Chapter 10.3, Problem 35E

To determine

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

Answer to Problem 35E

The given system of equation has infinitely no solutions.

Explanation of Solution

Given:

Equation given,

  $\begin{array}{l}x+4y-2z=-3\\ 2x-y+5z=12\\ 8x+5y+11z=30\end{array}$

Concept Used:

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

Calculation:

Consider first the given equations,

  $\begin{array}{l}x+4y-2z=-3\\ 2x-y+5z=12\\ 8x+5y+11z=30\end{array}$

Step1:

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

  $\left[\begin{array}{cccc}-2& 6& -2& -12\\ 1& -3& 2& 10\\ -1& 3& 2& 6\end{array}\right]$

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,

  $Row1/-2=NewRow1$

  $Row1/-2$

  $=$ $\left[\begin{array}{cccc}1& -3& 1& 6\\ 1& -3& 2& 10\\ -1& 3& 2& 6\end{array}\right]$

  $NewRow1$

  $=$ $\left[\begin{array}{cccc}1& -3& 1& 6\\ 1& -3& 2& 10\\ -1& 3& 2& 6\end{array}\right]$

The next row operation is,

  $Row2-Row1=NewRow2$

  $\left[\begin{array}{cccc}1& -3& 1& 6\\ 0& 0& 1& 4\\ -1& 3& 2& 6\end{array}\right]$

Step3:

The next row operation is to convert in row echelon form is,

  $Row3+Row1=NewRow3$

  $Row3+Row1=$ $\left[\begin{array}{cccc}1& -3& 1& 6\\ 0& 0& 1& 4\\ 0& 0& 3& 12\end{array}\right]$

The next operation to be performed is,

  $Row3/3=NewRow3$

  $Row3/3=$ $\left[\begin{array}{cccc}1& -3& 1& 6\\ 0& 0& 1& 4\\ 0& 0& 1& 4\end{array}\right]$

Now the last operation is,

  $Row2-Row3=NewRow2$

  $Row2-Row3$

  $\left[\begin{array}{cccc}1& -3& 1& 6\\ 0& 0& 0& 0\\ 0& 0& 1& 4\end{array}\right]$

In the second equation of the above matrix,

  $0=0$ . The equation will always be true irrespective of the values of $\left(x,y,z\right)$ .

Considering the last matrix the equation is written as,

  $x+4y-2x=-3$ .... $\left(1\right)$

  $y-z=-2$ .... $\left(2\right)$

As $z$ is not a leading variable it has infinitely many solutions.

The system is said to be dependent.

Conclusion:

Hence, the given system of equation has infinitely many solutions and is dependent.