Chapter 16, Problem 15CT

To determine

To find: The solution for the given system of equations.

Answer to Problem 15CT

The solution of the system is $\left(-\frac{1}{2},-1,-\frac{1}{2}\right)$ .

Explanation of Solution

Given information:

The given equations are.

  $\begin{array}{c}4x-3y=1\\ 6x-8z=1\\ y-2z=0\end{array}$ .

Calculation:

Creamer’s rule states that for a system of $n$ linear equations in $n$ variables, the solution is given by $x=\frac{D_x}{D},y=\frac{D_y}{D},\mathrm{...}$ where $D$ is determinant of the matrix of coefficient of the variables.

Such that $D\ne 0,$ and $D_x,D_y$ , are derived from $D$ by replacing the coefficients of $x,y$ respectively, by the constants.

If each element in one row or one column of a determinant is multiplied by a real number $k$ , then the determinant is multiplied by $k$ .

Solve the system of two equations in two variables.

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

Using the determinants.

  $\begin{array}{c}D=\left|\begin{array}{ccc}{a}_1& b_1& c_1\\ a_2& b_2& c_2\\ a_3& b_3& c_3\end{array}\right|\ne 0\\ D_x=\left|\begin{array}{ccc}{d}_1& b_1& c_1\\ d_2& b_2& c_2\\ d_3& b_3& c_3\end{array}\right|\\ D_y=\left|\begin{array}{ccc}{a}_1& d_1& c_1\\ a_2& d_2& c_2\\ a_3& d_3& c_3\end{array}\right|\\ D_z=\left|\begin{array}{ccc}{a}_1& b_1& d_1\\ a_2& b_2& d_2\\ a_3& b_3& d_3\end{array}\right|\end{array}$

Calculate the values of $D,D_x,D_y,D_z$ .

  $\begin{array}{c}D=\left|\begin{array}{ccc}4& -3& 0\\ 6& 0& -8\\ 0& 1& -2\end{array}\right|\ne 4\left(0+8\right)+3\left(-12-0\right)+0\left(6-0\right)=-4\\ D_x=\left|\begin{array}{ccc}1& -3& 0\\ 1& 0& -8\\ 0& 1& -2\end{array}\right|=1\left(0+8\right)+3\left(-2-0\right)+0\left(1-0\right)=2\\ D_y=\left|\begin{array}{ccc}4& 1& 0\\ 6& 1& -8\\ 0& 0& -2\end{array}\right|=4\left(-2+0\right)-1\left(-12-0\right)+0\left(0-0\right)=4\\ D_z=\left|\begin{array}{ccc}4& -3& 1\\ 6& 0& 1\\ 0& 1& 0\end{array}\right|=4\left(0-1\right)+3\left(0-0\right)+1\left(6-0\right)=2\end{array}$

Calculate the value of $x$ .

  $x=\frac{D_x}{D}=\frac{2}{-4}=-\frac{1}{2}$

Calculate the value of $y$ .

  $y=\frac{D_y}{D}=\frac{4}{-4}=-1$

Calculate the value of $z$ .

  $z=\frac{D_z}{D}=\frac{2}{-4}=-\frac{1}{2}$

Therefore, the solution of the system is $\left(-\frac{1}{2},-1,-\frac{1}{2}\right)$ .