Problem 40 - Precalculus Mathematics for Calculus

Question

Write the system of equations as a matrix equation. $$\left\\{\begin{aligned} 6 x-y+z &=12 \\ 2 x+z &=7 \\ y-2 z &=4 \end{aligned}\right.$$

Step-by-Step Solution

Answer

Matrix equation is \(AX = B\) with \(A = \begin{bmatrix} 6 & -1 & 1 \\ 2 & 0 & 1 \\ 0 & 1 & -2 \end{bmatrix}\) and \(B = \begin{bmatrix} 12 \\ 7 \\ 4 \end{bmatrix}\).

Step 1: Express the Linear Equations in Standard Form

The given system of equations are already in standard form:\[ 6x - y + z = 12 \]\[ 2x + 0y + z = 7 \]\[ 0x + y - 2z = 4 \]

Step 2: Identify Coefficient Matrix

To form the coefficient matrix, extract the coefficients of the variables \(x\), \(y\), and \(z\) from each equation. The coefficients form a matrix:\[A = \begin{bmatrix} 6 & -1 & 1 \ 2 & 0 & 1 \ 0 & 1 & -2 \end{bmatrix}\]

Step 3: Form the Variable Matrix

Represent the variables of the system \(x\), \(y\), and \(z\) as a matrix. This gives us:\[X = \begin{bmatrix} x \ y \ z \end{bmatrix}\]

Step 4: Identify the Constant Matrix

Identify the constants from the right-hand side of each equation, which forms another matrix:\[B = \begin{bmatrix} 12 \ 7 \ 4 \end{bmatrix}\]

Step 5: Write the Matrix Equation

Combine the matrices to form the matrix equation that represents the given system of equations. The equation can be written as:\[AX = B\]• Coefficient matrix \(A\): \( \begin{bmatrix} 6 & -1 & 1 \ 2 & 0 & 1 \ 0 & 1 & -2 \end{bmatrix} \)• Variable matrix \(X\): \( \begin{bmatrix} x \ y \ z \end{bmatrix} \)• Constant matrix \(B\): \( \begin{bmatrix} 12 \ 7 \ 4 \end{bmatrix} \)

Key Concepts

System of Equations

A system of equations is a collection of two or more equations with the same set of unknowns. In mathematics, solving these systems means finding values for the unknowns that will satisfy all the equations simultaneously. This concept appears frequently in various fields, like physics and economics, where multiple constraints must be satisfied at the same time.
For example, consider the system:

• \(6x - y + z = 12\)
• \(2x + z = 7\)
• \(y - 2z = 4\)

This system involves three equations with three unknowns \(x\), \(y\), and \(z\). The goal is to find the values of \(x\), \(y\), and \(z\) that make all three equations true simultaneously.

Coefficient Matrix

The coefficient matrix is derived directly from the coefficients of the unknowns in each equation of the system. Forming the coefficient matrix is a crucial step when solving a system of equations using matrix methods. It simplifies the problem by representing the system in a compact form.
For the given system, the coefficient matrix \(A\) is:

• First row: coefficients from \(6x - y + z = 12\) which are \([6, -1, 1]\)
• Second row: coefficients from \(2x + z = 7\) which are \([2, 0, 1]\)
• Third row: coefficients from \(y - 2z = 4\) which are \([0, 1, -2]\)

Thus, the coefficient matrix \(A\) is:\[A = \begin{bmatrix} 6 & -1 & 1 \ 2 & 0 & 1 \ 0 & 1 & -2 \end{bmatrix}\]The matrix effectively encapsulates all the information needed about the relationships between variables in the system.

Variable Matrix

The variable matrix is used to encapsulate all the unknown quantities we aim to solve for in a neat and ordered form. In the context of our system, the unknowns are \(x\), \(y\), and \(z\).
This matrix is basically a vertical list of these variables. For our example, the variable matrix \(X\) will be:\[X = \begin{bmatrix} x \ y \ z \end{bmatrix}\]By representing the unknowns in matrix form, it allows us to express the system of equations succinctly when combined with the coefficient matrix and constant matrix.
This approach is particularly advantageous for computations, especially when dealing with large systems.

Constant Matrix

The constant matrix represents the constants or the numbers that come after the equals sign in each equation of the system. This matrix contains the results when the system of equations is lined up and considered equal to zero on one side.
For our system, the constant matrix \(B\) is formed by taking:

• First equation: \(12\)
• Second equation: \(7\)
• Third equation: \(4\)

Thus, the constant matrix \(B\) is:\[B = \begin{bmatrix} 12 \ 7 \ 4 \end{bmatrix}\]This matrix plays a key role when using matrix equations to find solutions to the system. By expressing all constants in this form, we can easily apply techniques such as Gaussian elimination or using the inverse matrix to solve for the unknowns.