Problem 30 - Algebra for College Students

Question

For Problems 27-40, use the method of matrix inverses to solve each system. $$ \left(\begin{array}{rl} 6 x-y & =-14 \\ 3 x+2 y & =-17 \end{array}\right) $$

Step-by-Step Solution

Answer

The solution is \( x = -3 \) and \( y = -4 \).

Step 1: Write the System in Matrix Form

The given system of equations can be expressed in matrix form as \( AX = B \). Here, \( A \) is the coefficient matrix, \( X \) is the column matrix of variables, and \( B \) is the constant matrix. Thus, the system can be written as: \[A = \begin{pmatrix} 6 & -1 \ 3 & 2 \end{pmatrix}, \, X = \begin{pmatrix} x \ y \end{pmatrix}, \, B = \begin{pmatrix} -14 \ -17 \end{pmatrix}\]

Step 2: Find the Inverse of A

To solve for \( X \), we need to find the inverse of \( A \), denoted \( A^{-1} \). The formula for the inverse of a 2x2 matrix \( \begin{pmatrix} a & b \ c & d \end{pmatrix} \) is \( \frac{1}{ad-bc} \begin{pmatrix} d & -b \ -c & a \end{pmatrix} \). For matrix \( A \), the determinant \( (ad-bc) = (6 \times 2) - (3 \times -1) = 15 \) and its inverse is: \[A^{-1} = \frac{1}{15} \begin{pmatrix} 2 & 1 \ -3 & 6 \end{pmatrix}\]

Step 3: Multiply A-inverse by B to Solve for X

Now compute \( X = A^{-1}B \): \[X = \frac{1}{15} \begin{pmatrix} 2 & 1 \ -3 & 6 \end{pmatrix} \begin{pmatrix} -14 \ -17 \end{pmatrix}\] Perform the matrix multiplication: \[X = \frac{1}{15} \begin{pmatrix} (2 \times -14) + (1 \times -17) \ (-3 \times -14) + (6 \times -17) \end{pmatrix} = \frac{1}{15} \begin{pmatrix} -28 - 17 \ 42 - 102 \end{pmatrix} = \frac{1}{15} \begin{pmatrix} -45 \ -60 \end{pmatrix}\] Finally, divide each element by 15: \[X = \begin{pmatrix} -3 \ -4 \end{pmatrix}\]

Step 4: Verify the Solution

Substitute \( x = -3 \) and \( y = -4 \) back into the original equations to verify: 1. \( 6(-3) - (-4) = -18 + 4 = -14 \), which is correct. 2. \( 3(-3) + 2(-4) = -9 - 8 = -17 \), which is also correct. Both equations are satisfied, confirming the solution is correct.

Key Concepts

System of Equations

A system of equations consists of two or more equations containing common variables. In the context of a 2x2 system, this generally involves two linear equations with two unknowns. These systems can arise in numerous real-life contexts, such as predicting financial outcomes or solving engineering problems.

Here’s how a system of equations can be set up:

• Identify the variables common to both equations - usually represented by letters like \(x\) and \(y\).
• Construct equations that relate these variables - often forming a straight line when graphed.

Understanding how to solve systems of equations is foundational because it sets the stage for more complex mathematical concepts like matrices and vector spaces.

2x2 Matrix

A 2x2 matrix is a simple grid that consists of four numbers arranged into two rows and two columns. Each position in the matrix is significant and corresponds to a specific operation in mathematical expressions. For example, in a system of equations, a 2x2 matrix can represent the coefficients of the variables involved.

Consider the following matrix from the solution: \[A = \begin{pmatrix} 6 & -1 \ 3 & 2 \end{pmatrix}\]This matrix corresponds to the coefficients in the given system of equations.

The key properties of a 2x2 matrix include:

• Its determinant, which is calculated as \((ad-bc)\) for a matrix \( \begin{pmatrix} a & b \ c & d \end{pmatrix} \).

In this exercise, the determinant helps us find the inverse of the matrix, which is crucial for solving the system of equations using matrix inverses.

Matrix Multiplication

Matrix multiplication involves combining two matrices to produce a new matrix. This operation is essential when transitioning from the matrix equation \(AX = B\) to solve for \(X\), especially using matrix inverses. The process follows strict rules:

• To multiply a matrix \(A\) of size \(m \times n\) with another matrix \(B\) of size \(n \times p\), the number of columns in \(A\) must match the number of rows in \(B\).
• The resulting matrix will have dimensions \(m \times p\).
• Each element in the resulting matrix is calculated by taking the dot product of rows of the first matrix with columns of the second.

In the context of this exercise, we perform matrix multiplication between the inverse of a 2x2 matrix and another 2x1 matrix to find the values of \(x\) and \(y\). It's a powerful method not only because it provides a clear pathway to solutions, but also because it applies to broader mathematical problems beyond linear algebra.