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The system

$$\begin{cases}x-y+3z=-5\\5x+2y-6z=\alpha \\2x-y+\alpha z = -6 \end{cases}$$

for which $\alpha$ values the linear equation system:

  1. has no solution
  2. has one solution
  3. has more than one solution

I started to do Gauss elimination on it, but i have no idea what i am looking for and how to approach this, I'm stuck with the Gauss elimination.

My work so far: \begin{align} \left(\begin{array}{rrr|r} 1 & -1 & 3 & -5 \\ 5 & 2 & 6 & \alpha \\ 2 & -1 & \alpha & -6 \end{array}\right) &\leadsto \left(\begin{array}{rrr|r} 1 & -1 & 3 & -5 \\ 0 & 7 & -9 & \alpha + 25 \\ 2 & -1 & \alpha & -6 \end{array}\right) \\ &\leadsto \left(\begin{array}{rrr|r} 1 & -1 & 3 & -5 \\ 0 & 7 & -9 & \alpha+25 \\ 0 & 1 & \alpha-6 & 4 \end{array}\right) \\ &\leadsto \left(\begin{array}{rrr|r} 1 & 0 & \alpha + 3 & -1 \\ 0 & 7 & -9 & \alpha+25 \\ 0 & 1 & \alpha-6 & 4 \end{array}\right) \\ \end{align}

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  • $\begingroup$ Have you managed to put the system in echelon form ? $\endgroup$
    – user65203
    Feb 14, 2017 at 13:03
  • $\begingroup$ No , not really . That's the problem $\endgroup$
    – Eran
    Feb 14, 2017 at 13:04
  • 1
    $\begingroup$ Where are you stuck ? $\endgroup$
    – user65203
    Feb 14, 2017 at 13:05
  • 1
    $\begingroup$ Right here --> imgur.com/a/MdOX7 $\endgroup$
    – Eran
    Feb 14, 2017 at 13:10
  • 1
    $\begingroup$ I copied your image into the text of the question. Please check for typos. $\endgroup$
    – Xander Henderson ♦
    Feb 26, 2019 at 14:50

2 Answers 2

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Gaussian elimination: \begin{align} \left[\begin{array}{ccc|c} 1 & -1 & 3 & -5 \\ 5 & 2 & -6 & \alpha \\ 2 & -1 & \alpha & -6 \end{array}\right] &\to \left[\begin{array}{ccc|c} 1 & -1 & 3 & -5 \\ 0 & 7 & -21 & \alpha+25 \\ 0 & 1 & \alpha-6 & 4 \end{array}\right] &&\begin{aligned} R_2&\gets R_2-5R_1 \\ R_3&\gets R_3-2R_1\end{aligned} \\[6px]&\to \left[\begin{array}{ccc|c} 1 & -1 & 3 & -5 \\ 0 & 1 & -3 & (\alpha+25)/7 \\ 0 & 1 & \alpha-6 & 4 \end{array}\right] &&R_2\gets \tfrac{1}{7}R_2 \\[6px]&\to \left[\begin{array}{ccc|c} 1 & -1 & 3 & -5 \\ 0 & 1 & -3 & (\alpha+25)/7 \\ 0 & 0 & \alpha-3 & (3-\alpha)/7 \end{array}\right] &&R_3\gets R_3-R_2 \end{align} If $\alpha\ne3$, the system has unique solution.

If $\alpha=3$, the system has infinitely many solutions.

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Start with the matrix and data vector $$ \mathbf{A} = \left[ \begin{array}{rrr} 1 & -1 & 3 \\ 5 & 2 & -6 \\ 2 & -1 & \alpha \\ \end{array} \right], \ \alpha \in \mathbb{C}, \quad % b = \left[ \begin{array}{r} -5 \\ \alpha \\ -6 \end{array} \right] $$ The reduced row eschelon form is $$ \mathbf{E}_{A} = \left[ \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right]. $$ The matrix $\mathbf{A}$ has full rank $m=n=\rho=3$. The column vectors span $\mathbb{C}^{3}$.

Select option 2: a solution will always exist, and the solution will be unique. In fact it is $$ \mathbf{A} x = b \qquad \Rightarrow \qquad x = \frac{1}{7} \left[ \begin{array}{r} \alpha - 10 \\ \alpha + 22 \\ -1 \end{array} \right] . $$

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