Problem 27 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

(a) write each system of equations as a matrix equation and (b) solve the system of equations by using the inverse of the coefficient matrix. \(\begin{aligned} x+2 y+z &=b_{1} \\ x+y+z &=b_{2} \\ 3 x+y+z &=b_{3} \\\ \text { where } & \text { (i) } b_{1}=7, b_{2}=4, b_{3}=2 \\ \text { and } & \text { (ii) } b_{1}=5, b_{2}=-3, b_{3}=-1 \end{aligned}\)

Step-by-Step Solution

Answer

For system (i), the matrix equation is \(\begin{pmatrix} 1 & 2 & 1 \\ 1 & 1 & 1 \\ 3 & 1 & 1 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 7 \\ 4 \\ 2 \end{pmatrix}\) and the solution is \((x, y, z) = (1, 1, 2)\). For system (ii), the matrix equation is \(\begin{pmatrix} 1 & 2 & 1 \\ 1 & 1 & 1 \\ 3 & 1 & 1 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 5 \\ -3 \\ -1 \end{pmatrix}\) and the solution is \((x, y, z) = (3, 4, -1)\).

Step 1: System (i) as a matrix equation

First, write the system (i) as a matrix equation: \(\begin{pmatrix} 1 & 2 & 1 \\ 1 & 1 & 1 \\ 3 & 1 & 1 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 7 \\ 4 \\ 2 \end{pmatrix}\)

Step 2: System (ii) as a matrix equation

Next, write the system (ii) as a matrix equation: \(\begin{pmatrix} 1 & 2 & 1 \\ 1 & 1 & 1 \\ 3 & 1 & 1 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 5 \\ -3 \\ -1 \end{pmatrix}\)

Step 3: Find the inverse of the coefficient matrix

Now, calculate the inverse of the coefficient matrix: \(A=\begin{pmatrix} 1 & 2 & 1 \\ 1 & 1 & 1 \\ 3 & 1 & 1 \end{pmatrix}\) The inverse of the matrix A is: \(A^{-1}=\dfrac{1}{2}\begin{pmatrix} -1 & 1 & 1 \\ 3 & -1 & -1 \\ -1 & 1 & -1 \end{pmatrix}\)

Step 4: Solve system (i) using the inverse matrix

Now, we can find the solutions of the system (i) by multiplying the inverse matrix with the result matrix of system (i). \(\begin{pmatrix} x \\ y \\ z \end{pmatrix} = A^{-1} \begin{pmatrix} 7 \\ 4 \\ 2 \end{pmatrix} = \dfrac{1}{2} \begin{pmatrix} -1 & 1 & 1 \\ 3 & -1 & -1 \\ -1 & 1 & -1 \end{pmatrix} \begin{pmatrix} 7 \\ 4 \\ 2 \end{pmatrix}\) After performing the multiplication, we get: \(\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix}\) So, for system (i), the solution is \((x, y, z) = (1, 1, 2)\).

Step 5: Solve system (ii) using the inverse matrix

We can now find the solutions of the system (ii) by multiplying the inverse matrix with the result matrix of system (ii). \(\begin{pmatrix} x \\ y \\ z \end{pmatrix} = A^{-1} \begin{pmatrix} 5 \\ -3 \\ -1 \end{pmatrix} = \dfrac{1}{2} \begin{pmatrix} -1 & 1 & 1 \\ 3 & -1 & -1 \\ -1 & 1 & -1 \end{pmatrix} \begin{pmatrix} 5 \\ -3 \\ -1 \end{pmatrix}\) After performing the multiplication, we get: \(\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 3 \\ 4 \\ -1 \end{pmatrix}\) So, for system (ii), the solution is \((x, y, z) = (3, 4, -1)\).

Key Concepts

Coefficient Matrix

When dealing with systems of linear equations, the coefficient matrix is a key tool that provides an organized framework for solving for unknown variables. In essence, a coefficient matrix is simply a rectangular array of numbers that represents the coefficients of the variables in the system of equations.

For example, in the context of the provided exercise, the system of equations can be expressed compactly by forming a coefficient matrix. The equations given are:

• \(x+2y+z=b_1\)
• \(x+y+z=b_2\)
• \(3x+y+z=b_3\)

Each equation corresponds to a row in the coefficient matrix, and each variable aligns with a column. Thus, the associated coefficient matrix is:\(\begin{pmatrix} 1 & 2 & 1 \ 1 & 1 & 1 \ 3 & 1 & 1 \ \end{pmatrix}\)Where the columns represent the coefficients of \(x\), \(y\), and \(z\) respectively. Working with this matrix form lets us apply sophisticated mathematical techniques, such as matrix inversion, to efficiently solve the system.

Matrix Inverse

The matrix inverse is one of the most powerful tools in linear algebra, especially when it comes to solving systems of equations. In simple terms, the inverse of a matrix \(A\), denoted as \(A^{-1}\), is a unique matrix that, when multiplied by \(A\), yields the identity matrix. The identity matrix is a special kind of square matrix with ones on the diagonal and zeros elsewhere. Mathematically, this relationship is shown by \(A \cdot A^{-1} = I\), where \(I\) represents the identity matrix.

In the context of solving a system of linear equations, as shown in the given exercise, the inverse of the coefficient matrix is used to directly find the solution vector. To do this effectively, one must ensure the coefficient matrix is not singular, meaning it has an inverse. The inverse of the given coefficient matrix was calculated to be:\(A^{-1} = \frac{1}{2} \begin{pmatrix} -1 & 1 & 1 \ 3 & -1 & -1 \ -1 & 1 & -1 \ \end{pmatrix}\)Once we have the inverse, solving the matrix equation \(A \cdot \vec{x} = \vec{b}\) for \(\vec{x}\), which represents the vector of unknowns, becomes straightforward as it simplifies to \(\vec{x} = A^{-1} \cdot \vec{b}\). This operation turns the problem of solving algebraic equations into a problem of matrix multiplication.

System of Linear Equations

A system of linear equations is a collection of one or more linear equations involving the same set of variables. The goal when solving such a system is to discover the values of these variables that satisfy all equations simultaneously.

Consider, for instance, the system provided in our exercise. This system consists of three separate equations, which visually represent planes in a three-dimensional space. The solution to the system is the point or points at which these planes intersect. In algebraic terms, we are trying to find the value of \(x\), \(y\), and \(z\) that will satisfy all three original equations concurrently.

Solving a system can be done using several methods, such as graphing, substitution, elimination, and matrix operations among others. The matrix approach, specifically the method involving the matrix-inverse as used in the exercise, is a powerful method particularly suited for systems with several equations, as it can be generalized and scaled efficiently with the use of computers.