The systems of linear equations are in upper triangular form. Find all solutions of each system.

step 1 They want us to solve this system of linear equations here. So what we can do, since we already

The systems of linear equations are in upper triangular...

Key Concepts

Back-Substitution

Back-substitution is a method used to solve systems in upper triangular form. Starting from the last equation, which involves a single unknown, one solves for that variable and then substitutes its value into the previous equations. This step-by-step process simplifies solving the entire set of equations.

Systems of Linear Equations

A system of linear equations is a set of equations where each is linear in the unknowns. The goal is to find values for the unknowns that satisfy all equations simultaneously. This concept forms the basis for many applications in linear algebra and is central to methods for solving simultaneous constraints on variables.

Upper Triangular Form

An upper triangular form of a system of linear equations is one in which the coefficient matrix has all zeros below its main diagonal. This structure simplifies the resolution of the system because each equation contains fewer unknowns than the previous one, making it easier to systematically solve from the bottom up.

Transcript

00:01 They want us to solve this system of linear equations here.

00:05 So what we can do, since we already have it in this upper triangular form, we can just start by solving for z and then start back substituting and then solving for each of them that way.

00:15 So the first equation is 3z is equal to negative 1.

00:20 So we would divide each side by 3, and that's going to give us z is equal to negative 1 third.

00:27 Now to solve for y, we would plug z into that.

00:30 Second equation.

00:31 So this is going to be 3y plus 2 z is equal to 1.

00:36 And now we're going to replace z with negative 1 3rd.

00:40 So let's go ahead and remove that z there.

00:43 So it'll be negative 1 3rd.

00:46 Now i don't know about y 'all, but i don't like working with fractions.

00:50 So i'm going to multiply this whole thing by 3 to get rid of that 3 there.

00:56 So this would be 9y plus so negative 1 third times 3 would just be negative 1 so actually this would be negative 2 now and then we'd have equal to 3 so now i would add 2 over so i'll give us 9 y is equal to 5 divide each side by 9 and that gives us y is equal to 5 nights and now we would plug everything into that last equation so negative 4x plus 5 y is equal to 0 and all we would need to do is plug in y for this.

01:35 So we replace y with 5 nints, and then we start solving for x again.

01:47 So again, i don't want to have to work with fractions.

01:49 So let's multiply this whole thing by 9...

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