Question

For which value(s) of λ, do the pair of linear equations λx + y = λ2 and x + λy = 1 have
(i) no solution?
(ii) infinitely many solutions?
(iii) a unique solution?

NCERT Class X
Mathematics - Exemplar Problems

Chapter _PAIR OF LINEAR EQUATIONS IN TWO VARIABLES

Answer 1

Sol : If the a1x+b1Y+ c1 = 0,a2x+b2y+c2 = 0 system has unique solution then a1/a2 ≠ b1/b2 Given linear equations λx + y = λ2 and x + λy = 1 λ/ 1 ≠ 1 / λ ⇒ λ2 ≠ 1 ⇒ λ ≠ ±1 ∴ λ ≠ ±1 then it has unique solution. If the a1x+b1Y+ c1 = 0,a2x+b2y+c2 = 0 system has infinitely many solutions then a1/a2 = b1/b2 = c1/c2. λ/ 1 = 1 / λ = λ2 / 1 ∴ λ = ±1 then it has infinitely many solutions. If the a1x+b1Y+ c1 = 0,a2x+b2y+c2 = 0 system has no solution then a1/a2 = b1/b2 ≠ c1/c2. λ/ 1 = 1 / λ ≠ λ2 / 1 λ3 ≠ 1 ⇒ λ ≠ 1 ∴ λ = ±1 then it has no solution.

Answer 2

Answer:

(i) $\lambda=-1$

(ii) $\lambda=1$

(iii) All real values except $\lambda=1$.

Step-by-step explanation:

Given: $\begin{aligned}&\lambda x+y=\lambda^{2} \text { and } x+\lambda y=1 \\\end{aligned}$

To find value of $\lambda$ for no solution, infinitely many solution & unique solution.

The general equation of line is $ax+by+c=0$ on comparing given equation with general equation we have:

$a_{1}=\lambda, b_{1}=1, c_{1}=-\lambda^{2} \\&a_{2}=1, b_{2}=\lambda, c_{2}=-1$

(i) No solution

Condition for no solution is: $\mathrm{a}_{1} / \mathrm{a}_{2}=\mathrm{b}_{1} / \mathrm{b}_{2} \neq \mathrm{c}_{1} / \mathrm{c}_{2}$

$\lambda=1 / \lambda \neq \lambda^{2}$

so, $\lambda^{2}=1$;

and $\lambda^{2} \neq \lambda$

So, for $\lambda=-1$, system of linear equations has no solution.

(ii) infinitely many solution

Condition for infinitely many solution is: $a_{1} / a_{2}=b_{1} / b_{2}=c_{1} / c_{2}$

i.e. $\lambda=1 / \lambda=\lambda^{2}$

$\lambda=1 / \lambda$

$\lambda=1$

$\lambda=\lambda^{2}$ gives $\lambda=1,0$;

So, for $\lambda=1$ system has infinitely many solution.

(iii) unique solution

Condition for unique solution is: $\begin{aligned}&a_{1} / a_{2} \neq b_{1} / b_{2} \\\end{aligned}$

$&\text { so } \lambda \neq 1 / \lambda \\&\lambda^{2} \neq 1 \text {; } \\&\lambda \neq1 ;$

So, for all real values except $\lambda=1$ system has unique solution.

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