Chapter 10.3, Problem 25E

To determine

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

Answer to Problem 25E

The solution for given system of linear equation is $\left(-1,5,0\right)$ .

Explanation of Solution

Given:

Equation given,

  $\begin{array}{l}{x}_1+2x_2-x_3=9\\ 2x_1-x_3=-2\\ 3x_1+5x_2+2x_3=22\end{array}$

Concept Used:

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

Calculation:

Consider first the given equations,

  $\begin{array}{l}{x}_1+2x_2-x_3=9\\ 2x_1-x_3=-2\\ 3x_1+5x_2+2x_3=22\end{array}$

Step1:

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

  $\left[\begin{array}{cccc}1& 2& 1& 9\\ 2& 0& -1& -2\\ 3& 5& 2& 22\end{array}\right]$

Step 2:

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

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

Perform, $Row2-2*Row1=NewRow2$

  $Row2*2=\left[\begin{array}{cccc}2& 4& -2& 18\\ 2& 0& -1& -2\\ 3& 5& 2& 22\end{array}\right]$

  $Row2-2*Row1=\left[\begin{array}{cccc}1& 2& -1& 9\\ 0& -4& 1& -20\\ 3& 5& 2& 22\end{array}\right]$

  $NewRow2=$ $\left[\begin{array}{cccc}1& 2& -1& 9\\ 0& -4& 1& -20\\ 3& 5& 2& 22\end{array}\right]$

Step3:

The next row operation is,

  $Row3-3*Row1=NewRow3$

  $3*Row1=\left[\begin{array}{cccc}3& 6& -3& 27\\ 0& -4& 1& -20\\ 6& 5& 2& 22\end{array}\right]$

And,

  $Row3-3*Row1=\left[\begin{array}{cccc}1& 6& 1& 9\\ 0& 4& 1& 20\\ 0& 1& 2& 5\end{array}\right]$

  $NewRow3=$ $\left[\begin{array}{cccc}1& 6& 1& 9\\ 0& 4& 1& 20\\ 0& 1& 2& 5\end{array}\right]$

Step4:

The next row operation to be performed is,

  $Row2/-4=NewRow2$

  $Row2/-$

  $=\left[\begin{array}{cccc}1& 2& -1& 9\\ 0& 1& -\frac{1}{4}& 5\\ 0& -1& 5& -5\end{array}\right]$

After this, the next operation is,

  $Row3+Row2=NewRow3$

  $Row3+Row2$

  $=\left[\begin{array}{cccc}1& 2& -1& 9\\ 0& 1& -\frac{1}{4}& 5\\ 0& 0& \frac{19}{4}& 0\end{array}\right]$ \

Step5:

This step is called the row echelon form

Divide the above matrix ,

  $Row3by\frac{19}{4}$

The obtained matrix is,

  $\left[\begin{array}{cccc}1& 2& -1& 9\\ 0& 1& -\frac{1}{4}& 5\\ 0& 0& 1& 0\end{array}\right]$

The above matrix is in row echelon form, the equation written are,

  $\begin{array}{l}x+2y-z=9\\ y-\frac{1}{4}z=5\\ z=0\end{array}$

Step6:

The values of $\left(x,y,z\right)$ can be obtained using back substitution.

From equation $\left(3\right)$

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

To find the value of $y$ put $z=0$ in equation $\left(2\right)$

  $\begin{array}{l}y-\frac{1}{4}*\left(0\right)=5\\ y=5\\ \boxed{y=5}\end{array}$

To find the value of $x$ substitute the value of $y,z$ in equation $\left(1\right)$ .

  $\begin{array}{l}x+2*\left(5\right)-0=9\\ x+10=9\\ x=10-9\\ x=1\\ \boxed{x=1}\end{array}$

Conclusion:

Hence, the solution for given system is $\left(-1,5,0\right)$