When using Cramer's rule to solve a system of equations, the values of the calculated determinants reveal whether the system is dependent or inconsistent. If the main coefficient determinant is $$D = 0$$, the system does not have a single unique solution. If $$D = 0$$ and all of the variable determinants (such as $$D_x$$, $$D_y$$, and $$D_z$$) are also exactly zero, the system is consistent and dependent, meaning it has infinitely many solutions. However, if $$D = 0$$ and the variable determinants are not all equal to zero, the system is inconsistent and has no solution.

Claude

For a system of three linear equations represented as:
$$\left\{\begin{array}{l} a_1 x + b_1 y + c_1 z = k_1 \\\\ a_2 x + b_2 y + c_2 z = k_2 \\\\ a_3 x + b_3 y + c_3 z = k_3 \end{array}\right.$$

The solution $$(x, y, z)$$ can be determined using determinants:
$$x = \frac{D_x}{D}$$, $$y = \frac{D_y}{D}$$, and $$z = \frac{D_z}{D}$$

Where $$D$$ is the determinant of the coefficients of the variables:
$$D = \begin{vmatrix} a_1 & b_1 & c_1 \\\\ a_2 & b_2 & c_2 \\\\ a_3 & b_3 & c_3 \end{vmatrix}$$

$$D_x$$ is formed by replacing the $$x$$-coefficients with the constants $$k_1$$, $$k_2$$, and $$k_3$$:
$$D_x = \begin{vmatrix} k_1 & b_1 & c_1 \\\\ k_2 & b_2 & c_2 \\\\ k_3 & b_3 & c_3 \end{vmatrix}$$

$$D_y$$ is formed by replacing the $$y$$-coefficients with the constants:
$$D_y = \begin{vmatrix} a_1 & k_1 & c_1 \\\\ a_2 & k_2 & c_2 \\\\ a_3 & k_3 & c_3 \end{vmatrix}$$

$$D_z$$ is formed by replacing the $$z$$-coefficients with the constants:
$$D_z = \begin{vmatrix} a_1 & b_1 & k_1 \\\\ a_2 & b_2 & k_2 \\\\ a_3 & b_3 & k_3 \end{vmatrix}$$

Cramer's Rule for a System of Three Linear Equations

https://openstax.org/details/books/intermediate-algebra-2e

OpenStax Intermediate Algebra

Solve the system of equations using Cramer's rule: $$\left\{\begin{array}{l} 3x - 5y + 4z = 5 \\ 5x + 2y + z = 0 \\ 2x + 3y - 2z = 3 \end{array}ight.$$. First, evaluate the main determinant $$D$$ using the coefficients of the variables. By expanding by minors, we find $$D = -37$$. Next, evaluate $$D_x$$ by using the constants to replace the coefficients of $$x$$; expanding by minors yields $$D_x = -74$$. Then, evaluate $$D_y$$ by replacing the coefficients of $$y$$ with the constants, yielding $$D_y = 111$$. Finally, evaluate $$D_z$$ by replacing the coefficients of $$z$$ with the constants, which yields $$D_z = 148$$. Find the values of $$x$$, $$y$$, and $$z$$ by substituting into the formulas: $$x = \frac{D_x}{D} = \frac{-74}{-37} = 2$$, $$y = \frac{D_y}{D} = \frac{111}{-37} = -3$$, and $$z = \frac{D_z}{D} = \frac{148}{-37} = -4$$. Write the solution as the ordered triple $$(2, -3, -4)$$ and check that it is a solution to all three original equations.

Example: Solving a System of Three Equations Using Cramer's Rule

Apply Cramer's rule to solve a system of three linear equations, such as $$\left\{\begin{array}{l} 3x + 8y + 2z = -5 \\ 2x + 5y - 3z = 0 \\ x + 2y - 2z = -1 \end{array}ight.$$. First, evaluate the main determinant $$D$$ using the coefficients of the variables. Next, evaluate the determinants $$D_x$$, $$D_y$$, and $$D_z$$ by substituting the system's constants in place of the respective variable coefficients in each case. Then, calculate the values of the variables using the formulas $$x = \frac{D_x}{D}$$, $$y = \frac{D_y}{D}$$, and $$z = \frac{D_z}{D}$$. Finally, state the solution as an ordered triple and verify it by substituting the values back into all three original equations.

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