Chapter 10.3, Problem 33E

To determine

The system of linear equation is inconsistent or dependent. If the system is dependent to find the complete solution.

Answer to Problem 33E

The given system of equation has infinitely no solutions.

Explanation of Solution

Given:

Equation given,

  $\begin{array}{l}x-y+3z=3\\ 4x-8y+32z=24\\ 2x-3y+11z=4\end{array}$

Concept Used:

The concept to find the complete solution of the system using the row operations is used.

Calculation:

Consider first the given equations,

  $\begin{array}{l}x-y+3z=3\\ 4x-8y+32z=24\\ 2x-3y+11z=4\end{array}$

Step1:

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

  $\left[\begin{array}{cccc}1& -1& 3& 3\\ 0& -8& 32& 24\\ 2& -3& 11& 4\end{array}\right]$

Step 2:

In the second step reduce the augmented matrix to convert it into row echelon form,

Transform the augmented matrix into row echelon form.

The first row operation is,

  $Row2-4*Row1=NewRow2$

  $4*Row1$

  $=\left[\begin{array}{cccc}4& -4& 12& 12\\ 4& -8& 32& 24\\ 2& -3& 11& 4\end{array}\right]$

And,

  $Row2-4*Row1=\left[\begin{array}{cccc}1& -1& 3& 3\\ 0& -4& 20& 12\\ 2& -3& 11& 4\end{array}\right]$

  $=NewRow2$

  $NewRow2=$ $\left[\begin{array}{cccc}1& -1& 3& 3\\ 0& -4& 20& 12\\ 2& -3& 11& 4\end{array}\right]$

The next operation to obtain $NewRow3$ is,

  $Row3-2*Row1=NewRow3$

  $2*Row1=\left[\begin{array}{cccc}2& -2& 6& 6\\ 0& -4& 20& 12\\ 2& -3& 11& 4\end{array}\right]$ and,

  $Row3-2*Row1=\left[\begin{array}{cccc}1& -1& 3& 3\\ 0& -4& 20& 12\\ 2& -1& 5& -2\end{array}\right]=$ $NewRow3$

Step3:

The next step of the row operation is,

  $Row2/4$ to get $NewRow2$

  $NewRow2$

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

Now the next operation to be performed to get the $NewRow3$ is,

  $Row3+Row2=$ $\left[\begin{array}{cccc}1& -1& 3& 3\\ 0& 1& -5& -3\\ 0& 0& 0& -5\end{array}\right]$

  $NewRow3$ = $\left[\begin{array}{cccc}1& -1& 3& 3\\ 0& 1& -5& -3\\ 0& 0& 0& -5\end{array}\right]$

Now the matrix is obtained in the row echloen form, thus the process of Gaussian elimination is stopped.

The last row of the above matrix can be written in equation form as,

  $0x+0y+0z=-5$ or also can be written as $0=-5$ . Choosing any values for $\left(x,y,z\right)$ from the last equation it will never be a true statement.

Thus the equation does not have any solution.

Conclusion:

Hence, the given system of equation has no solutions.