Problem 36 - College Algebra

Question

write each matrix equation as a system of linear equations without matrices. $$ \left[\begin{array}{rrr} -1 & 0 & 1 \\ 0 & -1 & 0 \\ 0 & 1 & 1 \end{array}\right]\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{r} -4 \\ 2 \\ 4 \end{array}\right] $$

Step-by-Step Solution

Answer

The system of linear equations formed from the given matrix equation is: \n-1x + 0y + 1z = -4,\n0x -1y + 0z = 2,\n0x + 1y + 1z = 4

Step 1: Understanding Matrix Multiplication

Matrix multiplication works on the principle of dot product. Elements of each row of the first matrix are multiplied with elements of each column of the second matrix and added together to form one element of the resultant matrix. Since there's only one column in the second matrix here (x, y, z), the resultant matrix will also have one column.

Step 2: Forming the first equation

Take the first row of the left matrix and perform multiplication with the column [x, y, z]. This means, multiply -1 by x, 0 by y, 1 by z, and summing these values up. That gives -1x + 0y + 1z = -4.

Step 3: Forming the second equation

Similarly, perform the matrix multiplication for the second row gives : 0x -1y + 0z = 2.

Step 4: Forming the third equation

Lastly, perform the matrix multiplication for the third row gives: 0x + 1y + 1z = 4.

Key Concepts

Matrix Multiplication

Understanding matrix multiplication is crucial when dealing with systems of linear equations written in matrix form. In essence, matrix multiplication involves taking the dot product of rows from the first matrix and columns from the second matrix. This operation results in a new matrix. For each element in the resulting matrix, a row from the first matrix is multiplied by a column from the second matrix, and the products are summed up.

In the given exercise, we have a 3x3 matrix multiplied by a 3x1 matrix, which ultimately results in a 3x1 matrix. Since the second matrix contains a single column (the column vector with variables x, y, and z), the computation involves:

• Taking the first row \([-1, 0, 1]\) multiplied by \[ \begin{array}{l} x \ y \ z \end{array} \] gives the first element of the result,
• The second row \([0, -1, 0]\) provides the second element,
• And the third row \([0, 1, 1]\) generates the third element.

This process converts the matrix equation into individual scalar equations, one for each matrix row, thus forming a system of equations.

Linear Algebra

Linear algebra is the branch of mathematics concerning linear equations, matrices, and vector spaces. It serves as the foundation for numerous applications in science and engineering, particularly for solving systems of linear equations like the one examined here. A key aspect to grasp in linear algebra is how matrices can represent complex linear systems compactly.

A matrix, like our example array, holds coefficients of variables (in this case, x, y, z) used in linear equations. When multiplying matrices, linear algebra allows us to derive a set of equations that are equivalent to the original matrix equation.

• The matrix equation we started with is essentially a compact and efficient way to express multiple linear equations.
• These resulting equations retain the same solutions as the matrix equation, providing clarity when attempting to solve the system.

Understanding how to move between these representations is vital in taking advantage of linear algebra to simplify solving systems.

Algebraic Equations

Algebraic equations are mathematical statements of equality featuring variables and constants. They form the building blocks of the system of equations derived from the matrix equation. In the current exercise, translating the matrix equation to a system of linear equations allows us to express the relationships between the variables x, y, and z in terms of simple algebraic expressions.

Following the steps from matrix multiplication, the linear equations extracted are:

• 1. \(-x + z = -4\)
• 2. \(-y = 2\)
• 3. \(y + z = 4\)

These algebraic equations are significantly easier to process than manipulating the original matrices. The ultimate aim is to solve for the variable values that satisfy all equations simultaneously. This process demonstrates how algebraic equations help understand the represented relationships and find meaningful solutions to complex mathematical models.