Solve the system, if possible. 2 x-4 y+2 z =11 x+3 y-2 z =-9 4 x-2 y+z =7

Name: Problem 41, Chapter 6: College Algebra with Modeling and Visualization
Uploaded: 2021-09-01T10:44:16-08:00
Duration: 9 min 12 s
Channel: Ruchika Sangwan
Description: Solve the system, if possible. 2 x-4 y+2 z =11 x+3 y-2 z =-9 4 x-2 y+z =7

step 1 Here we are trying to solve the system of equations. So we start with making a matrix where firs

Solve the system, if possible. 2 x-4 y+2 z =11 x+3 y-2...

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Key Concepts

Row Echelon Form

Row echelon form is a simplified structure of an augmented matrix where all nonzero rows are above any rows of all zeros and each leading coefficient of a nonzero row is to the right of the leading coefficient of the row above it. Converting a system to this form simplifies the process of solving the equations.

Back Substitution

Back substitution is a technique used after reaching row echelon form in a system of equations. Starting from the bottom-most equation, which contains a single variable, one solves for that variable and then substitutes its value into the equations above. This step-by-step process continues until all variable values are found.

System of Linear Equations

A system of linear equations is a collection of equations where each equation is linear in the same set of variables. Solving the system involves finding values for the variables that satisfy all equations at the same time. This foundational concept in algebra appears in various subjects including pure math, physics, and engineering.

Gaussian Elimination

Gaussian elimination is a systematic method for solving systems of linear equations. It involves the use of row operations to transform the system’s augmented matrix into an upper triangular (or row echelon) form from which the solutions can be obtained easily through back substitution.

Transcript

00:01 Here we are trying to solve the system of equations.

00:05 So we start with making a matrix where first three columns are the coefficients of x, y and z.

00:17 1 3 minus 2 4 minus 2 1 and last column is the constant terms 11 minus 9 and 7.

00:31 To solve the system we need to get identity matrix for first three rows.

00:38 So let us start with getting a one here.

00:42 So r1 becomes r1 by 2.

00:49 So r1 by 2 is this divided by 2 divided by 2 divided by 2 and we get 1 minus 2 1 11 by 2 and all the other rows remain same minus 9 4 minus 2 1 and 7 then next we need 0's here so to get 0's here we do r2 is r2 minus r1 and r3 is r3 minus 4 times r1.

01:41 So first row remains same 1 minus 2, 11 by 2.

01:50 2.

01:51 2 2 minus 1, 3 minus minus 2 or plus 2 minus 2 minus 2 minus 1.

02:01 And minus 9 minus 11 by 2 so the second row is 0 5 minus 3 and minus 29 by 2 let us make a little more space here there you go now the last row is we need to simplify r3 minus 4 times r1, so 0, 6 minus 3 and minus 50.

02:56 So this is our matrix now.

03:00 Now we can simplify this row as divided by 3.

03:04 So let us do r3 as r3 by 3.

03:11 We get 1 minus 3.

03:13 2, 1, 11 by 2, 0 5, minus 3, minus 29 by 2, 0, 2, minus 1 and minus 5.

03:32 Now we need 0 here.

03:36 So let us add r1 plus r3...

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