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Problem 16: Discuss for all values of k the system of equations

Problem 16: Discuss for all values of k the system of equations

\begin{align*} 2x+3ky+(3k+4)z=0\\ x+(k+4)y+(4k+2)z=0\\ x+2(k+1)y+(3k+4)z=0 \end{align*}

Solution:

Consider the augmented matrix of this system and apply row operations.

\[\left[\begin{array}{c|c} A & B \end{array} \right] = \begin{align*} \left[\begin{array}{rrr|r} 2 & 3k & 3k+4& 0 \\ 1 & k+4 & 4k+2& 0 \\ 1 & 2k+2 & 3k+4& 0 \\ \end{array}\right] \xrightarrow{R1\leftrightarrow R2} \left[\begin{array}{rrr|r} 1 & k+4 & 4k+2& 0 \\ 2 & 3k & 3k+4& 0 \\ 1 & 2k+2 & 3k+4& 0 \\ \end{array}\right] \end{align*}\] \[\xrightarrow{R2\rightarrow R_2-2R_1,R3\rightarrow R_3-R_1 } \left[\begin{array}{rrr|r} 1 & k+4 & 4k+2& 0 \\ 0 & k-8 & -5k& 0 \\ 0 & k-2 & -k+2& 0 \\ \end{array}\right]\] \[\xrightarrow{R2\rightarrow \frac{1}{k-8}R_2} \left[\begin{array}{rrr|r} 1 & k+4 & 4k+2& 0 \\ 0 & 1 & \frac{-5k}{k-8}& 0 \\ 0 & k-2 & -k+2& 0 \\ \end{array}\right]\] \[\xrightarrow{R3\rightarrow R_3-(k-2)R_2} \left[\begin{array}{rrr|r} 1 & k+4 & 4k+2& 0 \\ 0 & 1 & \frac{-5k}{k-8}& 0 \\ 0 & 0 & (-k+2)-(k-2)\frac{-5k}{k-8}& 0 \\ \end{array}\right]\] \[\xrightarrow{R3\rightarrow (k-8)R_3} \left[\begin{array}{rrr|r} 1 & k+4 & 4k+2& 0 \\ 0 & 1 & \frac{-5k}{k-8}& 0 \\ 0 & 0 & (-k+2)(k-8)-(k-2)\frac{-5k}{k-8}& 0 \\ \end{array}\right]\\ \left[\begin{array}{rrr|r} 1 & k+4 & 4k+2& 0 \\ 0 & 1 & \frac{-5k}{k-8}& 0 \\ 0 & 0 & k^2-4& 0 \\ \end{array}\right]\]

This matrix is in row-echelon form. There are three cases according to $k^2 – 4$ is zero or not.

Case i: If $k = 2$, the last row in REF consists entirely of zeros, that is, \begin{align*} \left[\begin{array}{rrr|r} 1 & 6 & 10& 0 \\ 0 & 1 & \frac{10}{6}& 0 \\ 0 & 0 & 0& 0 \\ \end{array}\right] \end{align*}

we write down the corresponding system as \begin{align*} x+6y+10z=0\\ y+\frac{10}{6}z=0\\ 6y+10z=0\\ y=-\frac{5}{3}z x=0 \end{align*} This shows that in case $k=2$, the system has infinitely many solutions

Case ii: If $k = -2$, \begin{align*} \left[\begin{array}{rrr|r} 1 & 2 & -6& 0 \\ 0 & 1 & -1& 0 \\ 0 & 0 & 0& 0 \\ \end{array}\right] \end{align*}

we write down the corresponding system as \begin{align*} x+2y-6z=0\\ y-z=0\\ y=z\\ x=4z \end{align*} This shows that in case $k=-2$, the system has infinitely many solutions

Case iii: If $k^2-4 \ne 0$; that is, if $k\ne \pm 2$ in this case the system has trivial solution and $x=y=z=0$ This shows that in case $k^2-4 \ne 0$, the system has only one solution

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