Chapter 11, Problem 20CT

To determine

To find: The solution of the system of equations.

Answer to Problem 20CT

The solutions of the system of equations are $x=1,y=1$ and $z=4$ .

Explanation of Solution

Given information:

The system of the equations is

  $\{\begin{array}{c}4x-3y+2z=15\\ -2x+y-3z=-15\\ 5x-5y+2z=18\end{array}$

Calculation:

Consider the system of the equations as

  $\{\begin{array}{c}4x-3y+2z=15\\ -2x+y-3z=-15\\ 5x-5y+2z=18\end{array}$

Solve the determinant whose element consists of the coefficients of the variable in the above system of equation.

  $\begin{array}{c}D=\left|\begin{array}{ccc}4& -3& 2\\ -2& 1& -3\\ 5& -5& 2\end{array}\right|\\ =4\left(-13\right)+3\left(11\right)+2\left(5\right)\\ =-9\end{array}$

The determinant is nonzero hence calculate the matrix $D_x,D_y$ and $D_z$ .

Calculate the value of $D_x$ .

  $\begin{array}{c}{D}_x=\left|\begin{array}{ccc}15& -3& 2\\ -15& 1& -3\\ 18& -5& 2\end{array}\right|\\ =15\left(-13\right)+3\left(24\right)+2\left(57\right)\\ =9\end{array}$

Calculate the value of $D_y$ .

  $\begin{array}{c}{D}_y=\left|\begin{array}{ccc}4& 15& 2\\ -2& -15& -3\\ 5& 18& 2\end{array}\right|\\ =4\left(24\right)-15\left(11\right)+2\left(39\right)\\ =9\end{array}$

Calculate the value of $D_z$ .

  $\begin{array}{c}{D}_z=\left|\begin{array}{ccc}4& -3& 15\\ -2& 1& -15\\ 5& -5& 18\end{array}\right|\\ =4\left(-57\right)+3\left(39\right)+15\left(5\right)\\ =-36\end{array}$

Calculate the value of $x$ .

  $\begin{array}{c}x=\frac{D_x}{D}\\ =\frac{-9}{-9}\\ =1\end{array}$

Calculate the value of $y$ .

  $\begin{array}{c}y=\frac{D_y}{D}\\ =\frac{-9}{-9}\\ =1\end{array}$

Calculate the value of $z$ .

  $\begin{array}{c}z=\frac{D_z}{D}\\ =\frac{-36}{-9}\\ =4\end{array}$

Therefore, the solutions of the system of equations are $x=1,y=1$ and $z=4$ .