Chapter 10.3, Problem 41E

To determine

Solve the system of linear equations.

Answer to Problem 41E

The given system has no solution.

Explanation of Solution

Given:

Equation given,

  $\begin{array}{l}2x+y+3z=9\\ -x-7z=10\\ 3x+2y-z=4\end{array}$

Concept Used:

Concept of Gaussian elimination to solve system of linear equation is used.

Calculation:

Consider first the given equations,

  $\begin{array}{l}2x+y+3z=9\\ -x-7z=10\\ 3x+2y-z=4\end{array}$

Step $1$ :

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

  $\left[\begin{array}{cccc}2& 1& 3& 9\\ -1& 0& -7& 10\\ 3& 2& -1& 4\end{array}\right]$

Step $2$ :

In the second step convert the augmented matrix and use the Gauss Elimination method.

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

  $-R_2\leftrightarrow R_1$

  $-R_2\leftrightarrow R_1$

  $=\left[\begin{array}{cccc}1& 0& 7& -10\\ 2& 1& 3& 9\\ 3& 2& -1& 4\end{array}\right]$

The next row operation is,

  $R_2-2R_1\to R_2=$ $\left[\begin{array}{cccc}1& 0& 7& -10\\ 0& 1& -11& 29\\ 3& 2& -22& 34\end{array}\right]$

And,

  $R_3-3R_1\to R_3=$ $=\left[\begin{array}{cccc}1& 0& 7& -10\\ 0& 1& -11& 29\\ 0& 2& -22& 34\end{array}\right]$

Step $3$ :

The next row operation to be performed is,

Subtract $2*Row2$ from $Row3$

  $R_3-2R_2\to R_3$

  $=\left[\begin{array}{cccc}1& 0& 7& -10\\ 0& 1& -11& 29\\ 0& 0& 0& -24\end{array}\right]$

Step4:

The final matrix obtained is ,

  $\left[\begin{array}{cccc}1& 0& 7& -10\\ 0& 1& -11& 29\\ 0& 0& 0& -24\end{array}\right]$

The third row of the above matrix shows that $0=-24$ .Thus the system has no solution.

Conclusion:

Hence, the given system has no solution.