Chapter 11.2, Problem 26AYU

In Problems 27-38, 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 $x_1,x_2,x_3,x_4$ as variables. Determine whether the system is consistent or inconsistent. If it is consistent, give the solution.

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

To determine

To find: The system of equations corresponding to the given reduced row echelon augmented matrix and find the solution, if possible:

$\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}|\begin{array}{c}-4\\ 0\end{array}\right)$

Answer to Problem 26AYU

Solution:

$x+y=-4$

$x-y=-4$

$x=-4,y=0$

Explanation of Solution

Given:

$\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}|\begin{array}{c}-4\\ 0\end{array}\right)$

Calculation:

The system of equations corresponding to the given reduced row echelon augmented matrix is:

$x+y=-4$

$-x+y=4$

$\left(R_1=r_1+r_2;R_2=r_2-r_1\right)$

$x+y=-4$

$x-y=-4$

Solving above equation we get, $x=-4,y=0$.

Hence the system of equations is consistent.