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Problem 22: Investigate for what values a,b the equations

Problem 22: Investigate for what values a,b the equations

\begin{align*} x+2y+3z=4,\\ x+3y+4z=5,\\ x+3y+az = b, \end{align*} have i. no solution, ii. a unique solution and iii. an infinite number of solutions.

Solution:

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

\[\left[\begin{array}{c|c} A & B \end{array} \right] = \begin{align*} \left[\begin{array}{rrr|r} 1 & 2 & 3 & 4 \\ 1 & 3 & 4 & 5 \\ 1 & 3 & a & b \\ \end{array}\right] \end{align*}\] \[\xrightarrow{R2\rightarrow R_2-R_1,R3\rightarrow R_3-R_1} \left[\begin{array}{rrr|r} 1 & 2 & 3 & 4 \\ 0 & 1 & 1& 1 \\ 0 & 1 & a-3 & b-4 \\ \end{array}\right]\] \[\xrightarrow{R3\rightarrow R_3-R_2} \left[\begin{array}{rrr|r} 1 & 2 & 3 & 4 \\ 0 & 1 & 1& 1 \\ 0 & 0 & a-4 & b-5 \\ \end{array}\right]\]

Case i. $a = 4$ and $b \ne 5$ \begin{align*} \left[\begin{array}{rrr|r} 1 & 2 & 3 & 4 \\ 0 & 1 & 1& 1 \\ 0 & 0 & 0 & b-5 \\ \end{array}\right] \end{align*} showing that the equations are not consistent in this case and no solution is possible.

Case ii. $a \ne 4$ \begin{align*} \left[\begin{array}{rrr|r} 1 & 2 & 3 & 4 \\ 0 & 1 & 1& 1 \\ 0 & 0 & a-4 & b-5 \\ \end{array}\right]\\ \end{align*} \begin{align*} (a-4)z=b-5\\ z= \frac{(b-5)}{(a-4)} \end{align*} x,y, z has unique solution for all values of b if $a\ne 4$

Case iii. $a = 4$ and $b = 5$ \begin{align*} \left[\begin{array}{rrr|r} 1 & 2 & 3 & 4 \\ 0 & 1 & 1& 1 \\ 0 & 0 & 0 & 0 \\ \end{array}\right] \end{align*} x,y, z has infinite solutions

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