Q. 91 - Precalculus Enhanced with Graphing Utilities

Question

Make up a system of three linear equations containing three variables that has:

Part (a): No solution

Part (b): Exactly one solution

Part (c): Infinitely many solutions

Step-by-Step Solution

Answer

Part (a): The required system of three linear equations are given below,

$$\begin{gathered} x+3y+2z=1 \\ 3x+10y+10z=5 \\ 2x+8y+12z=9 \end{gathered}$$

Part (b): The required system of three linear equations are given below,

$$\begin{gathered} x-2y+3z=9 \\ -x+3y=-4 \\ 2x-5y+5z=17 \end{gathered}$$

Part (c): The required system of three linear equations are given below,

$$\begin{gathered} x+2y+z=8 \\ 3x+7y+6z=26 \\ 2x+4y+2z=16 \end{gathered}$$

Part (a) Step 1. Given information.

Consider the given question,

The linear equations should have no solution.

Assume the following system of three equations containing three variables,

$$\begin{gathered} x+3y+2z=1 \\ 3x+10y+10z=5 \\ 2x+8y+12z=9 \end{gathered}$$

Part (a) Step 2. Find the solution of the equations.

Write the augmented matrix of the three equations,

$\left[\begin{array}{cccc}1& 3& 2& |1\\ 3& 10& 10& |5\\ 2& 8& 12& |9\end{array}\right]$

By applying the row operations, we get the row-echelon form,

$\left[\begin{array}{cccc}1& 3& 2& |1\\ 0& 1& 4& |2\\ 0& 0& 0& |3\end{array}\right]$

Here, in the last row, all entries left to the vertical bar are zeros, so the system has no solution.

Part (a) Step 1. Given information.

Consider the given question,

The linear equations should has exactly one solution.

Assume the following system of three equations containing three variables,

$$\begin{gathered} x-2y+3z=9 \\ -x+3y=-4 \\ 2x-5y+5z=17 \end{gathered}$$

Part (b) Step 2. Find the solution of the equations.

Write the augmented matrix of the three equations,

$\left[\begin{array}{cccc}1& -2& 3& |9\\ -1& 3& 0& |-4\\ 2& -5& 5& |17\end{array}\right]$

By applying the row operations, we get the reduced row-echelon form,

$\left[\begin{array}{cccc}1& 0& 0& |1\\ 0& 1& 0& |-1\\ 0& 0& 1& |2\end{array}\right]$

Here, the system has exactly one solution.

Part (c) Step 1. Given information.

Consider the given question,

The linear equations should has infinitely many solutions.

Assume the following system of three equations containing three variables,

$$\begin{gathered} x+2y+z=8 \\ 3x+7y+6z=26 \\ 2x+4y+2z=16 \end{gathered}$$

Part (c) Step 2. Find the solution of the equations.

Write the augmented matrix of the three equations,

$\left[\begin{array}{cccc}1& 2& 1& |8\\ 3& 7& 6& |26\\ 2& 4& 2& |16\end{array}\right]$

By applying the row operations, we get the row-echelon form,

$\left[\begin{array}{cccc}1& 0& -5& |4\\ 0& 1& 3& |2\\ 0& 0& 0& |0\end{array}\right]$

Here, in the last row, all entries are zeros, so by back substitution we will get only two equations with three variables.

Therefore, the system has infinitely many solution.