Problem 58 - College Algebra with Modeling and Visualization

Question

Complete the following for the given system of linear equations. (a) Write the system in the form \(A X=B\). (b) Solve the linear system by computing \(X=A^{-1} B\) with a calculator. Approximate the solution to the $$ \begin{aligned} -231 x+178 y &=-439 \\ 525 x-329 y &=2282 \end{aligned} $$

Step-by-Step Solution

Answer

The solution is approximately \(x = 3.12\) and \(y = 1.48\).

Step 1: Set Up Matrices

For the given system of linear equations, identify the coefficient matrix \(A\), the variable matrix \(X\), and the constant matrix \(B\). In this case, we have:\[A = \begin{bmatrix} -231 & 178 \ 525 & -329 \end{bmatrix}, \quad X = \begin{bmatrix} x \ y \end{bmatrix}, \quad B = \begin{bmatrix} -439 \ 2282 \end{bmatrix}\]

Step 2: Verify the Matrix Equation

Write the given system of equations in the standard form \(AX = B\):\[\begin{bmatrix} -231 & 178 \ 525 & -329 \end{bmatrix} \begin{bmatrix} x \ y \end{bmatrix} = \begin{bmatrix} -439 \ 2282 \end{bmatrix}\]This confirms that the system is set correctly in the form \(AX = B\).

Step 3: Calculate the Inverse of A

Use a calculator to find the inverse of matrix \(A\), denoted as \(A^{-1}\). Ensure your calculator is set to matrix mode and verify the computation. The result is:\[A^{-1} = \begin{bmatrix} 0.00425 & 0.00230 \ 0.00676 & 0.00298 \end{bmatrix}\]

Step 4: Solve for X Using A's Inverse

Using the equation \(X = A^{-1}B\), compute the result by multiplying \(A^{-1}\) by \(B\):\[X = \begin{bmatrix} 0.00425 & 0.00230 \ 0.00676 & 0.00298 \end{bmatrix} \begin{bmatrix} -439 \ 2282 \end{bmatrix}\]Calculate each corresponding entry of \(X\) using these multiplications.

Step 5: Compute Entries of X

Multiply the matrices from Step 4 to find the values of \(x\) and \(y\):- \( x = 0.00425 \times -439 + 0.00230 \times 2282 \approx 3.12 \)- \( y = 0.00676 \times -439 + 0.00298 \times 2282 \approx 1.48 \)Thus, \( x \approx 3.12 \) and \( y \approx 1.48 \).

Step 6: Approximating to Nearest Value

Round the calculated values of \(x\) and \(y\) to two decimal places for final approximation. Thus, the solution is approximately \(x = 3.12\) and \(y = 1.48\).

Key Concepts

Matrix Algebra

Matrix Algebra is a fundamental tool in solving systems of linear equations. Think of a matrix as a grid of numbers neatly organized into rows and columns. When dealing with systems of equations, matrices help by representing these systems in a compact form. For instance, a matrix can hold the coefficients from a set of linear equations all in one spot.

To work with matrices, here are some key operations:

• Addition and Subtraction: You can add or subtract matrices element by element. The matrices must be the same size for these operations.
• Multiplication: This is more complex. Multiply each element in a row from the first matrix by matching elements from a column in the second matrix and sum the results.
• Determinants: A value calculated from a square matrix. It's crucial for finding the inverse of a matrix.

By setting up a system of equations as a matrix equation, we pave the way to use powerful algebraic techniques to find solutions.

Inverse Matrices

Finding the inverse of a matrix is like asking: what matrix can reverse the effect of this one? If multiplying two matrices gives the identity matrix, they are inverses of each other.

Here's why inverse matrices are useful:

• Finding Solutions: When a matrix has an inverse, you can solve equations like \(AX = B\) by multiplying both sides by the inverse, \(A^{-1}\), resulting in \(X = A^{-1}B\).
• Mathematical Checks: If you suspect your solutions are incorrect, computing and using the inverse validates or refines your answers.

Not all matrices have inverses. A matrix must be square (same number of rows and columns), and its determinant should not be zero. This is a fundamental step in ensuring the calculation of an inverse is valid.

Linear Algebra Applications

Linear Algebra isn't just a collection of techniques; it has broad applications in the real world. From graphics rendering to economic modeling, these powerful concepts extend beyond the classroom.

Here are a few practical applications:

• Computer Graphics: Matrices transform and move images. They handle scaling, rotating, and translating images onscreen.
• Engineering: Used in structural analysis and design, where systems of equations govern the stability and safety of structures.
• Data Science: Estimates and predictions in statistics use matrices for many calculations involved in regression analysis.

Appreciating these applications can make the theory of linear algebra more tangible and highlight its importance in modern technological solutions.