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For which c is there 1/0 solution?

mathmari

Well-known member
MHB Site Helper
Apr 14, 2013
4,713
Hey!! 😊

I am looking the following:

Solve for a fix number $c\in \mathbb{R}$ the following linear system of equations: $$\begin{cases}x_1-cx_2+(2c-1)x_3=-(c+1) \\ 3x_2+(5c+8)x_3=-(c-2)\end{cases}$$
For which values of $c$ is there one solution and for which values are there no solution?


I have done the following:

First we write this in matrix form:
\begin{equation*}\begin{pmatrix}\left.\begin{matrix}1 & -c & 2c-1 \\ 0 & 3 & 5c+8 \end{matrix}\right|\begin{matrix}-(c+1) \\ -(c-2)\end{matrix}\end{pmatrix}\end{equation*}
It is already in echelon form.

This at the second line we have "$3$" which doesn't depend on $c$, then the case "No solution"doesn't occur, right?

Since the second line cannotbe a multiple of the first one,we conclude that we always have One solution.

Is that correct?

:unsure:
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
9,591
Hey mathmari !!

From the second equation, we can write $x_2$ as a function of $x_3$, which will indeed have at least 1 solution due to the non-zero coefficient $3$.
After that we can always find $x_1$ from the first equation as a function of $x_2$ and $x_3$.
So we will always have infinitely many solutions, won't we? 🤔

Looking at it geometrically, we have the intersection of 2 planes.
Those are actual planes since the coefficients of each cannot all be zero, regardless of the value of $c$.
They can either coincide, or be parallel (no solutions), or intersect in a line.
Since the coefficients in the second line cannot be a multiple of the coefficients in the first line, we can conclude that those planes do not coincide, and they are not parallel either.
That leaves that the solution must be a line, which means we have infinitely many solutions. 🤔
 
Last edited:

mathmari

Well-known member
MHB Site Helper
Apr 14, 2013
4,713
From the second equation, we can write $x_2$ as a function of $x_3$, which will indeed have at least 1 solution due to the non-zero coefficient $3$.
After that we can always find $x_1$ from the first equation as a function of $x_2$ and $x_3$.
So we will always have infinitely many solutions, won't we? 🤔

Looking at it geometrically, we have the intersection of 2 planes.
Those are actual planes since the coefficients of each cannot all be zero, regardless of the value of $c$.
They can either coincide, or be parallel (no solutions), or intersect in a line.
Since the coefficients in the second line cannot be a multiple of the coefficients in the first line, we can conclude that those planes do not coincide, and they are not parallel either.
That leaves that the solution must be a line, which means we have infinitely many solutions. 🤔
I see!! Thank you!! (Handshake)