Chapter 11.3, Problem 33AYU

$\{\begin{array}{l}x+y-z=6\\ 3x-2y+z=-5\\ x+3y-2z=14\end{array}$

To determine

To find: The solution to the system of equations:

$x+y-z=6$

$3x-2y+z=-5$

$x+3y-2z=14$

Answer to Problem 33AYU

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

Explanation of Solution

Given:

$x+y-z=6$

$3x-2y+z=-5$

$x+3y-2z=14$

Formula used:

Cramer’s rule states that:

The solution of a system of three linear equations $a_{1}x+b_{1}y+c_{1}z=l;a_{2}x+b_{2}y+c_{2}z=m;a_{3}x+b_{3}y+c_{3}z=n$ is given by,

$x=\frac{D_x}{D};y=\frac{D_y}{D};z=\frac{D_z}{D}$

Where we have,

$D=\left|\begin{array}{ccc}{a}_1& b_1& c_1\\ a_2& b_2& c_2\\ a_3& b_3& c_3\end{array}\right|;D_x=\left|\begin{array}{ccc}l& b_1& c_1\\ m& b_2& c_2\\ n& b_3& c_3\end{array}\right|;D_y=\left|\begin{array}{ccc}{a}_1& l& c_1\\ a_2& m& c_2\\ a_3& n& c_3\end{array}\right|;D_z=\left|\begin{array}{ccc}{a}_1& b_1& l\\ a_2& b_2& m\\ a_3& b_3& n\end{array}\right|$

provided $D\ne 0$.

When $D=0$, the equations are inconsistent or dependent.

Calculation:

$x+y-z=6$

$3x-2y+z=-5$

$x+3y-2z=14$

Where we have:

$D=\left|\begin{array}{ccc}1& 1& -1\\ 3& -2& 1\\ 1& 3& -2\end{array}\right|=1\left(4-3\right)-1\left(-6-1\right)-1\left(9+2\right)=1\left(1\right)-1\left(-7\right)-1\left(11\right)=1+7-11=-3$

$D_x=\left|\begin{array}{ccc}6& 1& -1\\ -5& -2& 1\\ 14& 3& -2\end{array}\right|=6\left(4-3\right)-1\left(10-14\right)-1\left(-15+28\right)=6\left(1\right)-1\left(-4\right)-1\left(13\right)=6+4-13=-3$

$D_y=\left|\begin{array}{ccc}1& 6& -1\\ 3& -5& 1\\ 1& 14& -2\end{array}\right|=1\left(10-14\right)-6\left(-6-1\right)-1\left(42+5\right)=1\left(-4\right)-6\left(-7\right)-1\left(47\right)=-4+42-47=-9$

$D_z=\left|\begin{array}{ccc}1& 1& 6\\ 3& -2& -5\\ 1& 3& 14\end{array}\right|=1\left(-28+15\right)-1\left(42+5\right)+6\left(9+2\right)=1\left(-13\right)-1\left(47\right)+6\left(11\right)=-13-47+66=6$

So using Cramer’s rule we get,

$x=\frac{D_x}{D}=\frac{-3}{-3}=1$

$y=\frac{D_y}{D}=\frac{-9}{-3}=3$

$z=\frac{D_z}{D}=\frac{6}{-3}=-2$

Thus the solutions are $x=1;y=3;z=-2$.