Chapter 10.3, Problem 21E

To determine

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

Answer to Problem 21E

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

Explanation of Solution

Given:

Equation given,

  $\begin{array}{l}x+y+z=2\\ 2x-3y+2z=4\\ 4x+y-3z=1\end{array}$

Concept Used:

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

Calculation:

Consider first the given equations,

  $\begin{array}{l}x+y+z=2\\ 2x-3y+2z=4\\ 4x+y-3z=1\end{array}$

Step1:

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

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

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,

  $Row2-2*Row1=NewRow2$

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

Now perform, $Row2-NewRow1=NewRow2$

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

Step3:

Now subtract $Row3-4*Row1$ to get, $NewRow3$

  $Row3-4*Row1=NewRow3$

  $\left[\begin{array}{cccc}4& 4& 4& 8\\ 0& -5& 0& 0\\ 4& 1& -3& 1\end{array}\right]4*Row1$

  $\left[\begin{array}{cccc}1& 1& 1& 2\\ 0& -5& 0& 0\\ 0& -3& -7& -7\end{array}\right]Row3-4*Row1=NewRow3$

Step 4:

This step is called row-echelon form,

In the step 4 divide the $Row2$ by $-5$ to get $NewRow2$

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

Now the next operation to be performed is,

  $Row3+3*Row2=NewRow3$

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

  $\left[\begin{array}{cccc}1& 1& 1& 2\\ 0& 1& 0& 0\\ 0& 0& -7& -7\end{array}\right]Row3+3*Row2=NewRow3$

Now in the last step divide $Row3$ by $-7$ ,to get $Newrow3$

  $\left[\begin{array}{cccc}1& 1& 1& 2\\ 0& 1& 0& 0\\ 0/-7& 0/-7& -7/-7& -7/-7\end{array}\right]$ = $\left[\begin{array}{cccc}1& 1& 1& 2\\ 0& 1& 0& 0\\ 0& 0& 1& 1\end{array}\right]$

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

  $x+y+z=2$ ..... $\left(1\right)$

  $y=0$ ..... $\left(2\right)$

  $z=1$ ..... $\left(3\right)$

Step 5:

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

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

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

From equation $\left(2\right)$ to find $y$ ,

  $y=0$

  $\boxed{y=0}$

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

  $x+y+z=2$

  $\begin{array}{l}x+0+1=2\\ x+1=2\\ x=2-1\\ x=1\\ \boxed{x=1}\end{array}$

Conclusion:

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