Problem 70 - Intermediate Algebra

Question

Use Cramer's rule to solve each system of equations. If a system is inconsistent or if the equations are dependent, so indicate. $$ \left\\{\begin{array}{l} x-3 y+4 z-2=0 \\ 2 x+y+2 z-3=0 \\ 4 x-5 y+10 z-7=0 \end{array}\right. $$

Step-by-Step Solution

Answer

The solution is \(x = \frac{11}{7}\), \(y = \frac{5}{14}\), \(z = \frac{1}{2}\). The system is consistent and independent.

Step 1: Set Up the System

Write the given system of equations in the standard matrix form, separating the coefficients of the variables:\[A = \begin{bmatrix} 1 & -3 & 4 \ 2 & 1 & 2 \ 4 & -5 & 10 \end{bmatrix}, \quad B = \begin{bmatrix} 2 \ 3 \ 7 \end{bmatrix}\]where the system of equations is \(AX = B\).

Step 2: Calculate the Determinant of A

Calculate the determinant of matrix \(A\): \[|A| = \begin{vmatrix} 1 & -3 & 4 \ 2 & 1 & 2 \ 4 & -5 & 10 \end{vmatrix}\]Use the formula for a 3x3 determinant:\[|A| = 1(1 \times 10 - 2 \times (-5)) - (-3)(2 \times 10 - 4 \times 2) + 4(2 \times (-5) - 1 \times 4)\]Calculate step by step:\[|A| = 1(10 + 10) + 3(20 - 8) - 4( -10 - 4)\]\[|A| = 20 + 3 \times 12 - 4 \times (-14)\]\[|A| = 20 + 36 + 56 = 112\]

Step 3: Calculate Determinants for Each Variable

Calculate the determinants of matrices obtained by replacing one column in \(A\) with \(B\), denoted as \(A_x\), \(A_y\) and \(A_z\).- For \(A_x\):\[A_x = \begin{vmatrix} 2 & -3 & 4 \ 3 & 1 & 2 \ 7 & -5 & 10 \end{vmatrix}\]Calculating:\[|A_x| = 2(1 \times 10 - 2 \times (-5)) - (-3)(3 \times 10 - 2 \times 7) + 4(3 \times (-5) - 1 \times 7)\]\[|A_x| = 2(10 + 10) + 3(30 - 14) - 4( -15 - 7)\]\[|A_x| = 2 \times 20 + 3 \times 16 - 4 \times (-22)\]\[|A_x| = 40 + 48 + 88 = 176\]- For \(A_y\):\[A_y = \begin{vmatrix} 1 & 2 & 4 \ 2 & 3 & 2 \ 4 & 7 & 10 \end{vmatrix}\]Calculating:\[|A_y| = 1(3 \times 10 - 2 \times 7) - 2(2 \times 10 - 4 \times 7) + 4(2 \times 7 - 3 \times 4)\]\[|A_y| = 1 \times (30 - 14) - 2 \times (20 - 28) + 4 \times (14 - 12)\]\[|A_y| = 16 + 16 + 8 = 40\]- For \(A_z\):\[A_z = \begin{vmatrix} 1 & -3 & 2 \ 2 & 1 & 3 \ 4 & -5 & 7 \end{vmatrix}\]Calculating:\[|A_z| = 1(1 \times 7 - 3 \times (-5)) - (-3)(2 \times 7 - 3 \times 4) + 2(2 \times (-5) - 1 \times 4)\]\[|A_z| = 1 \times (7 + 15) + 3 \times (14 - 12) - 2 \times (-10 - 4)\]\[|A_z| = 22 + 6 + 28 = 56\]

Step 4: Solve for the Variables Using Cramer's Rule

Apply Cramer's rule to find the variables \(x\), \(y\), and \(z\):- \(x = \frac{|A_x|}{|A|} = \frac{176}{112} = \frac{11}{7}\)- \(y = \frac{|A_y|}{|A|} = \frac{40}{112} = \frac{5}{14}\)- \(z = \frac{|A_z|}{|A|} = \frac{56}{112} = \frac{1}{2}\)

Step 5: Conclusion

The solution to the system using Cramer's rule is \(x = \frac{11}{7}\), \(y = \frac{5}{14}\), and \(z = \frac{1}{2}\). Since the determinant \(|A| eq 0\), the system is consistent and independent.

Key Concepts

System of Equations

A system of equations is a collection of two or more equations with a common set of unknowns. In this exercise, the system involves three equations with three variables:

• Equation 1: \( x - 3y + 4z - 2 = 0 \)
• Equation 2: \( 2x + y + 2z - 3 = 0 \)
• Equation 3: \( 4x - 5y + 10z - 7 = 0 \)

The objective is to find the values of the unknowns \( x \), \( y \), and \( z \) that satisfy all the equations simultaneously. By doing so, we can determine if the system is consistent, meaning there is at least one solution, or inconsistent if there are no solutions.
When equations are dependent, they represent the same geometrical plane. If equations are independent, their solutions form points of intersection, which will be calculated using algebraic methods such as Cramer's rule.

Determinant Calculation

The determinant is a special number that can be calculated from a square matrix. It provides important properties about the matrix, such as whether solutions are possible. In Cramer's rule, determinant calculations help us verify the existence of a unique solution for a system of equations.
To find the determinant of a 3x3 matrix like matrix \( A \), you use this formula:

• First term: \( 1(1 \times 10 - 2 \times (-5)) \)
• Second term: \(- (-3)(2 \times 10 - 4 \times 2) \)
• Third term: \(4(2 \times (-5) - 1 \times 4) \)

Plugging in these values gives us the determinant \(|A| = 112\).
If the determinant is not zero, \(|A| eq 0\), it confirms that the system has a unique solution, and we can continue using Cramer's rule to find these solutions.

Matrix Algebra

Matrix algebra is used to organize and solve systems of linear equations efficiently. In this context, matrices allow us to represent and manipulate the coefficients of the system of equations in a structured form.
The matrix equation \( AX = B \) summarizes the system, where \( A \) is the matrix of coefficients, \( X \) is the column matrix of variables, and \( B \) is the constant matrix.
Once \( A \)'s determinant is known, you compute determinant replacements for each variable by substituting the constants from \( B \) into each column one at a time, deriving \( A_x \), \( A_y \), and \( A_z \).
Finally, Cramer's rule applies:

• \( x = \frac{|A_x|}{|A|} \)
• \( y = \frac{|A_y|}{|A|} \)
• \( z = \frac{|A_z|}{|A|} \)

In this exercise, these calculations yield the unique solutions \(( x = \frac{11}{7}, y = \frac{5}{14}, z = \frac{1}{2} )\), underpinning the powerful use of matrices in algebraic problem-solving.