Chapter 11.2, Problem 49AYU

In Problems $39-74$, solve each system of equations using matrices ( row operations). If the system has no solution, say that it is inconsistent.

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

To determine

The solution of the system of linear equations using matrices (row operations),

$\{\begin{array}{ccccccc}x& -& 4y& +& 2z& =& -9\\ 3x& +& y& +& z& =& 4\\ -2x& +& 3y& -& 3z& =& 7\end{array}$.

Answer to Problem 49AYU

Solution:

The solution of the system is $x=1,\text{}\text{}y=2$, and $z=-1$, or $\left(1,2,-1\right)$.

Explanation of Solution

Given information:

The system of linear equations, $\{\begin{array}{ccccccc}x& -& 4y& +& 2z& =& -9\\ 3x& +& y& +& z& =& 4\\ -2x& +& 3y& -& 3z& =& 7\end{array}$

Explanation:

To solve this system using row operation method, first write these equations in augmented matrix.

Therefore, matrix becomes

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

Interchanging the rows $R_1\leftrightarrow R_2$

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

Perform the row operation $R_2=r_2-\frac{1}{3}.r_1$,

$\Rightarrow \left[\begin{array}{cccc}3& 1& 1& 4\\ 0& -\frac{13}{3}& \frac{5}{3}& -\frac{31}{3}\\ -2& 3& -3& 7\end{array}\right].$

Perform the row operation $R_3=r_3+\frac{2}{3}\cdot r_1$,

$\Rightarrow \left[\begin{array}{cccc}3& 1& 1& 4\\ 0& -\frac{13}{3}& \frac{5}{3}& -\frac{31}{3}\\ 0& \frac{11}{3}& -\frac{7}{3}& \frac{29}{3}\end{array}\right].$

Perform the operation $R_3=r_3+\frac{11}{13}\cdot r_2$,

$\Rightarrow \left[\begin{array}{cccc}3& 1& 1& 4\\ 0& -\frac{13}{3}& \frac{5}{3}& -\frac{31}{3}\\ 0& 0& -\frac{12}{13}& \frac{12}{13}\end{array}\right].$

Perform the operation $R_3=-\frac{13}{12}\cdot r_3$,

$\Rightarrow \left[\begin{array}{cccc}3& 1& 1& 4\\ 0& -\frac{13}{3}& \frac{5}{3}& -\frac{31}{3}\\ 0& 0& 1& -1\end{array}\right].$

By performing $R_2=r_2-\frac{5}{3}\cdot r_3$,

$\Rightarrow \left[\begin{array}{cccc}3& 1& 1& 4\\ 0& -\frac{13}{3}& 0& -\frac{26}{3}\\ 0& 0& 1& -1\end{array}\right].$

Using $R_1=r_1-r_3$,

$\Rightarrow \left[\begin{array}{cccc}3& 1& 0& 5\\ 0& -\frac{13}{3}& 0& -\frac{26}{3}\\ 0& 0& 1& -1\end{array}\right].$

To make leading element of second row $1$, perform $R_2=-\frac{3}{13}{r}_2$,

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

By performing $R_1=r_1-r_2$

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

To make leading element of first row $1$, perform, $R_1=\frac{1}{3}\cdot r_1$,

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

Therefore, the solution of system of equations $\{\begin{array}{ccccccc}x& -& 4y& +& 2z& =& -9\\ 3x& +& y& +& z& =& 4\\ -2x& +& 3y& -& 3z& =& 7\end{array}$ is, $x=1,y=2,z=-1$ or using ordered triplet $\left(1,2,-1\right)$.