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Problem 21: Discuss for all values of $\lambda$, the system of equations

Problem 21: Discuss for all values of $\lambda$, the system of equations

\begin{align*} x+y+4z=6,\\ x+2y-2z=6,\\ \lambda x+y+z = 6, \end{align*} as regards existence and nature of solutions.

Solution:

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

\[\left[\begin{array}{c|c} A & B \end{array} \right] = \begin{align*} \left[\begin{array}{rrr|r} 1 & 1 & 4 & 6 \\ 1 & 2 & -2& 6 \\ \lambda & 1 & 1 & 6 \\ \end{array}\right] \end{align*}\] \[\xrightarrow{R2\rightarrow R_2-R_1,R3\rightarrow R_3-\lambda R_1} \left[\begin{array}{rrr|r} 1 & 1 & 4 & 6 \\ 0 & 1 & -6& 0 \\ 0 & 1-\lambda & 1-4\lambda & 6-6\lambda \\ \end{array}\right]\]

Therefore the coefficient matrix will be non-singular if and only if, \begin{align*} 1-4\lambda + 6-6\lambda \ne 0,\\ \lambda \ne \frac{7}{10}. \end{align*}

In Case $\lambda = \frac{7}{10}$, \begin{align*} \left[\begin{array}{rrr|r} 1 & 1 & 4 & 6 \\ 0 & 1 & -6& 0 \\ 0 & \frac{3}{10} & -\frac{18}{10} & \frac{18}{10} \\ \end{array}\right] \end{align*} \[\xrightarrow{R3\rightarrow R_3-\frac{3}{10}R_2} \left[\begin{array}{rrr|r} 1 & 1 & 4 & 6 \\ 0 & 1 & -6& 0 \\ 0 & 0 & 0 & \frac{18}{10} \\ \end{array}\right]\] showing that the equations are not consistent in this case and no solution is possible.

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