Gaussian Elimination with Parameters

Question

Can someone please help me with the solution of \begin{align}(\lambda+1)x+y&=\lambda+1\\ x+(\lambda+1)y&=1\\ x+y&=2\lambda+1\end{align}

Answer 1

Subtract the third from the second,

$$\lambda y=-2\lambda.$$

This is possible if either

• $\lambda=0$, then all three equations reduce to $x+y=1$;

• $y=-2$, giving

$$\begin{cases}(\lambda+1)x=\lambda+3,\\x=2\lambda+3\end{cases}$$ which requires

$$(\lambda+1)(2\lambda+3)=\lambda+3.$$

To summarize

$$\lambda=0\to (t,1-t),$$ $$\lambda=-2\to (-1,-2).$$

---

More systematic:

The system has a solution when the determinant

$$\begin{vmatrix}\lambda+1&1&\lambda+1\\1&\lambda+1&1\\1&1&2\lambda+1\end{vmatrix}=2l^3+4l^2$$ vanishes.

Then

• $\lambda=0\to$

$$\begin{cases}x+y&=1\\ x+y&=1\\ x+y&=1,\end{cases}$$

giving $y=1-x$,

• $\lambda=-2\to$

$$\begin{cases}-x+y&=-1\\ x-y&=1\\ x+y&=-3\end{cases},$$

giving $x=-1,y=-2$.

Answer 2

from the first equation we get $$y=\lambda+1-(\lambda+1)x$$ plugging this in the second equation we have $$x(-\lambda^2-2\lambda)=-\lambda^2-2\lambda$$ if $$-\lambda^2-2\lambda=0$$ we get $$\lambda=0$$ of $$\lambda=-2$$ in the first case we get infinity many Solutions. for the other case we get no Solutions. If $$\lambda\ne 0$$ or $$\lambda\ne -2$$ from here we get $$x=1,y=0$$ plugging this in the last equation we get $$1=\lambda+1$$ thus it must be $$\lambda=0$$ in the other case there are no Solutions.

Answer 3

If we sum the first and the second we get $$(\lambda+2)x+(\lambda+2)y=\lambda+2$$ If $\lambda\neq -2$ we can divide by $\lambda+2$ to get $$x+y=1$$ We also have $$x+y=2\lambda+1$$ So $\lambda=0$ and the system has infinitely many solutions which you can parametrize. If $\lambda=-2$ we get $-x+y=-1$ and $x+y=-3$ adding them together we get $y=-2$ and $x=-1$.

Answer 4

You can write that system more compact as $$ A u = b \iff \\ [A | b] = \left[ \begin{array}{cc|c} \lambda + 1 & 1 & \lambda + 1 \\ 1 & \lambda + 1 & 1 \\ 1 & 1 & 2\lambda + 1 \end{array} \right] \to \left[ \begin{array}{cc|c} 1 & 1 & 2\lambda + 1 \\ \lambda & 0 & -\lambda \\ 0 & \lambda & -2 \lambda \end{array} \right] $$ At this point we need to distinguish further between $\lambda = 0$ and $\lambda \ne 0$.

Case $\lambda = 0$: We proceed with the system $$ \left[ \begin{array}{cc|c} 1 & 1 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array} \right] $$ which has the line $L = \{ (x, y) \mid x + y = 1 \}$ as solution.

Case $\lambda \ne 0$: We continue with the system $$ \left[ \begin{array}{cc|c} 1 & 1 & 2\lambda + 1 \\ \lambda & 0 & -\lambda \\ 0 & \lambda & -2 \lambda \end{array} \right] $$ We are allowed to divide by $\lambda$ and get $$ \left[ \begin{array}{cc|c} 1 & 1 & 2\lambda + 1 \\ 1 & 0 & -1 \\ 0 & 1 & -2 \end{array} \right] \to \left[ \begin{array}{cc|c} 1 & 0 & -1 \\ 0 & 1 & -2 \\ 1 & 1 & 2\lambda + 1 \end{array} \right] \to \left[ \begin{array}{cc|c} 1 & 0 & -1 \\ 0 & 1 & -2 \\ 0 & 0 & 2\lambda + 4 \end{array} \right] $$ The equation of the last row can only be solved if $2\lambda + 4 = 0 \iff \lambda = -2$ leading to the solution $(x,y) = (-1,-2)$.

Summary

We found a single solution $(-1,-2)$ if $\lambda = -2$, a line of solutions $L = \{ (x,y) \mid x + y = 1 \}$ for $\lambda = 0$, for other values of $\lambda$ the system has no solutions.