Problem 56 - 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 $$ \begin{array}{rr} 31 x+18 y= & 64.1 \\ 5 x-23 y= & -59.6 \end{array} $$

Step-by-Step Solution

Answer

The solution is approximately \( x = 0.50 \) and \( y = 2.70 \).

Step 1: Express the System as a Matrix Equation

We can express the given system of equations in matrix form, \( A X = B \), where matrix \( A \) represents the coefficients of the variables, \( X \) is the column matrix of variables, and \( B \) is the column matrix of constants. The given equations are: \[ \begin{array}{rr} 31x + 18y &= 64.1 \ 5x - 23y &= -59.6 \end{array} \]Thus, the matrices are:\[ A = \begin{bmatrix} 31 & 18 \ 5 & -23 \end{bmatrix}, \quad X = \begin{bmatrix} x \ y \end{bmatrix}, \quad B = \begin{bmatrix} 64.1 \ -59.6 \end{bmatrix} \]

Step 2: Calculate the Inverse of Matrix A

To solve for \( X \), we need to compute the inverse of matrix \( A \), denoted as \( A^{-1} \). Calculate it using a calculator. The inverse of a 2x2 matrix \( \begin{bmatrix} a & b \ c & d \end{bmatrix} \) is given by: \[ A^{-1} = \frac{1}{ad - bc} \begin{bmatrix} d & -b \ -c & a \end{bmatrix} \]First, compute the determinant: \[ 31(-23) - 5(18) = -713 - 90 = -803 \]Then, the inverse matrix is:\[ A^{-1} = \frac{1}{-803} \begin{bmatrix} -23 & -18 \ -5 & 31 \end{bmatrix} \]

Step 3: Compute X = A⁻¹B

Now, multiply \( A^{-1} \) with \( B \) to find \( X \).\[ A^{-1} \times B = \frac{1}{-803} \begin{bmatrix} -23 & -18 \ -5 & 31 \end{bmatrix} \begin{bmatrix} 64.1 \ -59.6 \end{bmatrix} \]Perform the matrix multiplication:* The first row is \(-23 \times 64.1 + (-18) \times (-59.6)\), which simplifies to \(-23 \times 64.1 + 18 \times 59.6 = -1474.3 + 1072.8 = -401.5\)* The second row is \(-5 \times 64.1 + 31 \times (-59.6)\), which simplifies to \(-320.5 - 1847.6 = -2168.1\)So, we have:\[ X = \frac{1}{-803} \begin{bmatrix} -401.5 \ -2168.1 \end{bmatrix} \]

Step 4: Simplify to Find the Values of x and y

Finally, simplify to obtain the values for \( x \) and \( y \):\[ x = \frac{-401.5}{-803} \approx 0.5006 \]\[ y = \frac{-2168.1}{-803} \approx 2.6994 \]

Step 5: Conclusion

The values of variables are approximately \( x \approx 0.50 \) and \( y \approx 2.70 \). This means the solution to the system of equations is \( x = 0.50 \) and \( y = 2.70 \).

Key Concepts

Matrix Equation

Linear equations can often be expressed in the form of a matrix equation. This approach provides a structured way of solving the equations using linear algebra techniques. In a matrix equation, we represent the coefficients of the variables in equations as matrices. Here's how:

• Matrix \( A \): Contains coefficients of the variables. For example, if you have the system of equations \(31x + 18y = 64.1\) and \(5x - 23y = -59.6\), the matrix \(A\) would be \( \begin{bmatrix} 31 & 18 \ 5 & -23 \end{bmatrix} \).


• Matrix \( X \): Represents the variables. In this case, \(X = \begin{bmatrix} x \ y \end{bmatrix} \).


• Matrix \( B \): Contains the constant terms. Thus \(B = \begin{bmatrix} 64.1 \ -59.6 \end{bmatrix} \).

So, the system translates to \(A X = B\). By doing so, complex systems can be rearranged into grid-like matrices for computational solving.

Matrix Inverse

When it comes to solving the matrix equation \( A X = B \) for \( X \), finding the matrix inverse \( A^{-1} \) is the key step. An inverse matrix does for matrices what division does for numbers. When you have \( A^{-1} \), you can find \( X \) by calculating \( X = A^{-1} B \).
The inverse of a matrix is computed in a specific way, especially crucial for 2x2 matrices:

• Calculate the determinant (more on this in the next section).
• If the determinant is not zero, the inverse is computed as: \[ A^{-1} = \frac{1}{ad - bc} \begin{bmatrix} d & -b \ -c & a \end{bmatrix} \]

Not all matrices have inverses. Only matrices with a non-zero determinant are invertible. In practical terms, using tools like calculators is advisable to avoid error-prone manual computation.

Determinant

The determinant is a special value computed from a square matrix. It is crucial when calculating the inverse of a matrix. For a 2x2 matrix \( \begin{bmatrix} a & b \ c & d \end{bmatrix} \), the determinant is calculated as:
\[ ad - bc \]
If the determinant is zero, the matrix does not have an inverse. This is because division by zero is undefined, leading to a situation where the solutions may not exist, or the equations are dependent. For the given matrix \( A \):

• Determinant \(= 31(-23) - 5(18) = -803 \).


• Since \(-803 eq 0\), matrix \( A \) is invertible and we proceed to find \( A^{-1} \).

Understanding the determinant is vital since it affects the possibility of finding solutions using matrix inversion.

System of Equations

A system of equations consists of multiple equations that are solved together. The goal is to find values for the variables that satisfy all the equations simultaneously.
For linear systems like:

• \(31x + 18y = 64.1\)
• \(5x - 23y = -59.6\)

We aim to find \(x\) and \(y\) values. Traditional methods include substitution or elimination, but these become cumbersome with complex systems.
Using matrix algebra transforms these methods into straightforward calculations:

1. Express the system as \(A X = B\).
2. Compute \(A^{-1}\) and solve \(X = A^{-1} B \).

This approach leverages matrix operations, making it efficient for computers and ensuring accuracy in solving systems of linear equations.