Problem 34 - Algebra for College Students

Question

Subscript notation is frequently used for working with larger systems of equations. For Problems 31-34, use a matrix approach to solve each system. Express the solutions as 4-tuples of the form \(\left(x_{1}, x_{2}, x_{3}, x_{4}\right)\).\(\left(\begin{array}{rl}x_{1}+2 x_{2}-3 x_{3}+x_{4} & =-2 \\ -2 x_{1}-3 x_{2}+x_{3}-x_{4} & =5 \\ 4 x_{1}+9 x_{2}-2 x_{3}-2 x_{4} & =-28 \\\ -5 x_{1}-9 x_{2}+2 x_{3}-3 x_{4} & =14\end{array}\right)\)

Step-by-Step Solution

Answer

If the determinant is non-zero, express solution as \( X = A^{-1}B \).

Step 1: Write the System as a Matrix Equation

The system of equations can be expressed in matrix form as \( AX = B \), where \( A \) is the coefficient matrix, \( X \) is the variable matrix, and \( B \) is the constant matrix. Here,\[A = \begin{bmatrix}1 & 2 & -3 & 1 \-2 & -3 & 1 & -1 \4 & 9 & -2 & -2 \-5 & -9 & 2 & -3\end{bmatrix}, \ X = \begin{bmatrix} x_1 \ x_2 \ x_3 \ x_4 \end{bmatrix}, \ B = \begin{bmatrix} -2 \ 5 \ -28 \ 14 \end{bmatrix}.\]

Step 2: Determine the Inverse of Matrix A (if possible)

To solve \( AX = B \) using the matrix approach, find the inverse of matrix \( A \), which is denoted as \( A^{-1} \). For matrices that are invertible, \( A^{-1} \) exists. The solution to the system is given by \( X = A^{-1}B \). Calculate \( A^{-1} \) if \( \text{det}(A) eq 0 \).

Step 3: Check for the Determinant of A

Compute the determinant of matrix \( A \). If \( \text{det}(A) = 0 \), then matrix \( A \) does not have an inverse and other methods must be used to solve the system. Calculate:\[\text{det}(A) = \begin{vmatrix} 1 & 2 & -3 & 1 \ -2 & -3 & 1 & -1 \ 4 & 9 & -2 & -2 \ -5 & -9 & 2 & -3 \end{vmatrix}\]

Step 4: Solve for X Using Inverse Matrix

Assuming \( A^{-1} \) exists and the determinant is non-zero, calculate \( X = A^{-1}B \) to find the solution to the system of equations:- Compute \( A^{-1} \).- Multiply \( A^{-1} \) with \( B \) to obtain \( X \).

Step 5: Express the Solution in 4-Tuple Form

After solving for \( X \), express the variables \( x_1, x_2, x_3, \) and \( x_4 \) as a 4-tuple: \( (x_1, x_2, x_3, x_4) \). This gives the solution in the required form.

Key Concepts

Systems of Equations

A system of equations is a collection of multiple equations involving a number of variables. In the context of matrix algebra, systems of equations can be solved by rewriting the equations using matrices. This makes it easier to handle large and complex systems. Matrices offer a compact way of expressing and manipulating these equations.

In our example, we are dealing with four equations and four unknowns, which means we have a 4x4 matrix. Each row in the matrix represents one equation, and each column corresponds to one of the variables in the system. This setup can be succinctly written in the form of a single matrix equation:

• Coefficient matrix (denoted as \( A \)) includes all the coefficients from each equation.
• Variable matrix (denoted as \( X \)) represents the list of variables (i.e., \( x_1, x_2, x_3, \) and \( x_4 \)).
• Constant matrix (denoted as \( B \)) contains all the constants on the right side of the equations.

The aim is to solve for the variable matrix \( X \), which gives us the solution to the system.

Matrix Inversion

Matrix inversion is a crucial process in solving systems of linear equations using matrices. If we have the matrix equation \( AX = B \), we can solve for the variable matrix \( X \) by multiplying both sides with the inverse of \( A \), given by \( A^{-1} \).

The key steps are as follows:

• Calculate the inverse \( A^{-1} \). This step is only possible if matrix \( A \) is invertible. \( A \) is invertible if its determinant is not zero.
• Multiply the inverse matrix with the constant matrix \( B \) to isolate \( X \): \( X = A^{-1}B \).

This method requires that the dimensions of matrices \( A \) and \( B \) align correctly for multiplication. If \( A \) is a 4x4 matrix, then \( X \) and \( B \) should both have four rows.

Matrix inversion is a powerful technique but not always applicable due to the requirement of a non-zero determinant.

Determinants

Determinants are a scalar value that can be computed from a square matrix and hold significant importance in matrix algebra. They play a vital role in matrix inversion and in determining whether a matrix is invertible. The determinant of a matrix \( A \), denoted as \( \text{det}(A) \), can be zero or non-zero, which affects the matrix's invertibility.

For our 4x4 system, calculating \( \text{det}(A) \) helps to confirm whether \( A^{-1} \) can exist. If \( \text{det}(A) = 0 \), the system does not have a unique solution, and methods other than matrix inversion must be used.

• A non-zero determinant indicates that the matrix can be inverted and a unique solution to the system exists.
• A zero determinant indicates that the matrix is singular and cannot be inverted, implying the system may have no solution or infinitely many solutions.

Determinants provide important insights into the properties of matrices and the systems of equations they represent.

4-Tuples

In the context of this matrix algebra problem, solutions to the system of equations are represented as 4-tuples. A tuple is a mathematical term that signifies an ordered list of elements. A 4-tuple contains four components, each representing a solution to one of the variables in the system.

After solving the matrix equation \( X = A^{-1}B \), you obtain a column matrix displaying the values of \( x_1, x_2, x_3, \) and \( x_4 \). These are then formatted into a 4-tuple: \( (x_1, x_2, x_3, x_4) \), depicting the specific solution for each variable.

Using this format is beneficial because:

• It provides a neat and concise way to display the complete solution set.
• It highlights the order of the variables, which is essential for keeping track of each part of the solution.

4-tuples are an efficient method of conveying solutions to systems involving four variables, maintaining clarity and simplicity.