Problem 50 - College Algebra with Modeling and Visualization

Question

Complete the following. (A) Write the system in the form \(A X=B\). (B) Solve the system by finding \(A^{-1}\) and then using the equation \(\boldsymbol{X}=\boldsymbol{A}^{-1} \boldsymbol{B}\). (Hint: Some of your answers from Exercises \(15-28\) may be helpful.) $$ \begin{array}{c} x+3 y=-3 \\ 2 x+5 y=-2 \end{array} $$

Step-by-Step Solution

Answer

The solution to the system is \(x = 9\) and \(y = -4\).

Step 1: Identify Coefficients and Constants

We start by identifying the coefficients and constants from the system of linear equations. The system is given by:\[\begin{align*}1x + 3y &= -3, \2x + 5y &= -2.\end{align*}\]The coefficient matrix \(A\) is \(\begin{bmatrix} 1 & 3 \ 2 & 5 \end{bmatrix}\) and the constant matrix \(B\) is \(\begin{bmatrix} -3 \ -2 \end{bmatrix}\). The solution vector \(X\) is represented by \(\begin{bmatrix} x \ y \end{bmatrix}\).

Step 2: Formulate the System in Matrix Form

We write the system in the form \(AX = B\), where:\[A = \begin{bmatrix} 1 & 3 \ 2 & 5 \end{bmatrix}, \quad X = \begin{bmatrix} x \ y \end{bmatrix}, \quad B = \begin{bmatrix} -3 \ -2 \end{bmatrix}.\]This results in the matrix equation:\[\begin{bmatrix} 1 & 3 \ 2 & 5 \end{bmatrix} \begin{bmatrix} x \ y \end{bmatrix} = \begin{bmatrix} -3 \ -2 \end{bmatrix}.\]

Step 3: Find the Inverse of Matrix A

To solve for \(X\), we first need to find the inverse of matrix \(A\). The formula for the inverse of a 2x2 matrix \(\begin{bmatrix} a & b \ c & d \end{bmatrix}\) is:\[A^{-1} = \frac{1}{ad-bc} \begin{bmatrix} d & -b \ -c & a \end{bmatrix}.\]Applying this to \(A\):\[A^{-1} = \frac{1}{(1)(5) - (3)(2)} \begin{bmatrix} 5 & -3 \ -2 & 1 \end{bmatrix} = \frac{1}{5 - 6} \begin{bmatrix} 5 & -3 \ -2 & 1 \end{bmatrix} = \begin{bmatrix} -5 & 3 \ 2 & -1 \end{bmatrix}.\]

Step 4: Solve for X Using the Inverse

Now that we have \(A^{-1}\), we can solve \(AX = B\) using \(X = A^{-1} B\):\[X = \begin{bmatrix} -5 & 3 \ 2 & -1 \end{bmatrix}\begin{bmatrix} -3 \ -2 \end{bmatrix}.\]Performing the matrix multiplication:\[X = \begin{bmatrix} (-5)(-3) + (3)(-2) \ (2)(-3) + (-1)(-2) \end{bmatrix} = \begin{bmatrix} 15 - 6 \ -6 + 2 \end{bmatrix} = \begin{bmatrix} 9 \ -4 \end{bmatrix}.\]

Step 5: Conclusion: Interpret the Solution

The solution \(X\) gives us the values \(x = 9\) and \(y = -4\). This means that the point \((9, -4)\) satisfies both equations in the given system.

Key Concepts

Matrix Inversion

Matrix inversion is an essential tool in matrix algebra, especially when solving systems of linear equations. When you have a square matrix (same number of rows and columns), finding the inverse of this matrix can be thought of as identifying another matrix that, when multiplied by the original, gives the identity matrix. The identity matrix is like the number 1 for matrices—it doesn't change another matrix when you multiply by it.
Imagine you're trying to get back to the starting point after a series of moves in a game. The inverse matrix acts as the exact series of moves needed to return to the beginning. You can find the inverse of a 2x2 matrix \( A = \begin{bmatrix} a & b \ c & d \end{bmatrix} \) using the formula:
\[ A^{-1} = \frac{1}{ad-bc} \begin{bmatrix} d & -b \ -c & a \end{bmatrix} \]
If the determinant \((ad-bc)\) is zero, the matrix doesn't have an inverse, and you're stuck—just like getting lost without a map.
In our exercise, we found the inverse of the matrix \( \begin{bmatrix} 1 & 3 \ 2 & 5 \end{bmatrix} \) by using the formula above. This step gave us the key to solving the system of equations.

System of Linear Equations

A system of linear equations is simply a collection of two or more linear equations that share a set of variables. For example, in our original exercise, we are working with two equations that involve the variables \(x\) and \(y\). These equations are:

• \(x + 3y = -3\)
• \(2x + 5y = -2\)

The goal is to find values for \(x\) and \(y\) that satisfy both equations simultaneously. When solving these systems, one way is to express them in a matrix form, \(A X = B\), which allows us to see the entire system more clearly and provides a structure for further computations. Here, \(A\) is the matrix containing the coefficients of \(x\) and \(y\), \(X\) is the vector of the variables, and \(B\) is the constant vector.
Once translated into matrix algebra, solving the system often involves finding the inverse of \(A\) if it exists, making this an efficient method for handling complicated systems.

Matrix Multiplication

Matrix multiplication is the process used to combine two matrices. It's a bit like doing multiple sums of products, and it's critical for a wide range of applications, including solving systems of equations. In our exercise, after finding the inverse of matrix \(A\), the next step is to use this inverse to solve for \(X\) by multiplying it with \(B\).
The process involves taking rows from the first matrix and columns from the second, then multiplying the corresponding elements and adding them up to form each element of the result matrix. Here's a brief look at how this works. If you multiply \( A^{-1} \) and \( B \), each element of the resulting vector \( X \) comes from these sums:

• For each element in \(X\), sum the products of elements in the corresponding row of \(A^{-1}\) and column of \(B\). For example, \((1st\ row \times 1st\ column)\).

This operation may seem complex at first, but with practice, it becomes an intuitive way to combine matrices and explore the results. It ultimately provides the solution values for \(x\) and \(y\) we are seeking.