Reduce the given matrix to reduced rowechelon form and hence determine the rank of each matrix.

Reduce the given matrix to reduced rowechelon form and hence...

Key Concepts

Reduced Row Echelon Form

Reduced Row Echelon Form (RREF) is a specific form of a matrix where every nonzero row has a leading 1, and each leading 1 is the only nonzero entry in its column. This form simplifies the process of solving systems of linear equations and analyzing the structure of the matrix, as it clearly reveals the pivot positions and the dependencies among rows.

Elementary Row Operations

Elementary Row Operations are the fundamental techniques used to manipulate matrices, including row swapping, row multiplication by a nonzero scalar, and row addition or subtraction. These operations preserve the solution set of a related system of equations and are essential for transforming a matrix into its row echelon or reduced row echelon form.

Pivot Position

A pivot position in a matrix is the location corresponding to the leading 1 in a row of its reduced row echelon form. Identifying pivot positions is crucial because they indicate the linearly independent rows of the matrix and help in determining the rank, as well as solving systems of equations.

Matrix Rank

The rank of a matrix is defined as the number of linearly independent rows or columns in the matrix. It is determined by counting the pivot positions in its row echelon form. The rank is a fundamental concept in linear algebra as it provides insights into the solution space of linear systems and characterizes properties such as invertibility and dimensionality.

Transcript

00:01 Okay, so what we have here is a 4x4 matrix, where we are asked to find its reduced row echelon form and its rank.

00:11 So based from our previous lessons, we need to convert this first with its raw echelon form to determine its rank and do further operations to convert it to its reduced row echelon form.

00:21 So reduced row echelon form uses the same elementary row operations with raw echelon form, but has different criteria.

00:29 Yeah.

00:30 So we say that a matrix is in its reduced row, et cetera, and form if it satisfies the four requirements.

00:38 First, the first non -zero number and the first row is the number one.

00:42 Second, the second row also starts with the number one, which is further to the right than the leading entry in the first row.

00:51 So for every subsequent row, the number one must be further to the right.

00:56 So it should look like this resembling a staircase.

01:01 Okay, so third, the leading entry in each row must be the only non -zero in its column, and finally, any non -zero rows are placed at the bottom of the matrix.

01:13 So we can do this by doing series of elementary row operations, including interchanging one row with another, multiplying one row by a non -zero constant, and replacing one row with one row plus a constant times another row.

01:27 So for this example, what we're going to do first is to to transform it into its row echelon form.

01:35 So our first step is to swap r1 with r4.

01:49 So by doing that, we now have 4 negative 2, 38, 3, negative 2, 07, 2, negative 1, 07, 2, negative 1, 2, 4, and 1, negative 1, negative 1, 2.

02:16 Next, we want to cancel the leading coefficient in r2, and we do this by doing r2, we do this by doing r2 minus 3 4ths of r1.

02:34 By doing that, we now have 4, negative 2, 3 ,000.

02:43 8, 0, negative 1 half, negative 9 over 4, 1, 2, negative 1, 2 ,000, 2, 4, and 1, negative 1, negative 1, 2.

03:01 Next, we want to cancel the leading coefficient in r3, and we do this by doing r3 minus 1⁄2.

03:15 With that, we now have 4, negative 2, 3, 8, 0, negative 1 half, negative 9 over 4, 1, 0, 0, 0, 1 half, and finally, 1, negative 1, 2.

03:43 For our fourth step, i hope you're still following, what we want to do now is to cancel the leading coefficient in r4.

04:02 So we do this by doing r4 minus 1 fourth of r1.

04:12 Now we have 4, negative 2, 3, 8, 0, negative 1 1⁄2, negative 1⁄2, negative 9 over 4, and 1 ,000, 0, 1 half, 0, and then 0, negative 1 half, negative 1⁄2, and then 0.

04:45 Next again, we want to cancel the leading coefficient of r4, which is negative 1⁄2.

04:51 So what we're going to do now is r4 minus r2.

04:59 And with that, we now have 4, negative 2, 3, 8, 0, negative 1 half, negative 9 over 4, 1, 0, 0, 1⁄2 1⁄2, and then 0, 0, 0, 1⁄2, negative 1.

05:30 Our sixth step is to again cancel the leading coefficient in row 4.

05:38 So by canceling that, this will satisfy one of our criteria in row echelum form.

05:44 So we do this now by r4, by doing r4 minus r3.

05:58 Now we have 4, negative 2, 3, 8, 0, negative 1⁄2, negative 1⁄2, negative 1⁄2, negative 9 over 4, 1, 0, 0, 1 half, 0, then 0, 0, 0, 0, negative 1.

06:25 So this is already on its row echelon form, and what we want to do now is to get its rank.

06:35 So by definition, the rank is the maximum number of linearly independent vector in a matrix.

06:41 So this is equal to the number of non -zero rows in its row echelon matrix.

06:47 So if you are to look at our example, we have 1, 2, and 3 rows with non -zero element, which is row 1, row 2, and row 3.

06:56 Therefore, our rank is 3.

07:00 So we're not quite done yet since we need to get the reduced row echelon form of our matrix...

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