Row reduce the matrices in Exercises 3 and 4 to reduced echelon form. Circle the pivot positions

Row reduce the matrices in Exercises 3 and 4 to reduced...

Key Concepts

Row Reduction

Row reduction is the process of performing elementary row operations on a matrix to systematically simplify it into a form that is easier to analyze. These operations include swapping rows, multiplying a row by a nonzero scalar, and adding a multiple of one row to another row. This procedure is fundamental in solving systems of linear equations and analyzing the properties of the matrix such as its rank.

Elementary Row Operations

Elementary row operations are the basic actions performed on the rows of a matrix. They include row swapping, scalar multiplication of a row, and the replacement of a row by the sum of that row and a multiple of another row. These operations are instrumental in transforming the matrix into simpler forms like echelon or reduced row echelon form while preserving the solution set of the corresponding system of linear equations.

Row Echelon Form

A matrix is in row echelon form if it satisfies certain conditions: all nonzero rows are above any rows of all zeros, and the leading coefficient of a nonzero row is always to the right of the leading coefficient of the row above it. This form aids in back substitution when solving systems of linear equations.

Reduced Row Echelon Form

Reduced row echelon form (RREF) is a specialized form of the row echelon form in which, in addition to the echelon properties, every leading entry is 1 and is the only nonzero entry in its column. This unique form makes RREF particularly useful for identifying pivot positions and simplifying the analysis of the solution space of the system.

Pivot Positions

Pivot positions are the locations in a matrix that correspond to the leading entries (typically 1s in RREF) after the row reduction process. They are critical in determining the rank of a matrix and the independence of the rows or columns. Identifying pivot positions helps in characterizing the structure of the solutions to the associated system of linear equations.

Pivot Columns

Pivot columns are the columns in the original matrix that contain the pivot positions once the matrix has been transformed into row echelon or reduced row echelon form. These columns correspond to the variables that have leading ones in a system of equations and are essential in determining the basis for the column space of the matrix.

Transcript

00:01 When reducing matrices by reduced echelon form, we want to get it in a form such that we have ones in the leading diagonals when possible and zeros everywhere else.

00:16 So we perform row operations on each to get it reduced.

00:23 Let's go ahead and try doing that.

00:27 So our matrix is 1, 3, 5, 3, 5, 7, 5, 7, 9, 7, 9, 1.

00:44 All right, and i'm going to keep row 1 as is because there's already a 1 right in that first column of row 1.

00:54 But for row 2, rows 2 and 3, we can change it so that, alright, so this is me performing an operation on row 2.

01:06 So i'm going to say row 2 minus 3 times row 1 so we can get a 0 in that first column of row 2.

01:16 And for row 3 similarly, we're going to do row 3 minus 5 row 1 so that we can get a 0 right in that first column of row 3.

01:29 All right, and we're going to rewrite row 1 because we did not change anything.

01:32 So 1, 3, 5, 7, and these values now become whatever row 2 was minus 3 times row 1.

01:45 So we get 3 minus 3 times 1 which is 0, 5 minus 3 times 3 which is 5 minus 9.

01:56 5 minus 9 is negative 4.

02:01 7 minus 3 times 5 we get 7 minus 15 which is negative 8.

02:10 And then we have 9 minus 3 times 7 which is 9 minus 21.

02:23 9 minus 21 is negative 12.

02:27 Now for row 3 we have row 3 minus 5 times row 1.

02:33 So we have 5 minus 5, 0.

02:37 7 minus 5 times 3 is 7 minus 15.

02:42 7 minus 15 is negative 8.

02:48 Then we have 9 minus 5 times 5 which is 9 minus 25 and 9 minus 25 is negative 16.

02:58 Then we have 1 minus 5 times 7.

03:03 So we get 1 minus 35 which gives us negative 34.

03:10 Okay, and what i'm going to do is divide row 2 by 4 or by negative 4 or multiply by negative 4.

03:23 It doesn't matter which way you really say it, but it's always standard practice to say multiply by a fraction than to say divide.

03:33 So what i'm going to do is i'm going to change row 2 so that i'm multiplying row 2 by negative 1 half.

03:50 So negative 1 quarter, i'm sorry.

03:56 So negative 4 times negative 1 quarter just gives us 1.

04:04 Of course negative 1 quarter times 0 is 0.

04:06 That's why we have the 0 back there.

04:08 Negative 1 8, sorry, negative 8 times negative 1 quarter gives us 2.

04:18 Negative 12 times negative 1 quarter gives us 3.

04:25 Then we can rewrite this third row as we did not change anything here.

04:33 Negative 16, negative 34.

04:36 All right, so we can do two operations here...

Please give Ace some feedback