Chapter 10.3, Problem 20E

To determine

The solution of given system of linear equations having unique solutions using Gaussian elimination or gauss-Jordan elimination.

Answer to Problem 20E

The solution for given system is $\left(-1,4,0\right)$ .

Explanation of Solution

Given:

Equation given,

  $\begin{array}{l}x+y+6z=3\\ x+y+3z=3\\ x+2y+4z=7\end{array}$

Concept Used:

The concept of Gaussian elimination is used to concert the linear system into row echelon form.

Calculation:

Consider first the given equations,

  $\begin{array}{l}x+y+6z=3\\ x+y+3z=3\\ x+2y+4z=7\end{array}$

Step1:

In the first step first convert the given, equation into augmented matrix.

  $\left[\begin{array}{cccc}1& 1& 6& 3\\ 1& 1& 3& 3\\ 1& 2& 4& 7\end{array}\right]$

Step 2:

In the second step convert the augmented matrix into row echelon form.

For this row operation are performed. In the first row operation,

  $Row2-Row1=NewRow2$

  $\left[\begin{array}{cccc}1& 1& 6& 3\\ 0& 0& -3& 0\\ 1& 2& 4& 7\end{array}\right]$

Step3:

Now subtract $Row3$ from $Row1$ to get, $NewRow3$

  $Row3-Row1=NewRow3$

  $\left[\begin{array}{cccc}1& 1& 6& 3\\ 0& 0& -3& 0\\ 0& 1& -2& 4\end{array}\right]$

Now replace $Row2$ and $Row3$

  $\left[\begin{array}{cccc}1& 1& 6& 3\\ 0& 1& -2& 4\\ 0& 0& -3& 0\end{array}\right]$

Step 4:

This step is called row-echelon form,

In the step 4 divide the $Row3$ by $-3$ to get $NewRow3$

  $Row3/-3=NewRow3$

  $\left[\begin{array}{cccc}1& 1& 6& 3\\ 0& 1& -2& 4\\ 0& 0& 1& 0\end{array}\right]$

Now the matrix is obtained in the row echelon form , write the equations from the matrix,

  $x+y+6z=3$ ..... $\left(1\right)$

  $y-2z=4$ ..... $\left(2\right)$

  $z=0$ ..... $\left(3\right)$

Step 5:

The value of $x,y,z$ can be obtained using back substitution,

From equation $\left(3\right)$ value of $z$ can be obatined,

  $\begin{array}{l}z=0\\ \boxed{z=0}\end{array}$

Substitute the value of $z=0$ in equation $\left(2\right)$ to find $y$ ,

  $y-2z=4$

  $\begin{array}{l}y-2*\left(0\right)=4\\ y-0=4\\ y=4\\ \boxed{y=4}\end{array}$

To find $x$ Substitute the value of $z$ and $y$ in equation $\left(1\right)$

  $x+y+6z=3$

  $\begin{array}{l}x+4+0=3\\ x+4=3\\ x=3-4\\ x=-1\\ \boxed{x=-1}\end{array}$

Conclusion:

Hence, the solution for given system is $\left(-1,4,0\right)$ .