Chapter 10.3, Problem 24E

To determine

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

Answer to Problem 24E

The solution for given system of linear equation is $\left(3,1,2\right)$ .

Explanation of Solution

Given:

Equation given,

  $\begin{array}{l}2y+z=4\\ x+y=4\\ 3x+3y-z=10\end{array}$

Concept Used:

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

Calculation:

Consider first the given equations,

  $\begin{array}{l}2y+z=4\\ x+y=4\\ 3x+3y-z=10\end{array}$

Step1:

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

  $\left[\begin{array}{cccc}2& 0& 1& 4\\ 1& 1& 0& 4\\ 3& 3& -1& 10\end{array}\right]$

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, $R_1\to R_2$

  $R_1\to R_2=$ $\left[\begin{array}{cccc}2& 0& 1& 4\\ 1& 1& 0& 4\\ 3& 3& -1& 10\end{array}\right]\leftrightarrow \left[\begin{array}{cccc}1& 1& 0& 4\\ 2& 0& 1& 4\\ 3& 3& -1& 10\end{array}\right]$

Then again row operation , $R_3\to R_3-3*R_1$

  $R_3\to R_3-3*R_1$ = $\left[\begin{array}{cccc}1& 1& 0& 4\\ 2& 0& 1& 4\\ 3& 3& -1& 10\end{array}\right]\to \left[\begin{array}{cccc}1& 1& 0& 4\\ 0& 2& 1& 4\\ 0& 0& -1& -2\end{array}\right]$

After this perform operation on,

  $R_2\to \frac{R_2}{2}$

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

Step3:

In the step 3 again perform row operations,

First perform,

  $R_1-R_3=$ $\left[\begin{array}{cccc}1& 1& 0& 4\\ 0& 1& \frac{1}{2}& 2\\ 0& 0& -1& -2\end{array}\right]\to \left[\begin{array}{cccc}1& 1& -\frac{1}{2}& 2\\ 0& 1& \frac{1}{2}& 2\\ 0& 0& -1& -2\end{array}\right]$

Now,

  $R_1\to \left(-1\right)R_3$

  $=\left[\begin{array}{cccc}1& 1& -\frac{1}{2}& 2\\ 0& 1& \frac{1}{2}& 2\\ 0& 0& -1& -2\end{array}\right]\to \left[\begin{array}{cccc}1& 1& -\frac{1}{2}& 2\\ 0& 1& \frac{1}{2}& 2\\ 0& 0& 1& 22\end{array}\right]$

And ,

  $R_1\to R_1+\frac{1}{2}{R}_3=\left[\begin{array}{cccc}1& 0& -\frac{1}{2}& 2\\ 0& 1& \frac{1}{2}& 2\\ 0& 0& 1& 2\end{array}\right]\to \left[\begin{array}{cccc}1& 0& 0& 3\\ 0& 1& \frac{1}{2}& 2\\ 0& 0& 1& 2\end{array}\right]$

In the last row operation

  $R_2\to R_2-\frac{1}{2}{R}_3=\left[\begin{array}{cccc}1& 0& 0& 3\\ 0& 1& \frac{1}{2}& 2\\ 0& 0& 1& 2\end{array}\right]\to \left[\begin{array}{cccc}1& 0& 0& 3\\ 0& 1& 0& 1\\ 0& 0& 1& 2\end{array}\right]$

Step 4:

This step is called row-echelon form,

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

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

  $x=3$ ..... $\left(1\right)$

  $y=1$ ..... $\left(2\right)$

  $z=2$ ..... $\left(3\right)$

Thus the values of $\left(x,y,z\right)=\left(3,1,2\right)$

Conclusion:

Hence, the solution for given system is $\left(3,1,2\right)$