Problem 55 - Algebra 2

Question

Solve the system \(4 x-y=0,2 x+3 y=14\) by using inverse matrices.

Step-by-Step Solution

Answer

The solution is \(x = 1\) and \(y = 4\).

Step 1: Write the System in Matrix Form

We start by writing the system of equations as a matrix equation. The system \(4x - y = 0\) and \(2x + 3y = 14\) can be expressed as: \[A \cdot X = B\] where \(A = \begin{bmatrix} 4 & -1 \ 2 & 3 \end{bmatrix}\), \(X = \begin{bmatrix} x \ y \end{bmatrix}\), and \(B = \begin{bmatrix} 0 \ 14 \end{bmatrix}\).

Step 2: Find the Inverse of Matrix A

Next, we calculate the inverse of matrix \(A\), denoted \(A^{-1}\). The formula for the inverse of a 2x2 matrix \(\begin{bmatrix} a & b \ c & d \end{bmatrix}\) is \(\frac{1}{ad-bc} \begin{bmatrix} d & -b \ -c & a \end{bmatrix}\). For \(A = \begin{bmatrix} 4 & -1 \ 2 & 3 \end{bmatrix}\), the determinant \(ad - bc = (4)(3) - (-1)(2) = 14\). Thus, \[A^{-1} = \frac{1}{14} \begin{bmatrix} 3 & 1 \ -2 & 4 \end{bmatrix}\].

Step 3: Multiply to Solve for X

To find \(X\), we multiply both sides of the matrix equation \(A \cdot X = B\) by \(A^{-1}\). Thus, \(X = A^{-1} \cdot B\). Calculate this as follows: \[X = \frac{1}{14} \begin{bmatrix} 3 & 1 \ -2 & 4 \end{bmatrix} \cdot \begin{bmatrix} 0 \ 14 \end{bmatrix} = \frac{1}{14} \begin{bmatrix} 3(0) + 1(14) \ -2(0) + 4(14) \end{bmatrix} = \frac{1}{14} \begin{bmatrix} 14 \ 56 \end{bmatrix} = \begin{bmatrix} 1 \ 4 \end{bmatrix}.\]

Step 4: Interpret the Solution

Finally, from the matrix \(X = \begin{bmatrix} 1 \ 4 \end{bmatrix}\), we interpret the solution to the original system of equations as \(x = 1\) and \(y = 4\).

Key Concepts

Matrix Equation

A matrix equation is essentially a way to represent a system of linear equations using matrices so that it looks neat and compact. It is written in the form \(AX = B\), where \(A\) is the matrix of coefficients, \(X\) is the matrix representing the variables, and \(B\) is the constant matrix.

In the given exercise, the system consisting of two equations: \(4x - y = 0\) and \(2x + 3y = 14\) can be expressed in matrix form as \(\begin{bmatrix} 4 & -1 \ 2 & 3 \end{bmatrix} \cdot \begin{bmatrix} x \ y \end{bmatrix} = \begin{bmatrix} 0 \ 14 \end{bmatrix}\).

This makes it easier to apply mathematical operations like inverses, which can simplify finding the solution.

System of Equations

A system of equations is a collection of two or more equations with the same set of unknowns. The goal is to find values for these unknowns that satisfy all equations in the system simultaneously.

• Each equation in the system must be linear, meaning it forms a straight line when plotted on a graph.
• The solution is typically found by finding the point at which all lines intersect.

In this exercise, our system includes the equations \(4x - y = 0\) and \(2x + 3y = 14\). By representing these equations in matrix form, we transform the problem into a matrix equation which can be easier to solve systematically.

Determinant

The determinant is a special number that can be calculated from a square matrix. It provides key insights into the matrix's properties, such as whether it has an inverse.

• If the determinant is zero, the matrix does not have an inverse, implying that the system of equations might not have a unique solution.
• A non-zero determinant means the matrix is invertible and hence the system has a unique solution.

In the current context, the matrix \(A = \begin{bmatrix} 4 & -1 \ 2 & 3 \end{bmatrix}\) has a determinant calculated as \(ad - bc = 14\). This non-zero value indicates that the inverse exists, allowing further solution steps.

Solving Systems of Linear Equations

Solving systems of linear equations can be done through various methods, including substitution, elimination, and matrix operations like finding the inverse.

• Using inverse matrices is a powerful technique for solving systems, especially when the number of equations matches the number of variables.
• This approach involves first ensuring that the matrix of coefficients is invertible, then finding the inverse, and finally multiplying it by the constant matrix \(B\).

By solving \(AX = B\) for \(X\), we use the inverse to find the solution: \(X = A^{-1}B\). In our exercise, multiplications with the inverse matrix give us the final solution: \(x = 1\) and \(y = 4\). This method is straightforward once you get a good understanding of how matrix multiplication and inverses work.