Problem 20: For what values of the $\eta$ the equations

Problem 20: For what values of the $\eta$ the equations

\begin{align*} x+y+z=1\\ x+2y+4z=\eta\\ x+4y+10z = {\eta}^2? \end{align*} have a solution and solve them completely in each case.

Solution:

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

\[\left[\begin{array}{c|c} A & B \end{array} \right] = \begin{align*} \left[\begin{array}{rrr|r} 1 & 1 & 1 & 1 \\ 1 & 2 & 4& \eta \\ 1 & 4 & 10 & {\eta}^2 \\ \end{array}\right] \end{align*}\] \[\xrightarrow{R2\rightarrow R_2-R_1,R3\rightarrow R_3-R_1} \left[\begin{array}{rrr|r} 1 & 1 & 1 & 1 \\ 0 & 1 & 3& \eta-1 \\ 0 & 3 & 9 & {\eta}^2-1 \\ \end{array}\right]\] \[\xrightarrow{R3\rightarrow R_3-3R_2} \left[\begin{array}{rrr|r} 1 & 1 & 1 & 1 \\ 0 & 1 & 3& \eta-1 \\ 0 & 0 & 0 & {\eta}^2-3\eta +2 \\ \end{array}\right]\]

Now the given equations will be consistent if and only if, \begin{align*} {\eta}^2-3\eta +2 = 0\\ (\eta -2)(\eta -1)=0\\ \eta = 2 \;or \;\eta = 1. \end{align*}

Case i: $\eta = 2$, \begin{align*} \left[\begin{array}{rrr|r} 1 & 1 & 1 & 1 \\ 0 & 1 & 3& 1 \\ 0 & 0 & 0 & 0 \\ \end{array}\right] \end{align*} The above system of equations is equivalent to \begin{align*} y+3z=1,x+y+z=1\\ y=1-3z, x=2z. \end{align*} Thus $x=2k,y=1-3k,z=k$ constitute the general solution where $k$ is an aribtrary constant.

Case ii: $\eta = 1$, \begin{align*} \left[\begin{array}{rrr|r} 1 & 1 & 1 & 1 \\ 0 & 1 & 3& 0 \\ 0 & 0 & 0 & 0 \\ \end{array}\right] \end{align*} The above system of equations is equivalent to \begin{align*} y+3z=0,x+y+z=1\\ y=-3z, x=1+2z. \end{align*} Thus $x=1+2k,y=-3k,z=k$ constitute the general solution where $k$ is an aribtrary constant.