Problem 45 - Precalculus with Limits

Question

In Exercises 45-48, write the matrix in row-echelon form. (Remember that the row-echelon form of a matrix is not unique.) \( \left[\begin{array}{rr} 1 & 1 & 0 & 5 \\ -2 & -1 & 2 & -10 \\ 3 & 6 & 7 & 14 \\ \end{array}\right] \)

Step-by-Step Solution

Answer

The row-echelon form of the given matrix is: \( \left[\begin{array}{rr} 1 & 0 & 0 & 5 \ 0 & 1 & 0 & 0 \ 0 & 0 & 1 & -1 \ \end{array}\right] \)

Step 1: Initial Matrix

Looking at the given matrix: \( \left[\begin{array}{rr} 1 & 1 & 0 & 5 \ -2 & -1 & 2 & -10 \ 3 & 6 & 7 & 14 \ \end{array}\right] \) the first row is already in the right form with the leading 1.

Step 2: Manipulating Second Row

Now, to get a leading 1 in the second row, it should be added with twice the first one. Now, the second row becomes: \(-2 + 2*1, -1 + 2*1, 2+2*0, -10+2*5\), thus the new matrix becomes: \( \left[\begin{array}{rr} 1 & 1 & 0 & 5 \ 0 & 1 & 2 & 0 \ 3 & 6 & 7 & 14 \ \end{array}\right] \)

Step 3: Manipulating Third Row

To obtain a leading 1 in the third row, subtract the sum of 3 times the first row and three times the second row from the third row. The new third row becomes \((3-3*1, 6-3*1-3*1, 7-3*0-3*2, 14-3*5-3*0)\), thus the new matrix becomes: \( \left[\begin{array}{rr} 1 & 1 & 0 & 5 \ 0 & 1 & 2 & 0 \ 0 & 0 & 1 & -1 \ \end{array}\right] \)

Step 4: Simplifying Remaining Rows

The second row can be simplified by subtracting first row from it, and the first row can be simplified by subtracting the second row from it. The final row-echelon form of the matrix becomes: \( \left[\begin{array}{rr} 1 & 0 & 0 & 5 \ 0 & 1 & 0 & 0 \ 0 & 0 & 1 & -1 \ \end{array}\right] \)

Key Concepts

Linear Algebra

Linear algebra is a branch of mathematics that focuses on vector spaces and linear mappings between them. It is foundational for many areas of science and engineering. In the context of solving systems of linear equations, linear algebra provides powerful methods to analyze and compute solutions efficiently. For instance, matrices, which are rectangular arrays of numbers, represent coefficients in systems of equations.

Key concepts in linear algebra include vectors, which represent quantities having both magnitude and direction, and matrices, which can be used for transformations such as rotations, scalings, and shearing of spaces.

• Linear combinations involve sums of scalars multiplied by vectors.
• Linearity refers to the property of equations or transformations preserving addition and scalar multiplication.
• Matrix operations allow manipulation and transformation of data within the matrices.

This discipline underpins modern computational methods and algorithms, such as machine learning, computer graphics, and optimization problems.

Matrix Operations

Matrix operations consist of several manipulations that can be applied to matrices, which are central to linear algebra. These include addition, subtraction, and multiplication of matrices, as well as more complex manipulations like finding the inverse or determinant of a matrix.

Addition and subtraction involve the element-wise summation or difference between two matrices of the same dimension. Multiplication is more complex, involving the sum of products of elements from rows and columns. Here are some major operations:

• Scalar multiplication, where each element of the matrix is multiplied by a constant.
• Transpose operation, flipping a matrix over its diagonal.
• Determinant operation, providing a scalar value indicating the matrix's singularity or invertibility.

An important aspect of matrix operations in linear equations is the ability to transform the coefficient matrix into a simpler form like row-echelon form. This simplifies the process of finding solutions to the system of equations it represents.

Gaussian Elimination

Gaussian elimination is a systematic method used to solve systems of linear equations. It involves using a series of algorithmic steps to transform a matrix into row-echelon form, simplifying the system and making its solutions more apparent. This is particularly useful in linear algebra for finding solutions to systems of equations.

The process follows these steps:

• Initiate with the matrix representing your system and identify the leftmost column that is non-zero.
• Use row operations - swapping, multiplying, and adding - to create leading ones and zeros below them within the columns.
• Continue reducing each subsequent row until the matrix is in row-echelon or even reduced row-echelon form.

By transforming the original matrix into a simpler form, you can easily identify solutions or the need for additional strategies to solve the system, such as back-substitution. This method is foundational for many numerical analysis techniques, providing clear pathways to identify solutions efficiently.