Q. 31 - Precalculus Enhanced with Graphing Utilities

Question

The reduced row echelon form of a system of linear equations is given. Write the system of equations corresponding to the given matrix. Use x, y; or x, y, z; or x1, x2, x3, x4 as variables. Determine whether the system is consistent or inconsistent. If it is consistent, give the solution.  

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

Step-by-Step Solution

Answer

The system of equation is $x_1=1, x_2+x_4=2, x_3+2x_4=3$ and the system is consistent and the solution set is $x_1=1, x_2=2-x_4, x_3=3-2x_4, x_4 is any real number$

Step 1. Given Information

The given matrix is    

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

Step 2. Explanation

The given matrix has three rows. So, it represents a system of three linear equations. The 4 columns to the left of vertical bar indicate that the system has four variables.

Let the variables be $x_1, x_2, x_3 and x_4$. Then system of linear equation corresponding to the given matrix is as follows,

$$\begin{gathered} x_1=1, \\ {x}_2+x_4=2, \\ x_3+2x_4=3 \end{gathered}$$ 

Step 3. Explanation

A system of linear equation is said to be consistent when it has at least one solution otherwise it is inconsistent.

The bottom row of the matrix is equivalent to the equation $x_3+2x_4=3$ which has a solution. Thus, the system is consistent and the solution set is

$$\begin{gathered} x_1=1, \\ {x}_2=2-x_4, \\ {x}_3=3-2x_4, \\ {x}_4 \mathrm{is} \mathrm{any} \mathrm{real} \mathrm{number} \end{gathered}$$