# Determine the values of $\lambda$ for the following results

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.

• 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. – 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 $$det(A)\neq0$$.

$$(2-\lambda-\lambda^2)(1-\lambda)\neq0$$

So $$\lambda\neq1,-2$$

• 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. – raihan hossain Nov 15 '18 at 17:28
• You have an error in the very last line where you switched from $-2$ to $2$. – Ian Nov 15 '18 at 17:30
• @Ian Oh yeah. Thanks, I will correct it. – Jimmy Nov 15 '18 at 17:39
• @raihanhossain The method I applied to analyze the equations was Gaussian Elimination. You can read it and try to do it yourself. – Jimmy 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 $$\lambda\neq -2$$.

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