Chapter 11.2, Problem 61AYU

To determine

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

Answer to Problem 61AYU

Solution:

Consistent system of equations, one solution.

$x=\frac{1}{3},y=\frac{2}{3},z=1$

Explanation of Solution

Given:

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

$2x-y+z=1$

$4x+\text{ 2}y=\frac{8}{3}$

Calculation:

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 $z$.

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

$2x-y+z=1$

$4x+\text{ 2}y=\frac{8}{3}$

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

$\left[\begin{array}{ccc}3& 1& -1\\ 2& -1& 1\\ 4& 2& 0\end{array}|\begin{array}{c}\frac{2}{3}\\ 1\\ \frac{8}{3}\end{array}\right]$

$\approx \left[\begin{array}{ccc}1& \frac{1}{3}& -\frac{1}{3}\\ 2& -1& 1\\ 4& 2& 0\end{array}|\begin{array}{c}\frac{2}{9}\\ 1\\ \frac{8}{3}\end{array}\right];R_1=\frac{1}{3}{r}_1$

$\approx \left[\begin{array}{ccc}1& \frac{1}{3}& -\frac{1}{3}\\ 0& -\frac{5}{3}& \frac{5}{3}\\ 0& \frac{2}{3}& \frac{4}{3}\end{array}|\begin{array}{c}\frac{2}{9}\\ \frac{5}{9}\\ \frac{16}{9}\end{array}\right]R_2=-2R_1+r_2;R_3=-4R_1+r_3$

$\approx \left[\begin{array}{ccc}1& \frac{1}{3}& -\frac{1}{3}\\ 0& 1& -1\\ 0& 1& 2\end{array}|\begin{array}{c}\frac{2}{9}\\ -\frac{1}{3}\\ \frac{8}{3}\end{array}\right];R_2^{\text{'}}=-\frac{3}{5}{R}_2;R_3^{\text{'}}=\frac{3}{2}{R}_3$

$\approx \left[\begin{array}{ccc}1& \frac{1}{3}& -\frac{1}{3}\\ 0& 1& -1\\ 0& 0& 3\end{array}|\begin{array}{c}\frac{2}{9}\\ -\frac{1}{3}\\ 3\end{array}\right];R_3^{\text{'}\text{'}}=R_3^{\text{'}}-R_2\text{'}$

$\approx \left[\begin{array}{ccc}1& \frac{1}{3}& -\frac{1}{3}\\ 0& 1& -1\\ 0& 0& 1\end{array}|\begin{array}{c}\frac{2}{9}\\ -\frac{1}{3}\\ 1\end{array}\right];R_3^{\text{'}\text{'}\text{'}}=\frac{1}{3}{R}_3\text{'}\text{'}$

Thus we get the following equations:

$x+\frac{1}{3}y-\frac{1}{3}z=\frac{2}{9}$

$y-z=-\frac{1}{3}$

$z=1$

Solving we get,

$z=1$

$y-z=-\frac{1}{3}$

$y-1=-\frac{1}{3}$

$y=1-\frac{1}{3}$

$y=\frac{2}{3}$

And

$x+\frac{1}{3}y-\frac{1}{3}z=\frac{2}{9}$

$x+\frac{1}{3}\left(\frac{2}{3}\right)-\frac{1}{3}\left(1\right)=\frac{2}{9}$

$x+\frac{2}{9}-\frac{1}{3}=\frac{2}{9}$

$x=\frac{2}{9}-\frac{2}{9}+\frac{1}{3}$

$x=\frac{1}{3}$

Hence the solutions are: $x=\frac{1}{3},y=\frac{2}{3},z=1$.