Chapter 10, Problem 6T

(a)

To determine

The complete solution of given system of equations.

Answer to Problem 6T

The solution of the system is $\left(\frac{5}{2},\frac{5}{2},0\right)$.

Explanation of Solution

The given system is,

$\{\begin{array}{c}x-y+2z=0\\ 2x-4y+5z=-5\\ 2y-3z=5\end{array}$

Write the augmented matrix of the system.

$\left[\begin{array}{cccc}1& -1& 2& 0\\ 2& -4& 5& -5\\ 0& 2& -3& 5\end{array}\right]$

Apply row operation $R_2\to R_2-2R_1$ to put the augmented matrix in row-echelon form.

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

Apply row operation $R_3\to R_3+R_2$ in above matrix.

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

Apply row operation $R_2\to -\frac{1}{2}{R}_2$ and $R_3=-\frac{1}{2}{R}_3$  in above matrix.

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

This is the row-echelon form.

Write the above matrix in the system of equations.

$x-y+2z=0$ (1)

$y-\frac{1}{2}z=\frac{5}{2}$ (2)

$z=0$

Substitute $0$ for $z$ in equation (2) to find $y$.

$y=\frac{5}{2}$

Substitute $0$ for $z$ and $\frac{5}{2}$ for $y$ in equation (1) to find $x$.

$x=\frac{5}{2}$

Thus, the solution of the system is $\left(\frac{5}{2},\frac{5}{2},0\right)$.

(b)

To determine

The complete solution of the given system of equations.

Answer to Problem 6T

The system $\begin{array}{c}2x-3y+z=3\\ x+2y+2z=-1\\ 4x+y+5z=4\end{array}$ has no solution.

Explanation of Solution

The given system is,

$\{\begin{array}{c}2x-3y+z=3\\ x+2y+2z=-1\\ 4x+y+5z=4\end{array}$

Write the augmented matrix of the system.

$\left[\begin{array}{cccc}2& -3& 1& 3\\ 1& 2& 2& -1\\ 4& 1& 5& 4\end{array}\right]$

Apply row operation $R_1\leftrightarrow R_2$ to put the augmented matrix in row-echelon form.

$\left[\begin{array}{cccc}1& 2& 2& -1\\ 2& -3& 1& 3\\ 4& 1& 5& 4\end{array}\right]$

Apply row operation $R_2\to R_2-2R_1$ and $R_3\to R_3-4R_1$ in above matrix.

$\left[\begin{array}{cccc}1& 2& 2& -1\\ 0& -7& -3& 5\\ 0& -7& -3& 8\end{array}\right]$

Apply row operation $R_3\to R_3-R_2$ in above matrix.

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

Apply row operation $R_2\to -\frac{1}{7}{R}_2$ in above matrix.

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

Above matrix is in row echelon form but the coefficient of $z$ is zero.

No matter what the values of $x,y\text{and}z$, the last statement will never be a true statement.

Thus, the given system of equations has no solution.