Chapter 11.2, Problem 59AYU

To determine

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

Answer to Problem 59AYU

Solution:

Consistent system of equations, one solution.

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

Explanation of Solution

Given:

$x+2y-z=-3$

$2x-4y+z=-7$

$-2x+2y-3z=4$

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:

$x+2y-z=-3$

$2x-4y+z=-7$

$-2x+2y-3z=4$

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

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

$\approx \left[\begin{array}{ccc}1& 2& -1\\ 0& -8& 3\\ 0& 6& -5\end{array}|\begin{array}{c}-3\\ -1\\ -2\end{array}\right];R_2=-2r_1+r_2;R_3=2r_1+r_3$

$\approx \left[\begin{array}{ccc}1& 2& -1\\ 0& 1& -\frac{3}{8}\\ 0& 1& -\frac{5}{6}\end{array}|\begin{array}{c}-3\\ \frac{1}{8}\\ -\frac{1}{3}\end{array}\right];R_2^{\text{'}}=-\frac{1}{8}\cdot R_2;R_3^{\text{'}}=\frac{1}{6}{R}_3$

$\approx \left[\begin{array}{ccc}1& 2& -1\\ 0& 1& -\frac{3}{8}\\ 0& 0& -\frac{11}{24}\end{array}|\begin{array}{c}-3\\ \frac{1}{8}\\ -\frac{11}{24}\end{array}\right];R_3^{\text{'}\text{'}}=R_3^{\text{'}}-R_2\text{'}$

$\approx \left[\begin{array}{ccc}1& 2& -1\\ 0& 1& -\frac{3}{8}\\ 0& 0& 1\end{array}|\begin{array}{c}-3\\ \frac{1}{8}\\ 1\end{array}\right];R_3^{\text{'}\text{'}\text{'}}=-\frac{24}{11}{R}_3\text{'}\text{'}$

Thus we get the following equations:

$x+2y-z=-3$

$y-\frac{3}{8}z=\frac{1}{8}$

$z=1$

Solving we get,

$z=1$

$y-\frac{3}{8}\left(1\right)=\frac{1}{8}$

$y=\frac{1}{8}+\frac{3}{8}$

$y=\frac{4}{8}$

$y=\frac{1}{2}$

And

$x+2y-z=-3$

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

$x+1-1=-3$

$x=-3$

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