Question

Solve using Cramer's rule

$\left\{\begin{array}{l}3x+y-6z=-3\\ 2x+6y+3z=0\\ 3x+2y-3z=-6\end{array}\right.$

Short answer

$$\begin{gathered} \mathrm{The}\mathrm{solution}\mathrm{of}\mathrm{the}\mathrm{linear}\mathrm{equations}\mathrm{is}(x=-6, \\ y=3, \\ z=-2). \end{gathered}$$

Step-by-step solution

Step 1. Given information

The given linear equations are

$\left\{\begin{array}{l}3x+y-6z=-3\\ 2x+6y+3z=0\\ 3x+2y-3z=-6\end{array}\right.$

Step 2. First find the determinant D by using the coefficients of the variables.  

$$\begin{gathered} D=\left|\begin{array}{ccc}3& 1& -6\\ 2& 6& 3\\ 3& 2& -3\end{array}\right| \\ D=3\left(\right(6\times -3)-(3\times 2\left)\right)-1\left(\right(2\times -3)-(3\times 3\left)\right)-6\left(\right(2\times 2)-(6\times 3\left)\right) \\ D=3(-18-6)-1(-6-9)-6(4-18) \\ D=-72+15+84 \\ D=27 \end{gathered}$$

Step 3.  Evaluate the determinant Dx, Dy and Dz by using the constants to replace the coefficients of x, y and z. 

$$\begin{gathered} D_x=\left|\begin{array}{ccc}-3& 1& -6\\ 0& 6& 3\\ -6& 2& -3\end{array}\right| \\ {D}_x=-3\left(\right(6\times -3)-(3\times 2\left)\right)-1\left(\right(0\times -3)-(3\times -6\left)\right)-6\left(\right(2\times 0)-(6\times -6\left)\right) \\ {D}_x=-3(-18-6)-1\left(18\right)-6\left(36\right) \\ {D}_x=+72-18-216 \\ {D}_x=-162 \\ {D}_y=\left|\begin{array}{ccc}3& -3& -6\\ 2& 0& 3\\ 3& -6& -3\end{array}\right| \\ {D}_y=3\left(\right(0\times -3)-(3\times -6\left)\right)+3\left(\right(2\times -3)-(3\times 3\left)\right)-6\left(\right(-6\times 2)-(0\times 3\left)\right) \\ {D}_y=3\left(18\right)+3(-6-9)-6(-12) \\ {D}_y=54-45+72 \\ {D}_y=81 \\ {D}_z=\left|\begin{array}{ccc}3& 1& -3\\ 2& 6& 0\\ 3& 2& -6\end{array}\right| \\ {D}_z=3\left(\right(6\times -6)-(0\times 2\left)\right)-1\left(\right(2\times -6)-(0\times 3\left)\right)-3\left(\right(2\times 2)-(6\times 3\left)\right) \\ {D}_z=3(-36)-1(-12)-3(4-18) \\ {D}_z=-108+12+42 \\ {D}_z=-54 \end{gathered}$$

Step 4. Find x, y and z. 

$$\begin{gathered} x=\frac{D_x}{D}, \\ y=\frac{D_y}{D}\mathrm{and} \\ z=\frac{D_z}{D} \\ x=\frac{-162}{27}=-6 \\ y=\frac{81}{27}=3 \\ z=\frac{-54}{27}=-2 \end{gathered}$$

The solution of the linear equations is $(-6,3,-2)$.

By putting the values in the equations it satisfies the equations and the solution to these linear equations are $(-6,3,-2)$.

Step 5. Checking the equation

Putting the values
$$\begin{gathered} x=-6 \\ y=3 \\ z=-2 \end{gathered}$$
$$\begin{gathered} 3\mathrm{x}+\mathrm{y}-6\mathrm{z}=-3 \\ 3(-6)+3-6(-2)=-3 \\ -18+3+12=-3 \\ -3=-3 \\ \mathrm{true} \\ \mathrm{Now}\mathrm{Putting}\mathrm{the}\mathrm{value}\mathrm{of}\mathrm{x}=-6, \\ \mathrm{y}=3 \\ \mathrm{and}\mathrm{z}=-2 \\ \mathrm{in}\mathrm{the}\mathrm{equation}2\mathrm{x}+6\mathrm{y}+3\mathrm{z}=0 \\ 2(-6)+6\left(3\right)+3(-2)=0 \\ -12+18-6=0 \\ 0=0 \\ \mathrm{Putting}\mathrm{the}\mathrm{value}\mathrm{of}\mathrm{x}=-6, \\ \mathrm{y}=3 \\ \mathrm{and}\mathrm{z}=-2 \\ \mathrm{in}\mathrm{the}\mathrm{equation}3\mathrm{x}+2\mathrm{y}-3\mathrm{z}=-6 \\ 3(-6)+2\left(3\right)-3(-2)=-6 \\ -18+6+6=-6 \\ -6=-6 \end{gathered}$$

Hence these points satisfy the equations hence they are the solution to these equations.