Chapter 11.2, Problem 62AYU

To determine

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

Answer to Problem 62AYU

Solution:

Consistent system of equations, one solution.

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

Explanation of Solution

Given:

$x+y=1$

$2x-y+z=1$

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

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 $z$.

$x+y=1$

$2x-y+z=1$

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

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

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

$\approx \left[\begin{array}{ccc}1& 1& 0\\ 0& -3& 1\\ 0& 1& 1\end{array}|\begin{array}{c}1\\ -1\\ \frac{5}{3}\end{array}\right];R_2=-2r_1+r_2;R_3=r_3-r_1$

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

$\approx \left[\begin{array}{ccc}1& 1& 0\\ 0& 1& -\frac{1}{3}\\ 0& 0& \frac{4}{3}\end{array}|\begin{array}{c}1\\ \frac{1}{3}\\ \frac{4}{3}\end{array}\right];R_3^{\text{'}}=R_3-R_2$

$\approx \left[\begin{array}{ccc}1& 1& 0\\ 0& 1& -\frac{1}{3}\\ 0& 0& 1\end{array}|\begin{array}{c}1\\ \frac{1}{3}\\ 1\end{array}\right];R_3^{\text{'}\text{'}}=\frac{3}{4}{R}_3\text{'}$

Thus we get the following equations:

$x+y=1$

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

$z=1$

Solving we get,

$z=1$

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

$y-\frac{1}{3}\left(1\right)=\frac{1}{3}$

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

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

And

$x+y=1$

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

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

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

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