Q. 29 - 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.

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

Step-by-Step Solution

Answer

The system of equation is $x+2z=-1, y-4z=-2,0=0$and the system is consistent and the solution set is $\left\{(x,y,z)|x=-1-2z, y=-2+4z, z is any real number\right\}$

Step 1. Given Information

The given matrix is  

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

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,

$$\begin{gathered} x+2z=-1 \\ y-4z=-2 \\ 0=0 \end{gathered}$$

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 $x+2z=-1$and simplify,

$$\begin{gathered} x+2z-2z=-1-2z \\ x=-1-2z \end{gathered}$$

Next, consider $y-4z=-2$ and simplify,

$$\begin{gathered} y-4z+4z=-2+4z \\ y=-2+4z \end{gathered}$$

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 solution set is $\left\{(x,y,z)|x=-1-2z,y=-2+4z, z \mathrm{is} \mathrm{any} \mathrm{real} \mathrm{number}\right\}$