Chapter 7.4, Problem 52E

To determine

To write: the system of linear equation corresponding to the given augmented matrix and find the solution.

Answer to Problem 52E

The system of linear equation corresponding to the given augmented matrixis

  $\{\begin{array}{l}x+2z=-7\\ y+z=9\\ z=-3\end{array}$

The solution is given by $x=-1,y=12,z=-3$ .

Explanation of Solution

Given information:

An augmented matrix is given as

  $\left[\begin{array}{ccccc}1& 0& 2& ⋮& -7\\ 0& 1& 1& ⋮& 9\\ 0& 0& 1& ⋮& -3\end{array}\right]$

Concept used:

Co-efficient matrix of a system of linear equation of the form $\{\begin{array}{l}{a}_{1}x+b_{1}y=c_1\\ a_{2}x+b_{2}y=c_2\end{array}$ is given by

  $\left[\begin{array}{cc}{a}_1& b_1\\ a_2& b_2\end{array}\right]$

The augmented matrix of the system of linear equation is given by

  $\left[\begin{array}{cccc}{a}_1& b_1& ⋮& c_1\\ a_2& b_2& ⋮& c_2\end{array}\right]$

Consider the given matrix.

  $\left[\begin{array}{ccccc}1& 0& 2& ⋮& -7\\ 0& 1& 1& ⋮& 9\\ 0& 0& 1& ⋮& -3\end{array}\right]$

Now, the corresponding system of equation will be

  $\{\begin{array}{l}x+2z=-7\\ y+z=9\\ z=-3\end{array}$

Variables x , y , zhave been used with the coefficients.

Now, substitute back $z=-3$ in $y+z=9$ .

  $\begin{array}{c}y+z=9\\ y+\left(-3\right)=9\\ y-3=9\\ y=12\end{array}$

Now, substitute back $z=-3,y=12$ in $x+2z=-7$ .

  $\begin{array}{c}x+2z=-7\\ x+2\left(-3\right)=-7\\ x-6=-7\\ x=-1\end{array}$

Therefore, solution of the system of linear equation is

  $x=-1,y=12,z=-3$