Chapter 16.9, Problem 5WE

To determine

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

  $\begin{array}{c}x+y-z=0\\ 2x+y-2z=1\\ 3x-4z=1\end{array}$

Answer to Problem 5WE

The solution set of system of equations is $\left(3,-1,2\right)$ .

Explanation of Solution

Given information:

The system of equations are

  $\begin{array}{c}x+y-z=0\\ 2x+y-2z=1\\ 3x-4z=1\end{array}$

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

  $\begin{array}{c}x+y-z=0\\ 2x+y-2z=1\\ 3x-4z=1\end{array}$

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,

  $\begin{array}{c}D=\left|\begin{array}{ccc}1& 1& -1\\ 2& 1& -2\\ 3& 0& -4\end{array}\right|\\ =1\left(-4\right)-1\left(-8+6\right)-1\left(-3\right)\\ =-4+2+3\\ =1\end{array}$

And,

  $\begin{array}{c}{D}_x=\left|\begin{array}{ccc}0& 1& -1\\ 1& 1& -2\\ 1& 0& -4\end{array}\right|\\ =0-1\left(-4+2\right)-1\left(-1\right)\\ =2+1\\ =3\end{array}$

And,

  $\begin{array}{c}{D}_y=\left|\begin{array}{ccc}1& 0& -1\\ 2& 1& -2\\ 3& 1& -4\end{array}\right|\\ =1\left(-4+2\right)-0-1\left(2-3\right)\\ =-2+1\\ =-1\end{array}$

And,

  $\begin{array}{c}{D}_z=\left|\begin{array}{ccc}1& 1& 0\\ 2& 1& 1\\ 3& 0& 1\end{array}\right|\\ =1\left(1\right)-1\left(2-3\right)\\ =1+1\\ =2\end{array}$

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,

  $\begin{array}{c}x=\frac{D_x}{D}\\ =\frac{3}{1}\\ =3\end{array}$

And,

  $\begin{array}{c}y=\frac{D_y}{D}\\ =\frac{-1}{1}\\ =-1\end{array}$

And,

  $\begin{array}{c}z=\frac{D_z}{D}\\ =\frac{2}{1}\\ =2\end{array}$

Thus, the solution set of system of equations is $\left(3,-1,2\right)$ .