Problem 32 - College Algebra

Question

In Exercises \(29-32,\) write each linear system as a matrix equation in the form \(A X=B\), where \(A\) is the coefficient matrix and \(B\) is the constant matrix. $$\left\\{\begin{array}{c}x+4 y-z=3 \\\x+3 y-2 z=5 \\\2 x+7 y-5 z=12\end{array}\right.$$

Step-by-Step Solution

Answer

In matrix equation form, the system is \(A X = B\), where \(A = \begin{bmatrix} 1 & 4 & -1 \\ 1 & 3 & -2 \\ 2 & 7 & -5 \end{bmatrix}\), \(X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}\) and \(B = \begin{bmatrix} 3 \\ 5 \\ 12 \end{bmatrix}\).

Step 1: Identify Coefficient Matrix

The coefficient matrix \(A\) is composed of the coefficients of the variables. So after identifying the coefficients from each equation, you can form matrix \(A\), which would be a 3x3 matrix as we have three equaions and three variables. From the equations, the coeficients of \(x\), \(y\), and \(z\), you get \[A = \begin{bmatrix} 1 & 4 & -1 \\ 1 & 3 & -2 \\ 2 & 7 & -5 \end{bmatrix}\]

Step 2: Identify Constant Matrix

The constant matrix \(B\) consists of the constants in the system of linear equations. It is a 3x1 matrix (three rows, one column) as there are three equations. From the given equations, we have:\[B = \begin{bmatrix} 3 \\ 5 \\ 12 \end{bmatrix}\]

Step 3: Write the Matrix Equation

Finally, you form a matrix equation from matrices \(A\) and \(B\). The variable matrix \(X\) will consist of the variables from the system. This is typically written as: \[X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}\]The matrix equation \(A X = B \) hence becomes:\[\begin{bmatrix} 1 & 4 & -1 \\ 1 & 3 & -2 \\ 2 & 7 & -5 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 3 \\ 5 \\ 12 \end{bmatrix}\]

Key Concepts

Coefficient Matrix

A coefficient matrix is a key concept in linear algebra, particularly when dealing with systems of linear equations. It consists of the coefficients of the variables from the system, arranged in a rectangular array. In the context of our exercise, we're working with equations that involve the variables x, y, and z.

For the system given, with each equation structured as ax + by - cz = d, the coefficient matrix, represented by the capital letter A, is extracted by listing the coefficients in the same order they appear in the equations. Here's an illustrative breakdown:

• First equation: Coefficients of x, y, and z are 1, 4, and -1, respectively.
• Second equation: Coefficients of x, y, and z are 1, 3, and -2, respectively.
• Third equation: Coefficients of x, y, and z are 2, 7, and -5, respectively.

This forms our 3x3 matrix A, which is fundamental for transforming the system of equations into matrix format, enabling us to perform matrix operations to find the solution.

Constant Matrix

The constant matrix, denoted as B, comprises the constants from the right side of each equation in a system of linear equations. When we speak of constants, we refer to the numerical values that the equations are set equal to, without any variable attached to them.

In the given problem, the constants are 3, 5, and 12 from the respective equations. These values are then organized into a column matrix that forms matrix B:

• From the first equation: The constant is 3.
• From the second equation: The constant is 5.
• From the third equation: The constant is 12.

The structure is essential, as it preserves the order of equations, which must correspond to the order of variables and coefficients in the coefficient matrix A. This alignment is vital for the matrix equation A*X = B to correctly represent the system.

Linear Algebra

Linear algebra is an area of mathematics that is particularly concerned with vector spaces and linear mappings between these spaces. It includes the study of lines, planes, and subspaces, but is also concerned with properties common to all vector spaces.

In the realm of solving systems of equations, linear algebra introduces a powerful framework that extends far beyond the simple 2-variable cases often encountered in basic algebra. By using matrices and vectors, linear algebra allows for the systematic handling of systems with many variables and equations, providing both a visual and computational method of tackling problems.

The Matrix Factor

Through matrices, we can condense complex systems into a more manageable form, allowing us to apply various algorithms for finding solutions. The power of linear algebra lies in its ability to simplify and generalize; instead of working with individual equations, we manipulate entire matrices, applying processes like Gaussian elimination, matrix inversion, or eigendecomposition to reveal underlying patterns and solutions.

Solving Systems of Equations

Solving systems of equations is at the core of understanding the relationships between multiple linear equations. In the context of linear algebra, we aim to find values for the variables that satisfy all equations simultaneously. There are several methods to achieve this, but one of the most efficient is employing matrix equations.

Once we have our coefficient matrix A and our constant matrix B, as with the problem at hand, we can utilize various strategies to solve for the variable matrix X. For small-scale systems, this might involve straightforward methods like substitution or elimination. However, for larger and more complex systems, these methods become unwieldy.

Advantages of Matrix Methods

Matrix methods offer a streamlined approach. They allow us to use computer algorithms, which can rapidly perform calculations that would be too tedious or error-prone by hand. Among these, we have the row-reduction technique to reach reduced row echelon form, or the utilization of determinant and inverse matrix properties, if the coefficient matrix is square and invertible. By systematically applying linear algebra operations, we distill the vast possibilities down to the precise solutions, or in some cases, discover that no solution exists or that there are infinitely many solutions.