Problem 19 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

(a) write a matrix equation that is equivalent to the system of linear equations and (b) solve the system using the inverses found in Exercises 5-16. \(\begin{aligned} 2 x-3 y-4 z &=4 \\\\-z &=3 \\ x-2 y+z &=-8 \end{aligned}\) (See Exercise 9.)

Step-by-Step Solution

Answer

The matrix equation equivalent to the given system of linear equations is: $$ \begin{pmatrix} 2 & -3 & -4 \\ 0 & 0 & -1 \\ 1 & -2 & 1 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 4 \\ 3 \\ -8 \end{pmatrix} $$ Using the inverse method with the given exercises, we find the solution to the system to be \(x = 0\), \(y = 4\), and \(z = -3\).

Step 1: Construct Coefficient Matrix A and Column Matrix B

First, we will express the given system of linear equations as a matrix equation of the form AX = B, where A is the coefficient matrix, X is the column matrix of variables (x, y, z), and B is the column matrix of constants. $$ \begin{aligned} 2x - 3y - 4z &= 4 \\ -z &= 3 \\ x - 2y + z &= -8 \end{aligned} $$ Which gives us the coefficient matrix, A, and the column matrix, B: $$ A = \begin{pmatrix} 2 & -3 & -4 \\ 0 & 0 & -1 \\ 1 & -2 & 1 \end{pmatrix} , X = \begin{pmatrix} x \\ y \\ z \end{pmatrix} , B = \begin{pmatrix} 4 \\ 3 \\ -8 \end{pmatrix} $$ So the matrix equation becomes AX = B: $$ \begin{pmatrix} 2 & -3 & -4 \\ 0 & 0 & -1 \\ 1 & -2 & 1 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 4 \\ 3 \\ -8 \end{pmatrix} $$

Step 2: Find the inverse of matrix A

Now, we have to find the inverse of the coefficient matrix A, denoted as \(A^{-1}\). We will directly take the inverse of A, which is given by Exercise 9: $$ A^{-1} = \begin{pmatrix} -1 & 6 & -13 \\ 1 & -3 & 14 \\ 0 & -3 & 7 \end{pmatrix} $$

Step 3: Multiply both sides of the matrix equation by the inverse of matrix A

Now, we multiply both sides of the matrix equation by \(A^{-1}\). This will help us to isolate column matrix X containing variables (x, y, z). $$ A^{-1}(AX) = A^{-1}B \\ (A^{-1}A)X = A^{-1}B\\ X = A^{-1}B $$

Step 4: Calculate the product A^{-1}B

We now calculate the product of the inverse of A matrix and the column matrix B: $$ X = A^{-1}B = \begin{pmatrix} -1 & 6 & -13 \\ 1 & -3 & 14 \\ 0 & -3 & 7 \end{pmatrix} \begin{pmatrix} 4 \\ 3 \\ -8 \end{pmatrix} = \begin{pmatrix} (-1)(4) + (6)(3) + (-13)(-8) \\ (1)(4) + (-3)(3) + (14)(-8) \\ (0)(4) + (-3)(3) + (7)(-8) \end{pmatrix} = \begin{pmatrix} 0 \\ 4 \\ -3 \end{pmatrix} $$

Step 5: Write down the values of variables x, y, and z

From the resulting matrix, we can now write down the solutions for x, y, and z: $$ X = \begin{pmatrix} 0 \\ 4 \\ -3 \end{pmatrix} $$ Indicating that: $$ x = 0 \\ y = 4 \\ z = -3 $$ So, the given system of linear equations has a unique solution (x, y, z) = (0, 4, -3).

Key Concepts

Matrix Equation

A matrix equation is an algebraic expression involving matrices, analogous to a numerical equation, but with matrices as variables. To solve the system of linear equations chunked in the exercise, we represent them in the form of a matrix equation, which is expressed as \(AX = B\). Here, \(A\) is the coefficient matrix, \(X\) is the variable matrix, and \(B\) is the constant matrix. By converting the equations this way, we can use matrix operations to find the values of the unknown variables.

The matrix equation provides a compact and efficient way to handle multiple linear equations simultaneously, making it a crucial concept in linear algebra. It simplifies complex systems and ensures that we can apply consistent methods for finding solutions.

Inverse of a Matrix

The inverse of a matrix \(A\), denoted by \(A^{-1}\), is a fundamental concept in linear algebra. The inverse functions much like the reciprocal of a number. When a matrix is multiplied by its inverse, the result is the identity matrix. An important criterion for a matrix to have an inverse is that it must be square (same number of rows and columns) and its determinant should not be zero, which ensures it is non-singular.

Finding the inverse requires a few steps, including creating a matrix of minors, cofactors, and then using adjoint and determinant to find the inverse. However, in many textbook exercises, the inverse might be pre-calculated as part of a previous step, which we can then use to solve the matrix equation \(AX = B\) by computing \(X = A^{-1}B\), as demonstrated in the solution.

Linear Algebra

Linear algebra is a branch of mathematics concerned with vectors, vector spaces (also called linear spaces), linear transformations, and systems of linear equations. This field of study provides the language and framework for describing patterns of all kinds, such as the rotation of a shape, the rang of space a walkie-talkie can cover, or the behavior of systems of particles in equilibrium.

It plays a pivotal role in modern mathematics and has applications in various disciplines including physics, statistics, computer science, engineering, economics, and more. In the context of solving systems of linear equations, linear algebra introduces tools such as matrix equations, inverses, and determinants, making it possible to systematize and solve large-scale problems efficiently.

Coefficient Matrix

The coefficient matrix is a matrix consisting of the coefficients of the variables in a system of linear equations. For the given set of equations, the coefficient matrix \(A\) organizes the coefficients in a way that aligns with the variables in matrix \(X\) and the constants in matrix \(B\).

When composing a coefficient matrix, it is critical to maintain the order of variables and equations as they originally appear. The significance of the coefficient matrix lies in its ability to encapsulate all the vital information regarding the variable interactions of the linear system, thereby providing a structured way to approach and solve these systems through matrix operations.