Question

Match the following

$x+y+z=2$

$x+y+\lambda z=1$

$x+\mu y+2y=3$

(a) $\mu =1,\lambda =1$ (i) No solution

(b) $\mu =2,\lambda =3$ (ii) Unique Solution

(c) $\mu =1,\lambda =0$ (iii) Infinite Solutions

• A. a-ii; b – iii: c- ii
• B. a-i; b – ii: c- iii
• C. a-iii; b – iii: c- i
• D. a-i; b – iii: c- ii

Answer 1

The correct option is B

a-i; b – ii: c- iii

Let’s convert A into Echelon form of

A = $\left[\begin{array}{ccc}1& 1& 1\\ 1& 1& \lambda \\ 1& \mu & 2\end{array}\right]$

$R_3\to R_3–R_1,R_2\to R_2–R_1$ $\left[\begin{array}{ccc}1& 1& 1\\ 0& 0& \lambda -1\\ 0& \mu -1& 1\end{array}\right]$

$R_3\leftrightarrow R_2$ $\left[\begin{array}{ccc}1& 1& 1\\ 0& \mu -1& 1\\ 0& 0& \lambda -1\end{array}\right]$

Let’s convert C into echelon form

C = $\left[\begin{array}{cccc}1& 1& 1& 2\\ 1& 1& \lambda & 1\\ 1& \mu & 2& 3\end{array}\right]$

$R_2\to R_2–R_1$, $R_3\to R_3–R_1$

$\left[\begin{array}{cccc}1& 1& 1& 2\\ 0& 0& \lambda -1& -1\\ 0& \mu -1& 1& 1\end{array}\right]$

$R_2\leftrightarrow R_3$ $\left[\begin{array}{cccc}1& 1& 1& 2\\ 0& \mu -1& 1& 1\\ 0& 0& \lambda -1& -1\end{array}\right]$

When, $\mu$=1, $\lambda$=1

Rank of A = 2, Rank of C = 3

$\to$ No solutions possible

When, $\mu$=2, $\lambda$=3

Rank of A = 3, Rank of C = 3 = Number of unknowns

$\to$ unique solution

When, $\mu$=1, $\lambda$=0

Rank of A = 2, Rank of C = 2

2 < Number of unknowns

$\therefore$ Infinitie number of solutions.