Q. 30 - 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.    

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

Step-by-Step Solution

Answer

The system of equation is $x+4z=4, y+3z=2,0=0$and the system is consistent and the solution set is  $\left\{(x,y,z)|x=4-4z, y=2-3z, z is any real number\right\}$

Step 1. Given Information

The given matrix is   

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

Step 2. Explanation

The given matrix has three rows. So, it represents a system of three linear equations. The 3 columns to the left of vertical bar indicate that the system has three variables.

Let the variables be x, y and z. Then system of linear equation corresponding to the given matrix is as follows,

$$\begin{gathered} x+4z=4 \\ y+3z=2 \\ 0=0 \end{gathered}$$

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

Step 3. Calculation

Consider the equation $x+4z=4$and simplify,

$$\begin{gathered} x+4z-4z=4-4z \\ x=4-4z \end{gathered}$$

Next, consider $y+3z=2$ and simplify,

 $$\begin{gathered} y+3z-3z=2-3z \\ y=2-3z \end{gathered}$$

Here, we can see that values of x and y are independent in z. Thus, there will be infinitely many values for x and y for different values for z.

Thus, system is consistent and the solution set is $\left\{(x,y,z)|x=4-4z, y=2-3z, z \mathrm{is} \mathrm{any} \mathrm{real} \mathrm{number}\right\}$