Chapter 10.3, Problem 38E

To determine

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

Answer to Problem 38E

The system has no solution for the given system.

Explanation of Solution

Given:

Equation given,

  $\begin{array}{l}y-5z=7\\ 3x+2y=12\\ 3x+10z=80\end{array}$

Concept Used:

Calculation:

Consider first the given equations,

  $\begin{array}{l}y-5z=7\\ 3x+2y=12\\ 3x+10z=80\end{array}$

Step1:

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

  $\left[\begin{array}{cccc}0& 1& -5& 7\\ 3& 2& 0& 12\\ 3& 0& 10& 80\end{array}\right]$

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,

  $Row1\leftrightarrow Row3$ (Replace the row $1$ with row $3$ ,

  $\left[\begin{array}{cccc}3& 0& 10& 80\\ 3& 2& 0& 12\\ 0& 1& -5& 7\end{array}\right]$

The next row operation to be performed is,

  $Row2-Row1=NewRow2$

  $\left[\begin{array}{cccc}3& 0& 10& 80\\ 3& 2& 0& 12\\ 0& 1& -5& 7\end{array}\right]$

In the next operation, divide $Row1$ by $3$ and $Row2$ by $2$ to get the new $Row1$ and $Row2$ .

The matrix obtained is,

  $\left[\begin{array}{cccc}1& 0& \frac{10}{3}& \frac{80}{3}\\ 0& 1& -5& -34\\ 0& 1& -5& 7\end{array}\right]$

Now to obtain the $New$

  $Row3$ ,

Perform $Row3-Row2$

  $=NewRow3$

  $\left[\begin{array}{cccc}1& 0& \frac{10}{3}& \frac{80}{3}\\ 0& 1& -5& -34\\ 0& 0& 0& 41\end{array}\right]$

This matrix is now obtained in row- echelon form . Thus to stop the Gaussian elimination the last row of the matrix can be written in equation form as follows,

  $0x+0y+0z=41$ this equation is not true.

If any values of $x,y,z$ are chosen the last equation will be false. Thus the system does not have any solution.

Conclusion:

Hence, the given system of equation has no solution.