Chapter 11.2, Problem 50AYU

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

$\{\begin{array}{l}2x+y-3z=0\\ -2x+2y+z=-7\\ 3x-4y-3z=7\end{array}$

To determine

To find: The solution to the given system of equations using matrices.

Answer to Problem 50AYU

Solution:

$x=\frac{56}{13},y=-\frac{7}{13},z=\frac{35}{13}$

Consistent system.

Explanation of Solution

Given:

$2x+y-3z=0$

$-2x+2y+z=-7$

$3x-4y-3z=7$

Formula used:

To solve a system of three equations in $x,y$ and $z$ using matrices:

Step 1: Write the corresponding matrix associated with the system of equations.

Step 2: Use elementary row operations to get equivalent matrix of the form:

$\left[\begin{array}{ccc}1& a& b\\ 0& 1& c\\ 0& 0& 1\end{array}|\begin{array}{c}d\\ e\\ f\end{array}\right]$; where $a,b,c,d,e,f$ are constants.

Step 3: Solve for $x,y$ and $z$.

Calculation:

$2x+y-3z=0$

$-2x+2y+z=-7$

$3x-4y-3z=7$

The corresponding matrix associated with the above system of equations is:

$\left[\begin{array}{ccc}2& 1& -3\\ -2& 2& 1\\ 3& -4& -3\end{array}|\begin{array}{c}0\\ -7\\ 7\end{array}\right]$

$\approx \left[\begin{array}{ccc}1& \frac{1}{2}& -3\\ -2& 2& 1\\ 3& -4& -3\end{array}|\begin{array}{c}0\\ -7\\ 7\end{array}\right];R_1=\frac{1}{2}{r}_1$

$\approx \left[\begin{array}{ccc}1& \frac{1}{2}& -\frac{3}{2}\\ 0& 3& -2\\ 3& -4& -3\end{array}|\begin{array}{c}0\\ -7\\ 7\end{array}\right];R_2=2R_1+r_2$

$\approx \left[\begin{array}{ccc}1& \frac{1}{2}& -\frac{3}{2}\\ 0& 1& -\frac{2}{3}\\ 3& -4& -3\end{array}|\begin{array}{c}0\\ -\frac{7}{3}\\ 7\end{array}\right];R_2^{\text{'}}=\frac{1}{3}{R}_2$

$\approx \left[\begin{array}{ccc}1& \frac{1}{2}& -\frac{3}{2}\\ 0& 1& -\frac{2}{3}\\ 0& -\frac{11}{2}& \frac{3}{2}\end{array}|\begin{array}{c}0\\ -\frac{7}{3}\\ 7\end{array}\right];R_3=-3R_1+r_3$

$\approx \left[\begin{array}{ccc}1& \frac{1}{2}& -\frac{3}{2}\\ 0& 1& -\frac{2}{3}\\ 0& 0& -\frac{13}{6}\end{array}|\begin{array}{c}0\\ -\frac{7}{3}\\ -\frac{35}{6}\end{array}\right];R_3^{\text{'}}=\frac{11}{2}{R}_2+R_3$

$\approx \left[\begin{array}{ccc}1& \frac{1}{2}& -\frac{3}{2}\\ 0& 1& -\frac{2}{3}\\ 0& 0& 1\end{array}|\begin{array}{c}0\\ -\frac{7}{3}\\ \frac{35}{13}\end{array}\right];R_3^{\text{'}\text{'}}=-\frac{6}{13}{R}_3\text{'}$

Thus we get the following equations:

$x+\frac{1}{2}y-\frac{3}{2}z=0$

$y-\frac{2}{3}z=-\frac{7}{3}$

$z=\frac{35}{13}$

Solving we get,

$y-\frac{2}{3}\left(\frac{35}{13}\right)=-\frac{7}{3}$

$y-\frac{70}{39}=-\frac{7}{3}$

$y=-\frac{7}{3}+\frac{70}{39}$

$y=-\frac{21}{39}$ or $y=-\frac{7}{13}$

$x+\frac{1}{2}\left(-\frac{7}{13}\right)-\frac{3}{2}\left(\frac{35}{13}\right)=0$

$x-\frac{7}{26}-\frac{105}{26}=0$

$x=\frac{112}{26}$

$x=\frac{56}{13}$

Hence the solutions are: $x=\frac{56}{13},y=-\frac{7}{13},z=\frac{35}{13}$.