Problem 75 - Algebra 2

Question

Solve each system of equations by using inverse matrices. \(6 x+5 y=8\) \(3 x-y=7\)

Step-by-Step Solution

Answer

Solving using inverse matrices yields the solution (x, y) = (1, 1).

Step 1: Write the System in Matrix Form

To solve the system using inverse matrices, we first rewrite the system of equations in matrix form: \ \[ \begin{bmatrix} 6 & 5 \ 3 & -1 \end{bmatrix} \begin{bmatrix} x \ y \end{bmatrix} = \begin{bmatrix} 8 \ 7 \end{bmatrix} \] \ This can be expressed as \( AX = B \), where \( A \) is the coefficient matrix, \( X \) is the variable matrix, and \( B \) is the constant matrix.

Step 2: Calculate the Inverse of Matrix A

Find the inverse of matrix \( A \), which is \[ \begin{bmatrix} 6 & 5 \ 3 & -1 \end{bmatrix} \]. \ The inverse \( A^{-1} \) is calculated using the formula: \ \[ A^{-1} = \frac{1}{ad-bc} \begin{bmatrix} d & -b \ -c & a \end{bmatrix} \], \ where \( a = 6 \), \( b = 5 \), \( c = 3 \), and \( d = -1 \). \ First, calculate the determinant: \ \[ ad - bc = (6)(-1) - (5)(3) = -6 - 15 = -21 \] \ Then calculate the inverse: \ \[ A^{-1} = \frac{1}{-21} \begin{bmatrix} -1 & -5 \ -3 & 6 \end{bmatrix} = \begin{bmatrix} \frac{1}{21} & \frac{5}{21} \ \frac{3}{21} & -\frac{6}{21} \end{bmatrix} \].

Key Concepts

System of Equations

A system of equations is essentially a set of two or more equations with the same set of unknowns. In our exercise, the system is composed of two linear equations: \(6x + 5y = 8\) and \(3x - y = 7\). The primary goal when solving a system is finding values for the variables that satisfy all the equations simultaneously. This system involves two variables, \(x\) and \(y\), and we're looking for a solution pair \((x,y)\) that makes both equations true. For this, we can use different methods, such as graphing, substitution, elimination, or using matrix techniques like matrix inversion, which is the method we focus on here. The advantage of using matrices in equations is that it can simplify solving complex systems, especially when involving many variables and equations.

Matrix Inversion

Matrix inversion is a technique used to solve systems of equations, among other things. When you have a system of equations represented in matrix form as \(AX = B\), the inverse of the coefficient matrix \(A\), denoted as \(A^{-1}\), is a key player in finding the solution. Here's the basic idea:

• If \(A\) is a coefficient matrix and \(X\) is the variable matrix, solve for \(X\) by using the inverse: \(X = A^{-1}B\).
• To compute \(A^{-1}\), you need \(A\)'s determinant to be non-zero. If \(det(A) = 0\), the matrix cannot have an inverse, and this method won't work.

Matrix inversion is all about transforming a square matrix into another one such that when it's multiplied by the original matrix, it results in the identity matrix \(I\). This concept is vital for solving systems of equations using the matrix approach as it makes isolating the variable matrix \(X\) feasible.

Determinant of a Matrix

The determinant of a matrix is a special number that can tell us a lot about the matrix itself. For a 2x2 matrix, given by: \[\begin{bmatrix} a & b \ c & d \end{bmatrix}\]The determinant is calculated as \(ad - bc\). In the given exercise, our matrix is: \[\begin{bmatrix} 6 & 5 \ 3 & -1 \end{bmatrix}\]Here, the determinant is \((6)(-1) - (5)(3) = -6 - 15 = -21\). A non-zero determinant, such as -21 here, indicates the matrix is invertible, which means we can find its inverse and proceed to solve the system of equations.

• If the determinant is zero, the matrix is singular, meaning it lacks an inverse, and this approach cannot be applied.
• A determinant helps in understanding the properties of a matrix, like whether a set of vectors (rows or columns of the matrix) is linearly dependent or not.

Understanding the determinant is pivotal because it not only determines the possibility of matrix inversion but also has applications in calculus, geometry, and more.