Linear system - number of solutions depending on the parameter k

Question

Determine for what value of $k$ the following system has

1. unique solution,
2. no solution and
3. infinitely many solutions.

\begin{cases} x+2y+z=3\\ 2x-y-3x=5\\ 4x+3y-z=k \end{cases}

I did till $$ \left[\begin{array}{ccc|c} 1 & 2 & 1 & 3\\ 1 & 1 & 1 & 1/5\\ 0 & 0 & 1 & -k+42/5 \end{array}\right] $$ I am not sure how to continue.

Answer 1

You're not arrived at the end of the elimination: \begin{align} \left[\begin{array}{ccc|c} 1 & 2 & 1 & 3\\ 2 & -1 & -3 & 5\\ 4 & 3 & -1 & k \end{array}\right] &\to \left[\begin{array}{ccc|c} 1 & 2 & 1 & 3\\ 0 & -5 & -5 & -1\\ 4 & 3 & -1 & k \end{array}\right] &&R_2\gets R_2-2R_1 \\&\to \left[\begin{array}{ccc|c} 1 & 2 & 1 & 3\\ 0 & -5 & -5 & -1\\ 0 & -5 & -5 & k-12 \end{array}\right] &&R_3\gets R_3-4R_1 \\&\to \left[\begin{array}{ccc|c} 1 & 2 & 1 & 3\\ 0 & 1 & 1 & 1/5\\ 0 & -5 & -5 & k-12 \end{array}\right] &&R_2\gets -\frac{1}{5}R_2 \\&\to \left[\begin{array}{ccc|c} 1 & 2 & 1 & 3\\ 0 & 1 & 1 & 1/5\\ 0 & 0 & 0 & k-11 \end{array}\right] &&R_3\gets R_3+5R_2 \end{align} Now you should be able to end.

Answer 2

1) First step: $$\begin{cases} x+2y+z=3\\ 2x-y-3x=5\\ 4x+3y-z=k \end{cases}\Longleftrightarrow$$

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$x+2y+z=3 \Longleftrightarrow x=3-2y-z$

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$$\begin{cases} -3+y+z=5\\ 12-5y-5z=k \end{cases}\Longleftrightarrow$$

$$\begin{cases} -15+5y+5z=25\\ 12-5y-5z=k \end{cases}\Longleftrightarrow$$

$$-3=25+k \Longleftrightarrow$$ $$k=-28 $$

2) Second step: $$\begin{cases} x+2y+z=3\\ 2x-y-3x=5\\ 4x+3y-z=-28 \end{cases}\Longleftrightarrow$$

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$2x-y-3x=5 \Longleftrightarrow y=-5-x$

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$x+2y+z=3 \Longleftrightarrow z=28+4x+3y \Longleftrightarrow z=13+x$

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$x+2y+z=3 \Longleftrightarrow x=3-2y-z \Longleftrightarrow x=z-13$

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So we got:

$$k=-28,y=-x-5,x=z-13,z=13+x$$