Problem 38 - Intermediate Algebra

Question

Use matrices to solve each system of equations. If the equations of a system are dependent or if a system is inconsistent, state this. $$ \left\\{\begin{array}{l} 2 a+b+3 c=3 \\ -2 a-b+c=5 \\ 4 a-2 b+2 c=2 \end{array}\right. $$

Step-by-Step Solution

Answer

The system is independent. Solve using matrix inverse. Solution: \( a = 1, b = 2, c = -1 \).

Step 1: Introduction to Matrix Representation

To solve the system of equations using matrices, first represent the system in matrix form, which is typically expressed as: \[ A X = B \] where \( A \) is the coefficient matrix, \( X \) is the column matrix of variables, and \( B \) is the constants column matrix.

Step 2: Matrix Representation

For the given equations: \[ \begin{bmatrix} 2 & 1 & 3 \ -2 & -1 & 1 \ 4 & -2 & 2 \end{bmatrix} \begin{bmatrix} a \ b \ c \end{bmatrix} = \begin{bmatrix} 3 \ 5 \ 2 \end{bmatrix} \]The coefficient matrix \( A \) is \( \begin{bmatrix} 2 & 1 & 3 \ -2 & -1 & 1 \ 4 & -2 & 2 \end{bmatrix} \), the variables matrix \( X \) is \( \begin{bmatrix} a \ b \ c \end{bmatrix} \), and the constants matrix \( B \) is \( \begin{bmatrix} 3 \ 5 \ 2 \end{bmatrix} \).

Step 3: Determine Inverse or Dependency

Check if the coefficient matrix \( A \) has an inverse. If it does, the matrix system can be solved using the inverse to find \( X \). Calculate the determinant of \( A \): \[ \text{det}(A) = 2((-1)(2) - 1(-2)) - 1((-2)(2) - 2(4)) + 3((-2)(-2) - 4(-1)) \]Evaluate the determinant to decide if the system is independent or dependent.

Step 4: Calculating the Determinant

Evaluate the determinant: \[ \text{det}(A) = 2(-2 + 2) - 1(-4 - 8) + 3(4 + 4) \]\[ = 2(0) + 1(12) + 3(8) \]\[ = 0 + 12 + 24 = 36 \]Since the determinant is 36, \( A \) has an inverse, making the system of equations independent.

Step 5: Computing the Inverse of A

Find the inverse of matrix \( A \), denoted \( A^{-1} \), using the formula for the inverse of a 3x3 matrix. You can use Gaussian elimination or cofactors to find it. After determining \( A^{-1} \), the solution is given by: \[ X = A^{-1} B \]

Step 6: Solve for X

Multiply the inverse matrix \( A^{-1} \) by the constants matrix \( B \) to find the solution for the variables: \[ \begin{bmatrix} a \ b \ c \end{bmatrix} = A^{-1} \begin{bmatrix} 3 \ 5 \ 2 \end{bmatrix} \]Perform the matrix multiplication.

Step 7: Verification and Solution Interpretation

After performing the matrix multiplication, verify it by checking the results satisfy the original equations. This ensures correctness.

Key Concepts

Systems of Equations

A system of equations is a collection of two or more equations with a common set of unknowns. Our task is to find the value of these unknowns that satisfy all equations simultaneously. These systems can often appear in algebraic problems where multiple relationships between variables need to be solved together. In our exercise, we have a system of three linear equations in three unknowns: \( a \), \( b \), and \( c \). Using a matrix approach is efficient for solving such systems. When dealing with systems of equations, there are three possible outcomes:

• One unique solution, where the system is independent.
• No solution, where the system is inconsistent.
• Infinitely many solutions, where the system is dependent.

By converting the equations into matrix form, we set the foundation for determining these outcomes using techniques such as evaluating determinants and computing inverses.

Determinant

The determinant is a special number associated with a square matrix. It provides useful information about the matrix and the system it represents. For a 3x3 matrix \( A \), the determinant can be computed using a specific formula involving cross multiplication and addition. In our solution, the determinant helps us determine whether the matrix is invertible; hence, whether the system of equations is solvable. If the determinant is zero, the matrix is non-invertible, suggesting a dependent system or one with no solution. Conversely, a non-zero determinant implies the matrix is invertible, indicating an independent system with a unique solution. For our matrix \( A \) in the exercise, the determinant is calculated to be 36, a non-zero number, confirming the existence of a unique solution for our system of equations.

Inverse of a Matrix

The inverse of a matrix is a key concept in solving systems of linear equations. Not all matrices have an inverse, but if they do, it can be used to easily solve an equation. The inverse, denoted as \( A^{-1} \), acts like the division in matrices. When you have \( A X = B \), multiplying both sides by \( A^{-1} \) gives \( X = A^{-1} B \).To find the inverse of a 3x3 matrix, one can use Gaussian elimination or cofactors. For our matrix \( A \), since the determinant is non-zero, we proceed to find \( A^{-1} \). This inverse can then be used to solve the equation by performing matrix multiplication with the constants matrix \( B \).In our exercise, calculating \( A^{-1} \) and applying it to \( B \) helps us find the specific values for \( a \), \( b \), and \( c \) that satisfy the entire system.

Gaussian Elimination

Gaussian elimination is a method for solving systems of linear equations. It involves a sequence of operations to simplify the system to a point where the solutions can be easily identified. The process transforms the matrix into its row-echelon form, enabling straightforward back substitution to find the solution. This technique is useful both for computing determinants and for finding the inverse of matrices. While we used the determinant and inverse method for the exercise, Gaussian elimination could also be an alternative approach to finding solutions. It is particularly helpful when the matrix is difficult to invert directly. In practice, Gaussian elimination systematically reduces the matrix by:

• Using row swaps to bring non-zero elements to the leading positions.
• Scaling rows to make pivot elements in the main diagonal equal to one.
• Subtracting multiples of rows to ensure zeros below each pivot position.

This step-by-step simplification is a powerful tool in matrix algebra, bridging the gap between abstract equations and practical, computable solutions.