Chapter 16.9, Problem 3ST

To determine

To calculate: The solution set for the system of equations,

  $\begin{array}{c}2x+y-z=1\\ x+3y+z=10\\ 3x-2y-8z=-3\end{array}$

Answer to Problem 3ST

The solution set of system of equations is $\left(-3,5,-2\right)$ .

Explanation of Solution

Given information:

The system of equations are

  $\begin{array}{c}2x+y-z=1\\ x+3y+z=10\\ 3x-2y-8z=-3\end{array}$

Formula used:

For system of equations

  $\begin{array}{c}{a}_{1}x+b_{1}y+c_{1}z=d_1\\ a_{2}x+b_{2}y+c_{2}z=d_2\\ a_{3}x+b_{3}y+c_{3}z=d_3\end{array}$

The determinant $D=\left|\begin{array}{ccc}{a}_1& b_1& c_1\\ a_2& b_2& c_2\\ a_3& b_3& c_3\end{array}\right|,D_x=\left|\begin{array}{ccc}{d}_1& b_1& c_1\\ d_2& b_2& c_2\\ d_3& b_3& c_3\end{array}\right|,D_y=\left|\begin{array}{ccc}{a}_1& d_1& c_1\\ a_2& d_2& c_2\\ a_3& d_3& c_3\end{array}\right|\text{ and }{D}_z=\left|\begin{array}{ccc}{a}_1& b_1& d_1\\ a_2& b_2& d_2\\ a_3& b_3& d_3\end{array}\right|$ .

The values of x, y and z are $x=\frac{D_x}{D},y=\frac{D_y}{D}\text{ and }z=\frac{D_z}{D}$.

Calculation:

Consider the system of equations are

  $\begin{array}{c}2x+y-z=1\\ x+3y+z=10\\ 3x-2y-8z=-3\end{array}$

Recall that for system of equations

  $\begin{array}{c}{a}_{1}x+b_{1}y+c_{1}z=d_1\\ a_{2}x+b_{2}y+c_{2}z=d_2\\ a_{3}x+b_{3}y+c_{3}z=d_3\end{array}$

The determinant $D=\left|\begin{array}{ccc}{a}_1& b_1& c_1\\ a_2& b_2& c_2\\ a_3& b_3& c_3\end{array}\right|,D_x=\left|\begin{array}{ccc}{d}_1& b_1& c_1\\ d_2& b_2& c_2\\ d_3& b_3& c_3\end{array}\right|,D_y=\left|\begin{array}{ccc}{a}_1& d_1& c_1\\ a_2& d_2& c_2\\ a_3& d_3& c_3\end{array}\right|\text{ and }{D}_z=\left|\begin{array}{ccc}{a}_1& b_1& d_1\\ a_2& b_2& d_2\\ a_3& b_3& d_3\end{array}\right|$ .

Apply it,

  $\begin{array}{c}D=\left|\begin{array}{ccc}2& 1& -1\\ 1& 3& 1\\ 3& -2& -8\end{array}\right|\\ =2\left(-24+2\right)-1\left(-8-3\right)-1\left(-2-9\right)\\ =-44+11+11\\ =-22\end{array}$

And,

  $\begin{array}{c}{D}_x=\left|\begin{array}{ccc}1& 1& -1\\ 10& 3& 1\\ -3& -2& -8\end{array}\right|\\ =1\left(-24+2\right)-1\left(-80+3\right)-1\left(-20+9\right)\\ =-22+77+11\\ =66\end{array}$

And,

  $\begin{array}{c}{D}_y=\left|\begin{array}{ccc}2& 1& -1\\ 1& 10& 1\\ 3& -3& -8\end{array}\right|\\ =2\left(-80+3\right)-1\left(-8-3\right)-1\left(-3-30\right)\\ =-154+11+33\\ =-110\end{array}$

And,

  $\begin{array}{c}{D}_z=\left|\begin{array}{ccc}2& 1& 1\\ 1& 3& 10\\ 3& -2& -3\end{array}\right|\\ =2\left(-9+20\right)-1\left(-3-30\right)+1\left(-2-9\right)\\ =22+33-11\\ =44\end{array}$

The values of x, y and z are $x=\frac{D_x}{D},y=\frac{D_y}{D}\text{ and }z=\frac{D_z}{D}$ .

Apply it,

  $\begin{array}{c}x=\frac{D_x}{D}\\ =\frac{66}{-22}\\ =-3\end{array}$

And,

  $\begin{array}{c}y=\frac{D_y}{D}\\ =\frac{-110}{-22}\\ =5\end{array}$

And,

  $\begin{array}{c}z=\frac{D_z}{D}\\ =\frac{44}{-22}\\ =-2\end{array}$

Thus, the solution set of system of equations is $\left(-3,5,-2\right)$ .