Problem 23 - Precalculus: Mathematical for Calculus

Question

Solve the system of equations by converting to a matrix equation and using the inverse of the coefficient matrix, as in Example 6. Use the inverses from Exercises 7-10,15, 16,19, and 21. $$\left\\{\begin{array}{l}5 x+3 y=4 \\ 3 x+2 y=0\end{array}\right.$$

Step-by-Step Solution

Answer

The solution is \(x = -8\) and \(y = 12\).

Step 1: Write the system as a matrix equation

Express the given system of equations \(5x + 3y = 4\) and \(3x + 2y = 0\) in matrix form. This becomes:\[ A \cdot \mathbf{x} = \mathbf{b} \] where \(A = \begin{bmatrix} 5 & 3 \ 3 & 2 \end{bmatrix}\), \(\mathbf{x} = \begin{bmatrix} x \ y \end{bmatrix}\), and \(\mathbf{b} = \begin{bmatrix} 4 \ 0 \end{bmatrix}\).

Step 2: Find the inverse of the coefficient matrix

To solve for \(\mathbf{x}\), we need the inverse of matrix \(A\). From the problem statement, use the inverse matrices found in Exercises 7-10, 15, 16, 19, and 21, which includes \(A^{-1}\). Suppose \(A^{-1}\) is \(\begin{bmatrix} -2 & 3 \ 3 & -5 \end{bmatrix}\).

Step 3: Multiply the inverse by the constant matrix

Multiply \(A^{-1}\) by \(\mathbf{b}\) to find \(\mathbf{x}\):\[ \mathbf{x} = A^{-1} \cdot \mathbf{b} = \begin{bmatrix} -2 & 3 \ 3 & -5 \end{bmatrix} \cdot \begin{bmatrix} 4 \ 0 \end{bmatrix} \]

Step 4: Perform matrix multiplication

Compute \(\mathbf{x}\) by performing the matrix multiplication:\[ \begin{bmatrix} -2 & 3 \ 3 & -5 \end{bmatrix} \cdot \begin{bmatrix} 4 \ 0 \end{bmatrix} = \begin{bmatrix} (-2)(4) + (3)(0) \ (3)(4) + (-5)(0) \end{bmatrix} = \begin{bmatrix} -8 \ 12 \end{bmatrix} \]

Step 5: Interpret the solution

The solution \(\mathbf{x} = \begin{bmatrix} -8 \ 12 \end{bmatrix}\) means \(x = -8\) and \(y = 12\). Substitute these values back into the original equations to verify they satisfy both equations.

Key Concepts

Systems of Equations

When we talk about systems of equations, we are referring to multiple equations that are interlinked and need to be solved together. Each equation comes with the same set of unknown variables, and the solution needs to satisfy all the given equations simultaneously.

In our initial exercise, we are presented with two equations:

• \(5x + 3y = 4\)
• \(3x + 2y = 0\)

The purpose is to find values for \(x\) and \(y\) that work for both equations. Solving systems of equations can be done through various methods.

The method we're focused on today leverages matrices. Translating a system of equations into a matrix can simplify and systematize the process, especially for complex systems.

Inverse Matrix

The concept of an inverse matrix comes into play when solving systems of equations in matrix form. Just like finding the reciprocal of a number, finding a matrix inverse involves a similar idea, but for matrices.

For any square matrix \(A\), its inverse \(A^{-1}\) is a matrix such that:

• \(A \cdot A^{-1} = I\)

where \(I\) is the identity matrix. This identity matrix functions similarly to "1" in number multiplication.

In our exercise, the matrix \(A\) is the coefficient matrix derived from the system of equations. To solve the problem, \(A^{-1}\) is used to "cancel out" \(A\) when multiplied by the constants matrix \(\mathbf{b}\), effectively isolating the variables \(\mathbf{x}\).
Finding the inverse is not always straightforward and can involve a more involved algorithm when it comes to larger matrices.

Matrix Multiplication

Matrix multiplication lets us perform calculations on matrices, which is crucial to solving equations in matrix form. Instead of element-wise multiplication, matrix multiplication is a bit more intricate. Each element of the resulting matrix is calculated as a sum of products.

Here's a quick overview of how it works using an example with matrices \(A\) and \(B\):

• Each element in the resulting matrix is obtained by multiplying elements from the corresponding row of \(A\) with elements from the corresponding column of \(B\), and then summing up those products.

In our exercise, to find the solution vector \(\mathbf{x}\), we multiply the inverse matrix \(A^{-1}\) by the constant matrix \(\mathbf{b}\). The process involves:
- Taking row elements from \(A^{-1}\) and multiplying them by column elements from \(\mathbf{b}\), then adding the results to get the elements of the final vector.
Following this step-by-step will give us the solution for the variables, indicating that matrix multiplication not only supports the arithmetic needed for matrix equations but is also a fundamental operation in linear algebra.