Paper 1

UPSC 2014 Maths Optional Paper 1 Q2b-i — Step-by-Step Solution

10 marks · Section A

Solution of system of linear equations · Linear Algebra · asked 7× in 13 yrs · Read the full method →

Question

Investigate the values of $\lambda$ and $\mu$ so that the equations $x+y+z=6$, $x+2y+3z=10$, $x+2y+\lambda z=\mu$ have (1) no solution, (2) a unique solution, (3) an infinite number of solutions.

Technique

Standard “rank of coefficient vs. rank of augmented” classification via row reduction.

Solution

Strategy. Row-reduce the augmented matrix to echelon form; classify by comparing rank of coefficient matrix vs. rank of augmented matrix.

Step 1 — Row reduction

Augmented matrix:

$R_2\to R_2-R_1$, $R_3\to R_3-R_1$:

$R_3\to R_3-R_2$:

Step 2 — Classify by $(\lambda,\mu)$

Case (2): Unique solution. Rank of coefficient matrix = rank of augmented = 3.

This requires the $(3,3)$ entry $\lambda-3\ne 0$, i.e., $\lambda\ne 3$. Then $z=(\mu-10)/(\lambda-3)$, back-substitute for $y,x$.

Case (1): No solution. Rank coefficient $<$ rank augmented, i.e., the last row is $0=$ non-zero.

This requires $\lambda-3=0$ AND $\mu-10\ne 0$, i.e., $\lambda=3,\;\mu\ne 10$.

Case (3): Infinite solutions. Rank coefficient = rank augmented $=2<3$.

This requires the last row entirely zero: $\lambda-3=0$ AND $\mu-10=0$, i.e., $\lambda=3,\;\mu=10$.

Answer