Solve the system of linear equations using Gauss-Jordan elimination. x-2 y+3 z= 5

Solve the system of linear equations using Gauss-Jordan...

Key Concepts

System of Linear Equations

A system of linear equations is a set of equations where each equation is linear in nature, involving variables that appear with only first-degree exponents. Solving such systems means finding the values of the variables that satisfy all the equations simultaneously.

Augmented Matrix

An augmented matrix is a compact way to represent a system of linear equations. It consists of the coefficients of the variables in the system along with the constants from the right-hand side of the equations, separated by a divider, simplifying the application of matrix operations for solving.

Elementary Row Operations

Elementary row operations are the basic manipulations used on the rows of an augmented matrix, including row swapping, multiplying a row by a non-zero constant, and adding or subtracting multiples of one row from another. These operations help in transforming the matrix towards a simpler form without changing the solution set.

Gauss-Jordan Elimination

Gauss-Jordan elimination is a systematic method used to solve systems of linear equations by performing elementary row operations on an augmented matrix to reduce it to reduced row-echelon form (RREF). Once in RREF, the solutions to the system can be easily read off directly.

Reduced Row-Echelon Form (RREF)

Reduced row-echelon form is a specific form of a matrix where each leading entry is 1, each leading 1 is the only nonzero entry in its column, and each leading 1 occurs to the right of the leading 1 in the row above it. This form allows for a clear and direct interpretation of the solutions to the system of equations.

Transcript

00:01 I want to figure out how to solve each of these equations using gauss jordan elimination.

00:06 So first let me make an augmented matrix.

00:09 So this is going to correspond with each coefficient.

00:12 So 1, negative 2, 3, 5, 3, 6, negative 4, negative 12, negative 1, negative 4, 6, 16.

00:30 So notice that it corresponds to each coefficient, so 1, the negative.

00:34 The three and the five matches the coefficients in this first equation and so on with the second and third equation.

00:41 So what i want to do is i want to cancel out this three and this negative one.

00:48 So what i'm going to do is i'm going to multiply the first row by negative three and then i'm going to add it to row two.

01:01 That way this three will cancel out.

01:04 So what that'll look like is negative 3, 6, negative 9, negative 15.

01:16 Then this will be 3, 6, negative 4, negative 12.

01:26 So that'll get me 0, 12, negative 13, and then negative 27.

01:34 I actually don't need to multiply the first row by scalar to cancel this negative 1 in this last row.

01:39 So i'll just keep it as it is.

01:41 1, negative 2, 3, 5.

01:44 And negative 1, negative 4, 6, 16.

01:52 This will be 0, this will be negative 6, this will be 9, and this will be 21.

02:02 So now what i'm actually gonna do is i'm gonna swap row 2 and row 3.

02:07 Otherwise i'm gonna rewrite this augmented matrix, but row 2 and row 3 will be swapped.

02:15 So this stays the same, 1, negative 2, 3, 5, and then i'm going to have 0, negative 6, 9, 21.

02:27 And this will be 0, 12, negative 13, negative 27.

02:35 And these are swapped, this second and third row.

02:40 So now i want to cancel all this 12, so i'm going to multiply this negative 6 by 2.

02:45 So the second row is going to be multiplied by 2.

02:47 So it'll be negative 12, 18, and 42.

02:55 And i guess i'll put the zeros here in 0, 12, negative 13, negative 27.

03:04 So that will give me 0, 0, 0, 45, 42 minus 27 is 15.

03:19 So now the bottom row is going to look like this.

03:22 So i'll rewrite the augmented matrix once again.

03:25 So 1, negative 2, 3, 5, 0, negative 6, 9.

03:34 21, 0, 0, and 5, 15...

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