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- #1

- Apr 14, 2013

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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?