Problem 10 - Differential Equations and Linear Algebra

Question

use elementary row operations to reduce the given matrix to row-echelon form, and hence determine the rank of each matrix. $$\left[\begin{array}{rr} 2 & 1 \\ 1 & -3 \end{array}\right]$$.

Step-by-Step Solution

Answer

The row-echelon form of the given matrix is $$\left[\begin{array}{rr} 2 & 1 \\\ 0 & -5/2 \end{array}\right]$$ Since both rows are non-zero, the rank of the matrix is 2.

Step 1: Identify the pivot element

The pivot element is the first non-zero number in the first row of the matrix. In this case, the pivot element is 2. $$\left[\begin{array}{rr} 2 & 1 \\\ 1 & -3 \end{array}\right]$$

Step 2: Perform the first elementary row operation

To get a zero below the pivot element, we can multiply the first row by -1/2 and add it to the second row. This will create a new second row. $$\left[\begin{array}{rr} 2 & 1 \\\ 0 & -5/2 \end{array}\right]$$

Step 3: Identify the next pivot element

The next pivot element is the first non-zero number in the second row. In this case, it is -5/2.

Step 4: Perform the second elementary row operation

The matrix is already in row-echelon form, as all zero rows are at the bottom, and each leading entry is to the right of the row above it.

Step 5: Determine the rank of the matrix

The rank of a matrix is the number of non-zero rows in its row-echelon form. In this case, both rows are non-zero, so the rank of the given matrix is 2.

Key Concepts

Elementary Row Operations

Elementary row operations are fundamental tools in linear algebra used to manipulate matrices. Essentially, they consist of three types of operations that can be applied to the rows of a matrix. These are:

• Row swapping: Exchanging two rows of a matrix.
• Row multiplication: Multiplying all entries in a row by a non-zero constant.
• Row addition: Adding or subtracting the multiples of one row to another.

These operations are crucial as they are used to simplify matrices into more workable forms without changing the solution set of the associated linear system. For instance, in the textbook exercise presented, we see that by multiplying the first row by a scalar and adding it to the second, we alter the matrix but keep the system's solutions intact. Understanding how to apply these operations correctly is key to progressing in algebra courses and critical for more advanced mathematical topics.

Pivot Element

A pivot element in matrix algebra serves as an anchor around which we perform row operations to simplify a matrix. It is generally chosen as the first non-zero number in a row, starting from the top-left corner and moving to the right and down. Here's what you need to know about the pivot element:

• The pivot element is central in transforming a matrix into a row-echelon form (REF).
• It's used as a reference to create zeros in the column beneath it, helping in reducing the matrix to REF.

As observed in the provided solution, the first pivot element is 2. Using it, we perform elementary row operations to form zeros beneath this pivot, leading to a new second row. The next pivot, -5/2, is then used to further refine the matrix into REF, where each pivot is to the right of the one above it, signaling the proper structuring of REF.

Matrix Rank

The rank of a matrix is a fundamental concept, referring to the maximum number of linearly independent rows or columns in the matrix. In simpler terms, it indicates the number of dimensions in which the associated vectors span. Here are some essential points on matrix rank:

• It's determined after reducing the matrix to its row-echelon form.
• The rank equals the number of pivot elements in the REF, which corresponds to the non-zero rows.

The exercise solution shows the matrix's rank as 2, signifying that we have two linearly independent rows. This information reveals the matrix's ability to span a two-dimensional space. Understanding the rank of a matrix is especially critical in solving linear systems, as it helps in deciding the solutions' nature — whether they are unique, infinite, or if there's no solution at all.