Chapter 11.2, Problem 58AYU

To determine

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

Answer to Problem 58AYU

Solution:

Consistent system of equations, one solution.

$x=1,y=3,z=-2$

Explanation of Solution

Given:

$x-y+z=-4$

$2x-3y+4z=-15$

$5x+y-2z=12$

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:

$x-y+z=-4$

$2x-3y+4z=-15$

$5x+y-2z=12$

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

$\left[\begin{array}{ccc}1& -1& 1\\ 2& -3& 4\\ 5& 1& -2\end{array}|\begin{array}{c}-4\\ -15\\ 12\end{array}\right]$

$\approx \left[\begin{array}{ccc}1& -1& 1\\ 0& -1& 2\\ 0& 6& -7\end{array}|\begin{array}{c}-4\\ -7\\ 32\end{array}\right];R_2=-2r_1+r_2;R_3=-5r_1+r_3$

$\approx \left[\begin{array}{ccc}1& -1& 1\\ 0& 1& -2\\ 0& 6& -7\end{array}|\begin{array}{c}-4\\ 7\\ 32\end{array}\right];R_2\text{'}=-1.R_2$

$\approx \left[\begin{array}{ccc}1& -1& 1\\ 0& 1& -2\\ 0& 0& 5\end{array}|\begin{array}{c}-4\\ 7\\ -10\end{array}\right];R_3^{\text{'}}=-6R_2^{\text{'}}+R_3$

$\approx \left[\begin{array}{ccc}1& -1& 1\\ 0& 1& -2\\ 0& 0& 1\end{array}|\begin{array}{c}-4\\ 7\\ -2\end{array}\right];R_3^{\text{'}\text{'}}=\frac{1}{5}{R}_3\text{'}$

Thus we get the following equations:

$x-y+z=-4$

$y-2z=7$

$z=-2$

Solving we get,

$z=-2$

$y-2\left(-2\right)=7$

$y+4=7$

$y=7-4$

$y=3$

And

$x-y+z=-4$

$x-3-2=-4$

$x-5=-4$

$x=-4+5$

$x=1$

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