Chapter 10.3, Problem 26E

To determine

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

Answer to Problem 26E

The solution for given system is $\left(x,y,z\right)=\left(3,1,1\right)$ .

Explanation of Solution

Given:

Equation given,

  $\begin{array}{l}2x_1+x_2=7\\ 2x_1-x_2+x_3=6\\ 3x_1-2x_2+4x_3=11\end{array}$

Concept Used:

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

Calculation:

Consider first the given equations,

  $\begin{array}{l}2x_1+x_2=7\\ 2x_1-x_2+x_3=6\\ 3x_1-2x_2+4x_3=11\end{array}$

Step1:

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

  $\left[\begin{array}{cccc}2& 1& 0& 7\\ 2& -1& 1& 6\\ 3& -2& 4& 11\end{array}\right]$

Step 2:

In the second step elementary row operations are performed to transform the augmented matrix into row echelon form.

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

  $Row2-Row1=NewRow2$

  $Row2-Row1=\left[\begin{array}{cccc}2& 1& 0& 7\\ 0& -2& 1& -1\\ 3& -2& 4& 11\end{array}\right]$

Again the next row operation is,

  $Row1/2=Newrow1$

  $Row1/2=\left[\begin{array}{cccc}1& 0.5& 0& 3.5\\ 0& -2& 1& 1\\ 3& -2& 4& 11\end{array}\right]$

Step3:

In the Step3 again perform the row operations as follows,

  $Row3-3*Row1=NewRow3$

  $3*Row1=\left[\begin{array}{cccc}3& 1.5& 0& 10.5\\ 0& -2& 1& -1\\ 3& -2& 4& 11\end{array}\right]$

  $Row3-3*Row1=NewRow3=\left[\begin{array}{cccc}1& 0.5& 0& 3.5\\ 0& -2& 1& 1\\ 3& 3.5& 4& 0.5\end{array}\right]$

The next operation to be performed is,

  $Row2/-2=NewRow2$

  $Row2/-2=$ $\left[\begin{array}{cccc}3& 0.5& 0& 10.5\\ 0& 1& -0.5& 0.5\\ 3& -3.5& 4& 0.5\end{array}\right]$

Again the next elementary row operation is,

  $Row3+3.5*Row2=NewRow3$

  $3.5*Row2=\left[\begin{array}{cccc}1& 0.5& 0& 10.5\\ 0& 3.5& -1.75& 1.75\\ 0& -3.5& 4& 0.5\end{array}\right]$

  $Row3+3.5*Row2=NewRow3=\left[\begin{array}{cccc}1& 0.5& 0& 3.5\\ 0& 1& -0.5& 05\\ 0& 0& 2.25& 2.25\end{array}\right]$

Step 4:

This step is called row-echelon form,

In the step 4 divide the $Row3$ by $2.25$ to get $NewRow3$ .

  $\frac{Row3}{2.25}=NewRow3$

  $\frac{Row3}{2.25}=$ $NewRow3$

  $=\left[\begin{array}{cccc}1& 0.5& 0& 3.5\\ 0& 1& -0.5& 0.5\\ 0& 0& 1& 1\end{array}\right]$

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

  $x+0.5y=3.5$ ….. $\left(1\right)$

  $y-0.5z=0.5$ ….. $\left(2\right)$

  $z=1$ …… $\left(3\right)$

Step 5:

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

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

  $\begin{array}{l}z=1\\ \boxed{z=1}\end{array}$

Substitute the value of $z=1$ in equation $\left(2\right)$ to find $y$ ,

  $y-0.5z=0.5$

  $\begin{array}{l}y-0.5*\left(1\right)=0.5\\ y-0.5=0.5\\ y=0.5+0.5\\ y=1\\ \boxed{y=1}\end{array}$

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

  $x+0.5y=3.5$

  $\begin{array}{l}x+0.5*\left(1\right)=3.5\\ x+0.5=3.5\\ x=3.5-0.5\\ x=3\\ \boxed{x=3}\end{array}$

Conclusion:

Hence, the solution for given system is $\left(x,y,z\right)=\left(3,1,1\right)$ .