Q:Determine the values of $\lambda$ such that the following systems of linear equations have $(i)$ no solution ,$(ii)$More than one solution ,$(iii)$A unique solution. \begin{array}{c} x+\lambda y+z=1 \\ x+y+\lambda z=1 \\ \lambda x+y+z=1 \end{array}
My Approach:I used augmented matrix to determine the values of $\lambda$ $$ \left( \begin{array}{rrr|r} 1 & \lambda & 1 & 1 \\ 0 & 1 & \frac{1}{1+\lambda} & \frac{1}{1+\lambda} \\ 0 & 0 & \frac{-2-\lambda}{1+\lambda} & -\frac{1}{1+\lambda} \\ \end{array} \right)$$From here It's easy to determine the values of $\lambda =-2$ for case $(i)$.But i don't understand what for case $(ii),(iii)$.Any hints or solution will be appreciated.
Thanks in advance.

  • $\begingroup$ Something is not right. Are you sure the denominator is $1+\lambda$? Because $\lambda=1$ leads to case ($ii$), I would expect to see $1-\lambda$ in the denominator. $\endgroup$ – user593746 Nov 15 '18 at 17:00

A=$\begin{pmatrix}1 &\lambda & 1 & 1\\0 & 1-\lambda & \lambda-1 & 0\\0 & 0& 2-\lambda-\lambda^2 & 1-\lambda\end{pmatrix}$ is the correct matrix. You can't divide with 1+$\lambda$ because it isn't given anywhere that $\lambda\neq-1$.

Now for no solution $2-\lambda-\lambda^2=0$ and $1-\lambda\neq0$. So $\lambda =-2$ is correct.

For more than one solution rank(A)<3. So when $1-\lambda=0$ and $2-\lambda-\lambda^2=0$ more than one solution occurs. $\lambda=1$.

For unique solution, the matrix A must be invertible. So $det(A)\neq0$.


So $\lambda\neq1,-2$

  • $\begingroup$ Thanks a lot @Jimmy Sir.Maybe my row operation are incorrect.By the way it will be more helpful if you showed the correct Augmented matrix with proper row operation. $\endgroup$ – raihan hossain Nov 15 '18 at 17:28
  • 2
    $\begingroup$ You have an error in the very last line where you switched from $-2$ to $2$. $\endgroup$ – Ian Nov 15 '18 at 17:30
  • $\begingroup$ @Ian Oh yeah. Thanks, I will correct it. $\endgroup$ – Avanish Singh Nov 15 '18 at 17:39
  • $\begingroup$ @raihanhossain The method I applied to analyze the equations was Gaussian Elimination. You can read it and try to do it yourself. $\endgroup$ – Avanish Singh Nov 15 '18 at 17:43

As I said, your approach seems to have an error. I don't know where it came from since you didn't show how you got the augmented matrix. I am presenting a different solution.

Summing all the equations, we get $(\lambda+2)(x+y+z)=3$. Therefore, $\lambda=-2$ gives $0=3$, which is absurd. So, $\lambda=-2$ lands you in case ($i$).

For $\lambda=1$, you have three identical equations, which are all $x+y+z=1$. Hence, $\lambda=1$ lands you in case ($ii$).

I claim that if $\lambda\neq 1,-2$, then you have case ($iii$). We subtract the first equation from the second equation to get $(\lambda-1)(y-z)=0$. Because $\lambda\neq 1$, we have $y=z$. Similarly, $x=y$ (by subtracting the first and the last equation). Therefore, $x=y=z=\frac{1}{\lambda+2}$, which is a valid expression since $\lambda\neq -2$.

  • $\begingroup$ Thanks a lot @Zvi Sir. Maybe my operation are incorrect.Thanks again to correct me $\endgroup$ – raihan hossain Nov 15 '18 at 17:27

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