Problem 65 - Precalculus

Question

apply matrix algebra to solve the system of linear equations. $$4 x-9 y=-1$$ $$7 x-3 y=\frac{5}{2}$$

Step-by-Step Solution

Answer

The solution is \( x = \frac{1}{2} \) and \( y = \frac{1}{3} \).

Step 1: Write the System in Matrix Form

The given system of linear equations can be expressed in the matrix form as \( A\mathbf{x} = \mathbf{b} \). Here, \( A \) is the coefficient matrix, \( \mathbf{x} \) is the variable matrix, and \( \mathbf{b} \) is the constant matrix. Thus, we write:\[A = \begin{bmatrix} 4 & -9 \ 7 & -3 \end{bmatrix}, \quad \mathbf{x} = \begin{bmatrix} x \ y \end{bmatrix}, \quad \mathbf{b} = \begin{bmatrix} -1 \ \frac{5}{2} \end{bmatrix}\]

Step 2: Determine the Inverse of Coefficient Matrix A

To solve for \( \mathbf{x} \) using matrix algebra, we need the inverse of the coefficient matrix \( A \). Calculate the determinant of \( A \):\[ det(A) = 4(-3) - (-9)(7) = -12 + 63 = 51\]Then, the inverse of \( A \), denoted as \( A^{-1} \), is calculated as:\[A^{-1} = \frac{1}{51} \begin{bmatrix} -3 & 9 \ -7 & 4 \end{bmatrix}\]

Step 3: Multiply the Inverse Matrix by Constant Vector

Now multiply the inverse of the matrix \( A \) by the constant matrix \( \mathbf{b} \):\[\mathbf{x} = A^{-1} \mathbf{b} = \frac{1}{51} \begin{bmatrix} -3 & 9 \ -7 & 4 \end{bmatrix} \begin{bmatrix} -1 \ \frac{5}{2} \end{bmatrix}\]Perform the matrix multiplication:\[\mathbf{x} = \frac{1}{51} \begin{bmatrix} (-3)(-1) + 9\left(\frac{5}{2}\right) \ (-7)(-1) + 4\left(\frac{5}{2}\right) \end{bmatrix} = \frac{1}{51} \begin{bmatrix} 3 + 22.5 \ 7 + 10 \end{bmatrix} = \frac{1}{51} \begin{bmatrix} 25.5 \ 17 \end{bmatrix}\]

Step 4: Simplify the Solution

Finally, simplify the result to find \( \mathbf{x} = \begin{bmatrix} x \ y \end{bmatrix} \):\[ \mathbf{x} = \begin{bmatrix} \frac{25.5}{51} \ \frac{17}{51} \end{bmatrix} = \begin{bmatrix} \frac{1}{2} \ \frac{1}{3} \end{bmatrix}\]Therefore, \( x = \frac{1}{2} \) and \( y = \frac{1}{3} \).

Key Concepts

Inverse Matrix

An inverse matrix is essential when we want to solve a system of linear equations using matrix algebra, especially through a method known as matrix inversion. A matrix, in simple terms, is like a mathematical box filled with numbers arranged in rows and columns. If you have a square matrix \( A \), the inverse of this matrix, denoted as \( A^{-1} \), is a special kind of matrix that, when multiplied with \( A \), gives you the identity matrix. Here’s a bit more detail:

• The identity matrix is like the number 1 in algebra; it doesn’t change other numbers during multiplication.
• The way to think of it is: if you have \( A \) and want to "undo" \( A \), you use \( A^{-1} \).

For a 2x2 matrix, the inverse exists only if the determinant is not zero. The determinant is a special number you calculate from a matrix that can tell you a lot about the matrix, including whether it has an inverse. For example, if the determinant equals zero, the matrix has no inverse. In our example, the determinant was 51, which is not zero, so we could calculate the inverse matrix.

System of Linear Equations

A system of linear equations, like the one in our problem, consists of two or more equations that share some variables. The primary aim is to find values of these variables that satisfy all equations simultaneously. In our example, we have two equations:

• \(4x - 9y = -1\)
• \(7x - 3y = \frac{5}{2}\)

To solve these, we use a nice method involving matrices. It allows us to simplify complex systems and solve them efficiently. Essentially, by sticking the coefficients (the numbers in front of x and y) into a matrix form, we can apply matrix operations, like finding an inverse. Using a matrix to represent a system offers a structured, systematic approach. This approach scales up well, making it effective for much larger systems, where solving manually by substitution could be cumbersome and impractical.

Matrix Multiplication

Matrix multiplication is a fundamental operation that lets us perform various tasks in mathematics, including solving systems of equations through matrices. It’s slightly different from regular arithmetic multiplication because it involves a kind of dot-product approach.Let's break it down:

• You multiply elements from the rows of the first matrix by corresponding elements from the columns of the second matrix, then sum the products.
• This is repeated for each position in the new matrix.

In the system we analyzed, after finding the inverse matrix \( A^{-1} \), we multiplied it by the constant matrix \( \mathbf{b} \). This process is computing:\[ \mathbf{x} = A^{-1} \mathbf{b} \]This multiplication allows us to find the variable matrix \( \mathbf{x} \), which contains the solutions to our system of equations. The matrix operation effectively isolates \( x \) and \( y \), providing their specific values. It’s a powerful tool because it transforms a seemingly daunting system into a manageable procedure.