Problem 61 - 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 $$ \begin{array}{rr} 3 x-y+z= & 4.9 \\ 5.8 x-2.1 y & =-3.8 \\ -x & +2.9 z=3.8 \end{array} $$

Step-by-Step Solution

Answer

(a) A = \( \begin{bmatrix} 3 & -1 & 1 \\ 5.8 & -2.1 & 0 \\ -1 & 0 & 2.9 \end{bmatrix} \), (b) \( x \approx -0.031 \), \( y \approx 0.992 \), \( z \approx 1.345 \)

Step 1: Write the Coefficient Matrix A

Identify the coefficients of the variables from each equation to construct the matrix \( A \).The given system of equations is:1. \( 3x - y + z = 4.9 \)2. \( 5.8x - 2.1y = -3.8 \)3. \( -x + 2.9z = 3.8 \)The coefficient matrix \( A \) is:\[A = \begin{bmatrix} 3 & -1 & 1 \5.8 & -2.1 & 0 \-1 & 0 & 2.9 \end{bmatrix}\]

Step 2: Write the Variable Matrix X

Assign the variables of the system to matrix \( X \).The variable matrix \( X \) is:\[X = \begin{bmatrix} x \y \z \end{bmatrix}\]

Step 3: Write the Constants Matrix B

Extract the constant terms from each equation to create the matrix \( B \).The constants matrix \( B \) is:\[B = \begin{bmatrix} 4.9 \-3.8 \3.8 \end{bmatrix}\]

Step 4: Formulate the Equation A X = B

Now, express the system of equations in the matrix form \( AX = B \).Substitute the matrices:\[\begin{bmatrix} 3 & -1 & 1 \5.8 & -2.1 & 0 \-1 & 0 & 2.9 \end{bmatrix}\begin{bmatrix} x \y \z \end{bmatrix} = \begin{bmatrix} 4.9 \-3.8 \3.8 \end{bmatrix}\]

Step 5: Calculate the Inverse of Matrix A

Using a calculator or software that supports matrix operations, compute the inverse \( A^{-1} \) of the matrix \( A \).The inverse \( A^{-1} \) is computed using a calculator or software that supports matrix calculations.

Step 6: Solve for Matrix X

Multiply the inverse of matrix \( A \) by matrix \( B \) to solve for \( X \).\( X = A^{-1}B \).Using a calculator, multiply the matrices to find the approximate values of \( x, y, \text{ and } z \).

Step 7: Interpret the Solution

Using the calculated values, write down the solution for the variables:- \( x \approx -0.031 \)- \( y \approx 0.992 \)- \( z \approx 1.345 \)These values represent the approximate solution of the given system of equations.

Key Concepts

Matrix Operations

Matrix operations are foundational to solving systems of linear equations. They involve manipulating matrices to achieve a desired form or perform algebraic calculations. A matrix is simply a rectangular array of numbers, symbols, or expressions arranged in rows and columns.

In solving linear equations, matrix operations include:

• Addition and Subtraction: Matrices of the same dimension can be added or subtracted by adding or subtracting their corresponding elements.
• Scalar Multiplication: Each element of a matrix can be multiplied by a constant, known as a scalar.
• Matrix Multiplication: Involves multiplying rows by columns. You must ensure the number of columns in the first matrix matches the number of rows in the second.
• Matrix Transpose: The rows of a matrix are converted into columns and vice versa.

For our specific system of equations, we express them using these matrix operations in the form of \(A X = B\). Here, matrices \(A\) and \(B\) are constructed by extracting coefficients and constants from each equation respectively. This setup allows us to perform further operations to find \(X\), which holds the solutions.

Inverse Matrix

The inverse of a matrix is a concept similar to finding the reciprocal of a number. Just as the reciprocal of a number \(a\) is \(1/a\) and satisfies \(a \times (1/a) = 1\), the inverse of a matrix \(A\) is denoted \(A^{-1}\) and satisfies the equation \(A \times A^{-1} = I\), where \(I\) is the identity matrix.

Not all matrices have an inverse. A matrix must be square (same number of rows and columns) and have a non-zero determinant to possess an inverse.

For the system given in the problem, calculating \(A^{-1}\) is crucial. With a calculator, software, or algebraic methods, the inverse is used to isolate the variable matrix \(X\). This is done by:

• Ensuring \(A\) is square and calculating its determinant.
• Applying matrix operations to find \(A^{-1}\).

Once we have \(A^{-1}\), we multiply it by matrix \(B\) to get the solution \(X\). The operation follows \(X = A^{-1}B\), producing the values for the variables.

Solving Linear Equations

Solving linear equations using matrices provides a structured approach, especially for larger systems. Here, we focus on expressing the linear equations as \(A X = B\) and using the mathematical structure of matrices to find solutions.

This method is advantageous because it streamlines equations into compact forms, making it easier to solve using technology or by hand. The key steps in this process include:

• Formulating the system in matrix form: Represent equations as matrices \(A\) and \(B\).
• Finding the inverse if possible: Calculate \(A^{-1}\) to help transition from \(AX = B\) to \(X = A^{-1}B\).
• Computing the solution: Multiplying \(A^{-1}\) by \(B\) to uncover \(X\), the matrix holding the values of the original variables.

In our example, this approach yields approximate solutions for \(x, y,\) and \(z\) reflecting real-world measurements. It's efficient for interpreting linear relationships and has broad applications in fields like engineering and economics.