Question

In exercises, 1 through 10, find all solutions of the linear systems using elimination.Then check your solutions.

8.$$\begin{gathered} \left|x+2y+3z=0 \\ 4x+5y+6z=0 \\ 7x+8y+10z=0\right| \end{gathered}$$

Short answer

The system of equation has only trivial solution.The solution of system of equation is$x=0,y=0,z=0$

Step-by-step solution

Step 1:Transforming the equation

To get the solution, we will transform the value of x,y,z.

$$\begin{gathered} \left|x+2y+3z=0......\left(1\right) \\ 4x+5y+6z=0......\left(2\right) \\ 7x+8y+10z=0......\left(3\right)\right| \end{gathered}$$Into the form$$\begin{gathered} \left|x=.... \\ y=.... \\ z=... \\ \right| \end{gathered}$$

Eliminating the variables

In the given system of equations, we can eliminate the variables by adding or subtracting the equations.

In this system, we can eliminate the variable x from equation 2 and 3. First of all subtract the 4 times of equation 1 from equation2 and then subtract 7 times of first equation from equation 3.

$$\begin{gathered} \begin{array}{rcl}\left|x+2y+3z=0 \\ 4x-4x+5y-8y+6z-12z=0 \\ 7x-7x+8y-14y+10z-21z=0\right|& =& \left|x+2y+3z=0 \\ -3y-6z=0 \\ -6y-11z=0\right|\end{array} \end{gathered}$$

Now multiply the equation 2 by 2 and then subtract the equation 3 from equation 2.

$$\begin{gathered} \begin{array}{rcl}\left|x+2y+3z=0 \\ -6y-12z=0 \\ -6y-11z=0\right|& =& \left|x+2y+3z=0 \\ -3y-6z=0 \\ z=0\right|\end{array} \end{gathered}$$

Now put the value of z which is in the third equation to the first equation and second equation.

$$\begin{gathered} \begin{array}{rcl}\left|x+2\left(0\right)+3\left(0\right)=0 \\ y=0 \\ z=0\right|& =& \left|x=0 \\ y=0 \\ z=0\right|\end{array} \end{gathered}$$

Hence, we get value of$x=0,y=0,z=0$. Thus the system has only trivial solution.

Checking the solution

Now check the solution by putting the value of x, y and z in the given system of equation.

$\begin{array}{rcl}0+2\left(0\right)+3\left(0\right)& =& 0\\ 4\left(0\right)+5\left(0\right)+6\left(0\right)& =& 0\\ 7\left(0\right)+8\left(0\right)+10\left(0\right)& =& 0\\ & & \end{array}$

Hence, the solution of system of equation is$x=0,y=0,z=0$