Problem 9 - Precalculus

Question

Use one matrix inverse to solve the following systems of linear equations. $$ \left\\{\begin{aligned} 3 x+7 y &=26 \\ 5 x+12 y &=39 \end{aligned}\right. $$

Step-by-Step Solution

Answer

The solution is \( x = 39 \), \( y = -13 \).

Step 1: Write the System in Matrix Form

First, express the system of linear equations as a matrix equation. Define the coefficient matrix \( A \), the variable matrix \( X \), and the constant matrix \( B \) like this:\[ A = \begin{bmatrix} 3 & 7 \ 5 & 12 \end{bmatrix}, \quad X = \begin{bmatrix} x \ y \end{bmatrix}, \quad B = \begin{bmatrix} 26 \ 39 \end{bmatrix} \] The matrix equation is then \( AX = B \).

Step 2: Find the Inverse of Matrix A

Calculate the inverse of the matrix \( A \). For a 2x2 matrix \( \begin{bmatrix} a & b \ c & d \end{bmatrix} \), the inverse is given by \[ A^{-1} = \frac{1}{ad - bc} \begin{bmatrix} d & -b \ -c & a \end{bmatrix} \]Here, \( a = 3 \), \( b = 7 \), \( c = 5 \), \( d = 12 \). Therefore:\[ A^{-1} = \frac{1}{(3)(12) - (7)(5)} \begin{bmatrix} 12 & -7 \ -5 & 3 \end{bmatrix} = \frac{1}{36 - 35} \begin{bmatrix} 12 & -7 \ -5 & 3 \end{bmatrix} = \begin{bmatrix} 12 & -7 \ -5 & 3 \end{bmatrix} \]

Step 3: Multiply Inverse of A with B

To solve for \( X \), multiply the inverse of matrix \( A \) by matrix \( B \): \[ X = A^{-1}B = \begin{bmatrix} 12 & -7 \ -5 & 3 \end{bmatrix} \begin{bmatrix} 26 \ 39 \end{bmatrix} \]Perform the matrix multiplication:- First row: \( 12 \times 26 + (-7) \times 39 = 312 - 273 = 39 \)- Second row: \( -5 \times 26 + 3 \times 39 = -130 + 117 = -13 \)This gives:\[ X = \begin{bmatrix} 39 \ -13 \end{bmatrix} \]

Step 4: Conclude Values of x and y

From the matrix \( X = \begin{bmatrix} x \ y \end{bmatrix} \), we identify that \( x = 39 \) and \( y = -13 \). These are the solutions to the system of equations.

Key Concepts

Understanding Systems of Linear Equations

Systems of linear equations consist of two or more linear equations involving the same set of variables. The goal is typically to find a solution for the variables that satisfy all equations simultaneously. For instance, let's consider two equations:

• Equation 1: \( 3x + 7y = 26 \)
• Equation 2: \( 5x + 12y = 39 \)

We have two variables, \(x\) and \(y\), and each equation describes a line in a 2-dimensional space. The solution to the system is the point where these lines intersect.To solve these types of systems, you can use methods like substitution, elimination, or matrix operations. Using matrices is particularly efficient for systems with multiple equations. In this approach, we express the system in a compact matrix form which simplifies calculations and provides a clear overview of how the equations relate to one another.

Exploring Matrix Multiplication

Matrix multiplication is a fundamental operation used in solving systems of linear equations, among other applications. When multiplying matrices, each element in the resulting matrix is calculated as the sum of products of elements from a row of the first matrix and a column of the second matrix.Here's how matrix multiplication works step-by-step:

• Consider two matrices: \( A = \begin{bmatrix} a & b \ c & d \end{bmatrix} \) and \( B = \begin{bmatrix} e \ f \end{bmatrix} \).
• To multiply \( A \) by \( B \), calculate the elements of the resulting matrix \( C \). The first element is \( ae + bf \), and the second is \( ce + df \). Thus, \( C = \begin{bmatrix} ae + bf \ ce + df \end{bmatrix} \).

This process is used not only for solving systems of linear equations but also in various applications like computer graphics and economic forecasting, making its understanding indispensable.

Demystifying Inverse Matrix Calculation

An inverse matrix is akin to the reciprocal of a number. When you multiply a matrix by its inverse, you get the identity matrix, which acts like the number 1 in matrix arithmetic. Not all matrices have inverses, only those that are non-singular, meaning they have a non-zero determinant.For a 2x2 matrix \( A = \begin{bmatrix} 3 & 7 \ 5 & 12 \end{bmatrix} \), the process to find the inverse \( A^{-1} \) involves:

• Calculating the determinant \( \text{det}(A) = ad - bc \), where \( a = 3 \), \( b = 7 \), \( c = 5 \), and \( d = 12 \) resulting in \( 36 - 35 = 1 \).
• Using the formula for the inverse: \[ A^{-1} = \frac{1}{\text{det}(A)} \begin{bmatrix} d & -b \ -c & a \end{bmatrix} = \begin{bmatrix} 12 & -7 \ -5 & 3 \end{bmatrix} \].

The inverse matrix allows us to solve the matrix equation \( AX = B \) by transforming it into \( X = A^{-1}B \), which directly finds the variables, providing an efficient solution to the system of equations.