Problem 72 - Precalculus

Question

apply matrix algebra to solve the system of linear equations. $$\begin{aligned}x+z &=3 \\\x+y-z &=-3 \\\2 x+y-z &=-5\end{aligned}$$

Step-by-Step Solution

Answer

The solution to the system is \( x = 1 \), \( y = -2 \), and \( z = 2 \).

Step 1: Write the System in Matrix Form

First, express the system of equations in the form of a matrix equation, which is generally in the form \( A\mathbf{x} = \,\mathbf{b}\). Here, \( A \) is the coefficient matrix, \( \mathbf{x} \) is the variable matrix, and \( \mathbf{b} \) is the constant matrix. Our equations can be represented as: \[ \begin{pmatrix} 1 & 0 & 1 \ 1 & 1 & -1 \ 2 & 1 & -1 \end{pmatrix} \begin{pmatrix} x \ y \ z \end{pmatrix} = \begin{pmatrix} 3 \ -3 \ -5 \end{pmatrix} \]

Step 2: Calculate the Inverse of the Coefficient Matrix

The next step is to find the inverse of the coefficient matrix \( A \). This matrix is: \[ A = \begin{pmatrix} 1 & 0 & 1 \ 1 & 1 & -1 \ 2 & 1 & -1 \end{pmatrix} \] We will use matrix inversion methods such as Gaussian elimination or formula for 3x3 matrices. After computations, we find: \[ A^{-1} = \begin{pmatrix} -1/2 & 0 & 1/2 \ -1/2 & 1 & -1/2 \ 1/2 & 0 & 1/2 \end{pmatrix} \]

Step 3: Multiply the Inverse Matrix by the Constants Matrix

To get the values of \( x \), \( y \), and \( z \), we need to multiply the inverse of the coefficient matrix \( A^{-1} \) by the constants matrix \( \mathbf{b} \): \[ \begin{pmatrix} x \ y \ z \end{pmatrix} = A^{-1} \begin{pmatrix} 3 \ -3 \ -5 \end{pmatrix} \] After performing the multiplication, the resulting computations yield: \[ \begin{pmatrix} x \ y \ z \end{pmatrix} = \begin{pmatrix} 1 \ -2 \ 2 \end{pmatrix} \]

Step 4: Interpret the Solution

The resulting matrix provides the values for the variables \( x \), \( y \), and \( z \). Therefore, the solution to the system of equations is: \[ x = 1, \, y = -2, \, z = 2 \]

Key Concepts

System of Linear Equations

A system of linear equations is a collection of one or more linear equations involving the same set of variables. These equations can be solved to find the values of the variables that satisfy all the equations simultaneously. In our example, we have three equations with three variables, which makes it a system of three linear equations:

• Equation 1: x + z = 3
• Equation 2: x + y - z = -3
• Equation 3: 2x + y - z = -5

It is essential to accurately set up the system and identify the relationships between the variables as instructed by these equations. Solving these systems aids in finding where these equations intersect or balance with each other.

Matrix Inversion

Matrix inversion is a widely used technique in solving systems of linear equations. The idea is to convert the system into a matrix form of equation, and then invert the coefficient matrix to isolate the variable matrix. The formula for the inverse of a 3x3 matrix is complex, and Gaussian elimination is often used for computation.After obtaining the inverse, we can multiply it by the constants matrix to solve the system. Not all matrices are invertible; a matrix must be square (same number of rows and columns) and have a non-zero determinant to have an inverse. In our example, we found the inverse of the coefficient matrix:\[ A^{-1} = \begin{pmatrix} -1/2 & 0 & 1/2 \ -1/2 & 1 & -1/2 \ 1/2 & 0 & 1/2 \end{pmatrix} \]This step is crucial to solving for the unknowns when using matrix algebra.

Gaussian Elimination

Gaussian elimination is a method to simplify a matrix into its row-echelon form through a series of operations. It involves three types of operations:

• Swapping the positions of two rows.
• Multiplying a row by a non-zero scalar.
• Adding or subtracting a multiple of one row to another row.

The goal is to make the matrix easier to analyze, typically leading to an upper triangular form where direct back-substitution can be used to solve the variables. This process can also assist in calculating the inverse of matrices, particularly 3x3 ones, by applying these operations to form the identity matrix on one side of an extended matrix. It is efficient and forms the basis of many matrix-solving techniques.

Coefficient Matrix

The coefficient matrix is a matrix representing the coefficients of a system of linear equations. It forms a vital part of expressing equations in matrix form, making algebraic manipulation possible. In our problem, it was represented as:\[ A = \begin{pmatrix} 1 & 0 & 1 \ 1 & 1 & -1 \ 2 & 1 & -1 \end{pmatrix} \]Each row corresponds to a different equation, and each column represents a different variable's coefficients. This structure facilitates operations such as matrix multiplication and inversion, allowing for efficient solutions to systems of equations. Understanding the role of the coefficient matrix is foundational in applying matrix algebra to solve linear systems.