Question
Which value(s) of $\lambda,$ do the pair of linear equations $\lambda\text{x}+\text{y}=\lambda^2$ and $\text{x}+\lambda\text{y}=1$ have:
No solution?
No solution?
✓
Answer
The given pair of linear equations is
$\lambda\text{x}+\text{y}=\lambda^2$ and $\text{x}+\lambda\text{y}=1$
Here,
$\text{a}_1=\lambda,\text{b}_1=1,\text{c}_1=-\lambda^2$
$\text{a}_2=1,\text{b}_2=\lambda,\text{c}_2=-1$
For no solution,
$\frac{\text{a}_1}{\text{a}_1}=\frac{\text{b}_1}{\text{b}_2}\neq\frac{\text{c}_1}{\text{c}_2}$
$\Rightarrow\frac{\lambda}{1}=\frac{1}{\lambda}\neq\frac{-\lambda^2}{-1}$
$\Rightarrow\lambda^2-1=0$
$\Rightarrow(\lambda-1)(\lambda+1)=0$
$\Rightarrow\lambda=1,-1$
Here, we take only $\lambda=-1$ because at $\lambda=1$ the system of linear equations has infinitely many solutions.
$\lambda\text{x}+\text{y}=\lambda^2$ and $\text{x}+\lambda\text{y}=1$
Here,
$\text{a}_1=\lambda,\text{b}_1=1,\text{c}_1=-\lambda^2$
$\text{a}_2=1,\text{b}_2=\lambda,\text{c}_2=-1$
For no solution,
$\frac{\text{a}_1}{\text{a}_1}=\frac{\text{b}_1}{\text{b}_2}\neq\frac{\text{c}_1}{\text{c}_2}$
$\Rightarrow\frac{\lambda}{1}=\frac{1}{\lambda}\neq\frac{-\lambda^2}{-1}$
$\Rightarrow\lambda^2-1=0$
$\Rightarrow(\lambda-1)(\lambda+1)=0$
$\Rightarrow\lambda=1,-1$
Here, we take only $\lambda=-1$ because at $\lambda=1$ the system of linear equations has infinitely many solutions.
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