Problem 52 - Precalculus with Limits

Question

In Exercises 51-58, use an inverse matrix to solve (if possible) the system of linear equations. \(\begin{cases} 18x + 12y = 13 \\ 30x + 24y = 23 \end{cases}\)

Step-by-Step Solution

Answer

The solution to the system of linear equations is x = 13/6 and y = 1/4

Step 1: Convert the system into a matrix equation

We write the given system of equations as a matrix equation. Here it is \( \begin{bmatrix} 18 & 12 \\ 30 & 24 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 13 \\ 23 \end{bmatrix} \). This is in the form AX = B, where A is the matrix of coefficients, X is the column matrix of variables and B is the column matrix of constants.

Step 2: Find the determinant of the matrix A

The determinant of a 2x2 matrix \( \begin{bmatrix} a & b \\ c & d \end{bmatrix} \) is given by \( det(A) = ad - bc \). In our case, \( det(A) = 18×24 - 12×30 = 432 - 360 = 72 \). Since the determinant is not equal to 0, an inverse of matrix A exists and so the system of linear equations has a unique solution.

Step 3: Compute the inverse of matrix A

The inverse of a 2x2 matrix \( \begin{bmatrix} a & b \\ c & d \end{bmatrix} \) is given by \( \frac{1}{det(A)} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix} \). So, the inverse of our matrix A is \( \begin{bmatrix} 24 & -12 \\ -30 & 18 \end{bmatrix} \) multipled by \( \frac{1}{72} \), which is \( \begin{bmatrix} 1/3 & -1/6 \\ -5/12 & 1/4 \end{bmatrix} \).

Step 4: Solve for X

The solution of the system of equations is given by X = A^{-1}B. Hence, we multiply the inverse of A we found in the previous step by B to find X, which is \( \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 1/3 & -1/6 \\ -5/12 & 1/4 \end{bmatrix} \begin{bmatrix} 13 \\ 23 \end{bmatrix} = \begin{bmatrix} 13/6 \\ 1/4 \end{bmatrix} \). So, the solution of the original system of linear equations is x = 13/6 , y = 1/4.

Key Concepts

Determinant

The determinant is a special number that can be calculated from a square matrix. It's crucial because it helps determine whether a matrix can be inverted. For a 2x2 matrix, like the one in our problem, the determinant formula is quite simple: for a matrix \( \begin{bmatrix} a & b \ c & d \end{bmatrix} \), the determinant is \( ad - bc \).

This value tells us a lot about the matrix. If the determinant is zero, the matrix does not have an inverse, meaning we cannot use it to solve systems of equations. However, if it is not zero, as in our problem where the determinant was 72, the matrix is invertible.

An invertible matrix leads us to a unique solution to the system of equations. Thus, the determinant acts as a gatekeeper, informing us whether we can proceed to use the inverse matrix for solving equations.

System of Linear Equations

A system of linear equations is a collection of one or more linear equations involving the same set of variables. These are equations that map to straight lines when graphed. For our example, the two equations with variables \( x \) and \( y \) make up a system:

• \(18x + 12y = 13\)
• \(30x + 24y = 23\)

The objective is to find the values of \( x \) and \( y \) that satisfy both equations simultaneously.

One effective method to solve such systems is by using matrices. We transform the system into a matrix equation \( AX = B \). In this matrix equation, \( A \) is the matrix of coefficients, \( X \) is the column of variables, and \( B \) is the column of constants. By finding the matrix \( A \)'s inverse, we multiply the inverse with \( B \) to find \( X \), which holds the values of \( x \) and \( y \).

This approach is systematic and powerful, especially when dealing with larger systems, as it allows for using computers to solve complex systems efficiently.

Matrix Multiplication

Matrix multiplication is a process used frequently when solving systems of equations with matrices. It's slightly different from the regular arithmetic multiplication you might be used to. In essence, to multiply two matrices, you multiply rows by columns, and sum the products.

In our problem, once we have found the inverse of matrix \( A \), we perform matrix multiplication to find the solution \( X \). We take our inverse matrix \( A^{-1} \) and the matrix \( B \) containing our constants from the equations. The multiplication will yield a new matrix \( X \), which contains the values of the variables \( x \) and \( y \).

The operation is crucial because it allows us to transform the system back from matrix form to understandable numeric values that solve the original equations. Remember, the order matters in multiplication: \( A^{-1}B \) might not give the same result as \( BA^{-1} \). So it's important to follow the order as dictated by the system of equations.

Understanding this concept requires practice, especially with recognizing how each element of the row in the first matrix relates to a column in the second matrix. It's this interplay of elements that determines the output of the matrix multiplication.