Problem 14 - 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. $$\left[\begin{array}{cc} 2 & -1 \\ 3 & 2 \\ 2 & 5 \end{array}\right]$$.

Step-by-Step Solution

Answer

The row-echelon form of the given matrix is: $$\left[\begin{array}{cc} 2 & -1 \\ 0 & \frac{7}{2} \\ 0 & 6 \end{array}\right]$$ There are no zero rows and there are leading 1s in the first two rows, so the rank of the matrix is 2.

Step 1: Get a zero in the element in the second row and first column

To get a zero in the element in the second row and first column, we can subtract 1.5 times the first row from the second row. Let's perform the operation: $$R_2 = R_2 - \frac{3}{2}R_1$$ $$\left[\begin{array}{cc} 2 & -1 \\ 0 & \frac{7}{2} \\ 2 & 5 \end{array}\right]$$

Step 2: Get a zero in the element in the third row and first column

To get a zero in the element in the third row and first column, we can subtract the first row from the third row. Let's perform the operation: $$R_3 = R_3 - R_1$$ $$\left[\begin{array}{cc} 2 & -1 \\ 0 & \frac{7}{2} \\ 0 & 6 \end{array}\right]$$

Step 3: Determine the rank of the matrix

The row-echelon form of the matrix is: $$\left[\begin{array}{cc} 2 & -1 \\ 0 & \frac{7}{2} \\ 0 & 6 \end{array}\right]$$ This matrix has a leading 1 in the first and second rows. There are no zero rows, so the rank of the matrix is the number of rows with a leading 1, which in this case is 2. Therefore, the rank of the given matrix is 2.

Key Concepts

Matrix Rank

The rank of a matrix is a fundamental concept in linear algebra that represents the maximum number of linearly independent row or column vectors in the matrix. This is crucial as it tells us about the dimensions of the vector space spanned by its rows or columns. Essentially, the rank gives us insight into the properties and potential solutions of linear systems.

• A matrix with full rank means its rank is equal to the number of its rows or columns, indicating maximum linear independence.
• A lower rank suggests the presence of linear dependency among the vectors.
• Knowing the rank helps in determining whether a system of linear equations has a single solution, no solution, or infinitely many solutions.

For our example, the matrix \[ \begin{bmatrix} 2 & -1 \ 3 & 2 \ 2 & 5 \end{bmatrix} \] has rank 2, indicating two linearly independent rows. This was determined after transforming it into row-echelon form and observing the non-zero rows.

Elementary Row Operations

Elementary row operations are the tools that allow us to manipulate matrices into row-echelon form or even reduced row-echelon form. There are three main types of these operations:

• Row Swapping (Interchanging): This involves swapping two rows within a matrix without altering the solutions of the system of equations it represents.
• Row Multiplication: You can multiply all elements of a row by a non-zero scalar, which changes the scale of the row but not its proprieties of solutions.
• Row Addition: Adding or subtracting a multiple of one row to or from another row helps create zeros and simplify the matrix.

These operations maintain the equivalence of the matrix in terms of solutions, making them incredibly useful for simplifying matrices. In the problem solution, we used row addition to create zeros below the leading ones.

Reduced Row Echelon Form

Reduced row-echelon form (RREF) takes row-echelon form a step further by not only having leading 1s in rows, but also ensuring that every leading 1 is the only non-zero entry in its column. This makes RREF a powerful matrix form, as it expresses the matrix in its simplest form.

• Every leading entry is 1, and is the only non-zero entry in its column.
• Every zero row, if present, is at the bottom of the matrix.
• Leading 1s move strictly to the right as you move down the rows.

While our problem only reached row-echelon form where some rows still had additional non-zero entries in the columns of the leading 1s, RREF would further simplify the matrix. This further simplification is not always necessary to determine the rank but is crucial for solving linear systems completely or finding a basis for a space.