Chapter 10.3, Problem 34E

To determine

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

Answer to Problem 34E

The given system has solution $\left(3t+2,t,4\right)$ where $t$ is the real number.

Explanation of Solution

Given:

Equation given,

  $\begin{array}{l}-2x+6y-2z=-12\\ x-3y+2z=10\\ -x+3y+2z=6\end{array}$

Concept Used:

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

Calculation:

Consider first the given equations,

  $\begin{array}{l}-2x+6y-2z=-12\\ x-3y+2z=10\\ -x+3y+2z=6\end{array}$

Step $1$ :

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 by using elementary row transformation to convert it into row echelon form,

Transform the augmented matrix into row echelon form.

The first row operation is,

  $Row1/-2=NewRow1$

  $Row1/-2=\left[\begin{array}{cccc}1& -3& 1& 6\\ 1& -3& 2& 10\\ -1& 3& 2& 6\end{array}\right]$

Step $3$ :

The next step of the row operation is,

  $Row2-Row1=NewRow2$

  $Row2-Row1=$ $\left[\begin{array}{cccc}1& -3& 1& 6\\ 0& 0& 1& 4\\ -1& 3& 2& 6\end{array}\right]$

  $NewRow2$

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

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

  $Row3+Row1=NewRow3$

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

  $NewRow3$ = $Row3+Row1=\left[\begin{array}{cccc}1& -3& 1& 6\\ 0& 0& 1& 4\\ 0& 0& 3& 12\end{array}\right]$

Step $4$ :

The next row operation to be performed is,

  $Row2-Row3=NewRow2$

  $Row2-Row3=\left[\begin{array}{cccc}1& -3& 1& 6\\ 0& 0& 0& 0\\ 0& 0& 1& 4\end{array}\right]$

  $NewRow2$

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

From the above matrix the third row has equation,

  $0=0$ . This equation is true irrespective of the values that are used for $\left(x,y,z\right)$ .

As the equation does not have any information it can be neglected from the matrix.

So, the remaining equations are,

  $x-3y+z=6$ ..... $\left(1\right)$

  $z=4$ ..... $\left(2\right)$

As, $y$ is not a leading variable. It has infinitely many solutions.

Thus the system is dependent.

Step $5$ :

As the system has infinitely many solutions based on the above matrix in which $z=4$ from the second equation.As, $y$ has infinitely many solutions put $y=t$ and solve to get the value of $x$

  $x-3y+z=6$ ..... $\left(1\right)$

Substitute the value of $y,z$ in above equation.

  $\begin{array}{l}x-3*\left(t\right)+z=6\\ x-3t+4=6\\ x=3t+2y,z\\ \boxed{3t+2,t,4}\end{array}$

Conclusion:

Hence, the given system has solution $\left(3t+2,t,4\right)$ where $t$ is the real number.