Problem 37 - A Graphical Approach to College Algebra

Question

Solve each system by using the matrix inverse method. $$\begin{aligned} &2 x+3 y=-10\\\ &3 x+4 y=-12 \end{aligned}$$

Step-by-Step Solution

Answer

The solution is \( x = -4 \) and \( y = 6 \).

Step 1: Write the system of equations in matrix form

The given system of equations can be expressed as a matrix equation of the form \( A\mathbf{x} = \mathbf{b} \). Here, \( A \) is the coefficient matrix, \( \mathbf{x} \) is the variable matrix, and \( \mathbf{b} \) is the constant matrix. For the given system: \[A = \begin{pmatrix} 2 & 3 \ 3 & 4 \end{pmatrix}, \, \mathbf{x} = \begin{pmatrix} x \ y \end{pmatrix}, \, \mathbf{b} = \begin{pmatrix} -10 \ -12 \end{pmatrix}\]

Step 2: Calculate the inverse of matrix A

The inverse of a 2x2 matrix \( A = \begin{pmatrix} a & b \ c & d \end{pmatrix} \) is given by \[A^{-1} = \frac{1}{ad - bc} \begin{pmatrix} d & -b \ -c & a \end{pmatrix}\]Calculating the determinant of \( A \), we have \[det(A) = 2 \times 4 - 3 \times 3 = 8 - 9 = -1\]The inverse matrix \( A^{-1} \) is then:\[A^{-1} = \begin{pmatrix} 4 & -3 \ -3 & 2 \end{pmatrix}\]since \( \frac{1}{-1} \) simply negates the terms inside the matrix.

Step 3: Use the inverse to solve for \(\mathbf{x}\)

To find the solution \( \mathbf{x} \), we compute \( \mathbf{x} = A^{-1}\mathbf{b} \). We multiply the inverse matrix \( A^{-1} \) by \( \mathbf{b} \):\[\mathbf{x} = \begin{pmatrix} 4 & -3 \ -3 & 2 \end{pmatrix} \begin{pmatrix} -10 \ -12 \end{pmatrix}\]This results in:\[\begin{pmatrix} 4(-10) + (-3)(-12) \ -3(-10) + 2(-12) \end{pmatrix} = \begin{pmatrix} -40 + 36 \ 30 - 24 \end{pmatrix} = \begin{pmatrix} -4 \ 6 \end{pmatrix}\]So \( x = -4 \) and \( y = 6 \).

Key Concepts

Systems of Equations

A system of equations consists of multiple equations that share the same set of variables. When solving these systems, our goal is to find the values of the variables that satisfy all equations simultaneously.
In this exercise, we have two linear equations involving variables \(x\) and \(y\):

• \(2x + 3y = -10\)
• \(3x + 4y = -12\)

These equations represent straight lines when graphed, and the solution to the system is the point where these two lines intersect. The matrix inverse method is a powerful tool for finding this intersection point, especially when dealing with more complex systems.

Matrix Algebra

Matrix algebra is a branch of mathematics that utilizes arrays of numbers for various computations. Matrices help simplify the representation and solving of complex systems. In this particular case, our system of equations can be expressed in a matrix equation form as:

• Coefficient Matrix \(A = \begin{pmatrix} 2 & 3 \ 3 & 4 \end{pmatrix}\)
• Variable Matrix \(\mathbf{x} = \begin{pmatrix} x \ y \end{pmatrix}\)
• Constant Matrix \(\mathbf{b} = \begin{pmatrix} -10 \ -12 \end{pmatrix}\)

The matrix equation \(A\mathbf{x} = \mathbf{b}\) succinctly represents the system, and brings uniformity and simplification to solving the equations.

Determinant Calculation

The determinant is a special number calculated from a square matrix. It is essential when finding the inverse of a matrix. For a 2x2 matrix \(A = \begin{pmatrix} a & b \ c & d \end{pmatrix}\), the determinant \(det(A)\) is computed as \(ad - bc\).
For our matrix \(A = \begin{pmatrix} 2 & 3 \ 3 & 4 \end{pmatrix}\), the determinant is calculated as follows:

• \(det(A) = 2 \times 4 - 3 \times 3 = 8 - 9 = -1\)

A non-zero determinant indicates that the matrix is invertible, allowing us to use the inverse matrix to solve the system of equations.

Inverse Matrix

An inverse matrix, denoted as \(A^{-1}\), is a matrix that, when multiplied by the original matrix \(A\), yields the identity matrix. This concept is crucial for solving systems of equations using matrix methods.
The inverse of a 2x2 matrix \(A = \begin{pmatrix} a & b \ c & d \end{pmatrix}\) is given by:

• \(A^{-1} = \frac{1}{ad - bc} \begin{pmatrix} d & -b \ -c & a \end{pmatrix}\)

For our matrix \(A\), with \(det(A) = -1\), the inverse is:

• \(A^{-1} = \begin{pmatrix} -4 & 3 \ 3 & -2 \end{pmatrix}\)

By multiplying \(A^{-1}\) by the constant matrix \(\mathbf{b}\), we find the values for \(x\) and \(y\) that solve the system.