Problem 34 - Precalculus Mathematics for Calculus

Question

Use a calculator that can perform matrix operations to solve the system, as in Example 7 . $$\left\\{\begin{array}{l} 3 x+4 y-z=2 \\ 2 x-3 y+z=-5 \\ 5 x-2 y+2 z=-3 \end{array}\right.$$

Step-by-Step Solution

Answer

Use matrix multiplication to find \( x = -1 \), \( y = 0 \), \( z = 1 \).

Step 1: Write the System of Equations as a Matrix

Represent the system of equations in matrix form. The system can be written as the equation \( A\mathbf{x} = \mathbf{b} \), where \( A \) is the coefficient matrix, \( \mathbf{x} \) is the variable matrix, and \( \mathbf{b} \) is the constant matrix.\[ A = \begin{bmatrix} 3 & 4 & -1 \ 2 & -3 & 1 \ 5 & -2 & 2 \end{bmatrix}, \quad \mathbf{x} = \begin{bmatrix} x \ y \ z \end{bmatrix}, \quad \mathbf{b} = \begin{bmatrix} 2 \ -5 \ -3 \end{bmatrix} \]

Step 2: Calculate the Inverse of Matrix A

Use a calculator to find the inverse of the coefficient matrix \( A \), denoted as \( A^{-1} \). This step requires that matrix \( A \) is invertible.

Step 3: Compute the Solution Matrix

Multiply the inverse of matrix \( A \) by matrix \( \mathbf{b} \) to find \( \mathbf{x} \). Perform the matrix multiplication: \[ \mathbf{x} = A^{-1} \mathbf{b} \] This will provide the values of \( x \), \( y \), and \( z \).

Step 4: Interpret the Results

From step 3, you will get the numerical values for \( x \), \( y \), and \( z \). These values are the solutions to the system of equations.

Key Concepts

Solving Systems of Equations

When faced with solving systems of linear equations, one efficient method is using matrix operations. Systems of equations consist of two or more equations that share variables. The goal is to find the value for each variable that satisfies all equations simultaneously. In our given example, we have a system with three equations and three unknowns, which can often seem challenging to solve manually.By converting these equations into a matrix form, we simplify the problem significantly. This involves representing the coefficients of the variables in a matrix known as the coefficient matrix, the variables themselves in a variable matrix, and the constants in a constant matrix.

• The coefficient matrix in the exercise is \( A = \begin{bmatrix} 3 & 4 & -1 \ 2 & -3 & 1 \ 5 & -2 & 2 \end{bmatrix} \).
• The variable matrix is \( \mathbf{x} = \begin{bmatrix} x \ y \ z \end{bmatrix} \).
• The constant matrix is \( \mathbf{b} = \begin{bmatrix} 2 \ -5 \ -3 \end{bmatrix} \).

Once written in this matrix form as \( A \mathbf{x} = \mathbf{b} \), solving for the variables \( x, y, \) and \( z \) becomes a process of matrix operations, including finding the inverse of a matrix and performing matrix multiplication.

Inverse of a Matrix

The inverse of a matrix is a crucial concept in linear algebra, especially when solving systems of equations. Not every matrix is invertible, but when a matrix does have an inverse, it can be a powerful tool for finding solutions.For a square matrix \( A \), the inverse is denoted as \( A^{-1} \), and the property that defines the inverse is: \[ A A^{-1} = A^{-1} A = I \] Here, \( I \) is the identity matrix, which acts as the multiplicative identity in matrix multiplication, similar to "1" in scalar multiplication.Finding the inverse involves a series of steps or using a calculator capable of performing matrix operations. For our system’s coefficient matrix \( A \), we need to find \( A^{-1} \) in order to solve the equation \( A\mathbf{x} = \mathbf{b} \) by converting it to \( \mathbf{x} = A^{-1}\mathbf{b} \).When you multiply \( A^{-1} \) with \( \mathbf{b} \), you exploit this property to isolate the variables in \( \mathbf{x} \). This process essentially "undoes" the effect of \( A \) on the variable matrix, providing the solution vector \( \mathbf{x} \).

Matrix Multiplication

Matrix multiplication is an essential operation in handling matrices, including solving systems of equations. It is different from regular multiplication and involves specific rules.To multiply matrices, the number of columns in the first matrix must match the number of rows in the second matrix. In our scenario, after finding \( A^{-1} \), we perform the multiplication \( A^{-1} \mathbf{b} \) to solve for the variables.

• Each element in the resulting matrix \( \mathbf{x} \) is derived from the sum of products of corresponding elements from a row in \( A^{-1} \) and a column in \( \mathbf{b} \).
• This operation will result in a matrix that gives us the values for \( x, y, \) and \( z \), effectively solving the system of equations.

Understanding matrix multiplication is crucial because it helps in performing the operation accurately and comprehending why it allows us to solve such systems. Familiarity with matrix multiplication enables deeper insights into linear algebra concepts and broadens the tools available for mathematical problem-solving.