Problem 11 - Algebra and Trigonometry

Question

A matrix is given. (a) Determine whether the matrix is in row-echelon form. (b) Determine whether the matrix is in reduced row-echelon form. (c) Write the system of equations for which the given matrix is the augmented matrix. $$ \left[\begin{array}{rrr}{1} & {0} & {-3} \\ {0} & {1} & {5}\end{array}\right] $$

Step-by-Step Solution

Answer

(a) Yes, in row-echelon form. (b) Yes, in reduced row-echelon form. (c) System: \( x = -3 \), \( y = 5 \).

Step 1: Understand Row-Echelon Form

A matrix is in row-echelon form if it satisfies the following conditions: 1. All non-zero rows are above any zero rows. 2. The leading entry of each non-zero row is 1 (known as leading 1). 3. Each leading 1 is to the right of any leading 1s above it. The given matrix \( \begin{bmatrix} 1 & 0 & -3 \ 0 & 1 & 5 \end{bmatrix} \) follows all these conditions because the first leading 1 is in the first column and the second leading 1 is in a column to the right of it. There are no zero rows.

Step 2: Evaluate Reduced Row-Echelon Form

A matrix is in reduced row-echelon form if: 1. It is in row-echelon form. 2. The leading 1 in each row is the only non-zero entry in its column. The given matrix \( \begin{bmatrix} 1 & 0 & -3 \ 0 & 1 & 5 \end{bmatrix} \) meets these criteria since each leading 1 (in respective positions of the rows) is the only non-zero entry in its column.

Step 3: Interpret Augmented Matrix

An augmented matrix can be written as a system of equations, where each row represents an equation. The given matrix \( \begin{bmatrix} 1 & 0 & -3 \ 0 & 1 & 5 \end{bmatrix} \) corresponds to the system: \( x = -3 \) and \( y = 5 \), where the matrix is of form \( \begin{bmatrix} 1 & 0 & | & -3 \ 0 & 1 & | & 5 \end{bmatrix} \), representing the equations with \( x \) and \( y \) as variables and the last column as constants.

Key Concepts

Reduced Row-Echelon Form

A matrix is considered to be in reduced row-echelon form (RREF) if it satisfies certain conditions. It not only follows the criteria for row-echelon form but has additional requirements:
1. The matrix is already in row-echelon form.
2. The leading entry (1) of each row is the only non-zero number in its column.In other words, once a matrix is simplified to have leading ones, each leading one should stand alone in its column. This means all other entries in the leading one’s column should be zeros.
For example, in the matrix: \[ \begin{bmatrix} 1 & 0 & -3 \ 0 & 1 & 5 \end{bmatrix} \] Each leading 1 (as seen in the first and second rows) is unique in its column. The rest of the entries in those columns are zero. Thus, this matrix qualifies as being in reduced row-echelon form. This refined form is especially useful for easily identifying solutions to corresponding systems of equations.

Augmented Matrix

An augmented matrix is a convenient way to represent a system of linear equations. It combines both the coefficients of the variables and the constants from the right side of the equations into one matrix. This presentation helps streamline the application of matrix operations to solve systems of equations.For the given matrix, \[ \begin{bmatrix} 1 & 0 & -3 \ 0 & 1 & 5 \end{bmatrix} \], it shows an augmented form where the last column contains the constants. Writing this as a system of equations, it translates to:

• 1st equation: \( x + 0 \cdot y = -3 \) or simply \( x = -3 \).
• 2nd equation: \( 0 \cdot x + y = 5 \) or simply \( y = 5 \).

Each row in the matrix represents a separate equation of the system, allowing us to easily read off the relationships between the variables and their assigned values.

System of Equations

A system of equations is a set of equations with multiple variables. The goal is to find values for the variables that satisfy all the equations simultaneously. In the context of matrices, each row of the augmented matrix corresponds to one equation in the system.For the matrix \[ \begin{bmatrix} 1 & 0 & -3 \ 0 & 1 & 5 \end{bmatrix} \], as an augmented matrix, the system of equations is represented as:

• \( x = -3 \)
• \( y = 5 \)

The solution to the system is straightforward because each equation directly gives the value of a variable. Typically, solving such systems involves methods like substitution, elimination, or matrix operations like those used to obtain the row-echelon or reduced row-echelon forms. In this specific case, since the matrix is already in reduced row-echelon form, the solution is immediately apparent from the equations \( x = -3 \) and \( y = 5 \). This clarity demonstrates why converting a matrix to this form can be incredibly useful when solving linear systems.