Solve each system. { 4 a+b+2 c-3 d=-16 3 a-3 b+c-4 d=-20 a-2 b-5 c-d=4 5 a

step 1 All right, so this one is challenging. So here you have four equations with four variables. So i

Solve each system. { 4 a+b+2 c-3 d=-16 3 a-3 b+c-4...

Key Concepts

Systems of Linear Equations

These consist of multiple linear equations involving the same set of variables. The goal is to find the values of the variables that satisfy all the equations simultaneously.

Matrix Representation

A system of linear equations can be expressed in matrix form using an augmented matrix, which encapsulates both the coefficients of the variables and the constants. This representation facilitates systematic operations and solution techniques.

Gaussian Elimination

This method involves performing a series of row operations on the augmented matrix to transform it into an upper triangular or row-echelon form, which simplifies the process of solving for the variables.

Back Substitution

Once a system has been transformed to an upper triangular form through elimination, back substitution is used to solve for the variables starting from the last equation and working upwards.

Uniqueness of Solutions

Analyzing the system, often through determinants or the structure of the row-reduced matrix, helps to determine whether the system has a unique solution, infinitely many solutions, or no solution.

Transcript

00:01 All right, so this one is challenging.

00:04 So here you have four equations with four variables.

00:09 So it's going to get confusing.

00:13 So what we're going to do is just slowly eliminate equations until we get down to, i mean slowly eliminate variables until we get down to one variable.

00:26 And then from there kind of work backwards.

00:30 Okay.

00:30 Okay, so what i am looking at is equation one and two.

00:41 What i can easily eliminate is b's and c's.

00:46 In my case, i'm going to eliminate the b, just because they are opposite symbols right now, and so i can just add them and i like to add.

00:56 So i'm going to multiply this by three, equation one by three.

01:01 I get 12a plus 3b plus 6c, subtract 9d, equals negative 48.

01:14 And i'm going to take equation 2 just how it is.

01:30 All right.

01:32 And i'm going to add them, like i said.

01:35 Because the bs have the same coefficient, the different symbols, they will be eliminated when i add them.

01:43 So 12 plus 3, 15a, or bs again are eliminated, 6 plus 1, 7c, negative 9 plus negative 4, negative 13, negative 48 plus negative 20, negative 68.

02:04 So, okay, we have eliminated bs.

02:08 We're going to have to take two other equations and eliminate bs as well.

02:15 So i'm going to look at my last two because one's a positive four and one's a negative two.

02:22 So i just need to multiply this by two.

02:28 So this was equation one multiplied by three here, taken with equation two.

02:41 I want to eliminate this.

02:46 So it doesn't get as confusing.

02:49 Okay.

02:49 Now i'm taking equation.

02:58 I'm going to take equation four first, just 5a, since i don't have to do anything with it.

03:05 4b, plus 3c, subtract d equals negative 10, and i'm going to take equation 3 multiplied by 2.

03:19 I get 2a, subtract 4b, subtract 10, c, subtract 10, c, subtract 2d equals 8.

03:35 Just double checking 2a, subtract 4b, check 10c, subtract 2d equals 8.

03:42 Okay, and i'm going to add these, because my b, which i wanted to focus on, same coefficient, different symbol.

03:52 5 plus 2, 7a, these are eliminated, 3 plus negative 10.

03:59 10, negative 7, negative 1, subtract 2, negative 3, negative 10 plus 8 is negative 2.

04:12 Okay, so we have eliminated b.

04:23 Now look at the equation i have here and the equation i have here.

04:27 I already have something in common.

04:29 I have the sevens.

04:30 So i'm just going to write this down here.

04:39 So i'm just rewriting our new one that we got up there.

04:45 And i'm just going to add them because i'm going to be able to now eliminate c, one step closer.

04:53 So 15 plus 722, c's negative 3 subtract 13.

05:02 It's like adding them.

05:05 Negative 2 subtracts 68 or added to negative 68 is negative 70.

05:13 Okay.

05:15 So i am down to two variables and i wish i could solve it at this point, but i can't.

05:22 Right? i need to do that whole thing, everything i just did so that i can get another equation down to two variables.

05:31 And then i can add or subtract them.

05:36 Okay, so i am moving it up.

05:41 So this one we will come back to.

05:46 Okay, the next thing i did was, and i know you can't see my original, i'll keep scrolling up.

05:53 But i took equation one up here, and this time i multiplied it by two.

06:05 Okay? because i want to then take it with equation three and eliminate the b.

06:09 So i have 8a plus 2b plus 4c, subtract 6d equals negative 32.

06:26 And i'm going to take it with equation 3.

06:38 So this is equation 1 multiplied by 2 this time, taken with equation 3.

06:44 So again, i can, just like last time, eliminate bees.

06:51 So let's do it.

06:54 So 8 plus 1, 9a, these are eliminated.

07:01 4 plus negative 5, negative 1c, negative 6 plus negative 1, 70, negative 32 plus 4, negative 28.

07:15 Okay.

07:19 The good news is that i can take this equation that i had up here and, oh, the cs aren't quite the same.

07:39 But remember, c is what i wanted to eliminate next.

07:42 So what do i need to multiply this by? right now it's a negative 1c to match it up with this positive c or 7c up there.

07:52 7.

07:54 Yes, very large numbers, but that's okay.

07:57 7 times 9, 63a, 7 times negative 1, negative 7c, 7 times negative 7 times negative 7, negative 49d, and 7 times negative 28, negative 196.

08:16 Now again, i'm going to take this equation up here, the 15a plus 7c ,000, subtract 13d equals negative 68 and i am going to add these right because this remember my second step here was i eliminated c's so if i add these i will eliminate c's again okay that's 78 a large numbers negative 49 plus a negative 13 is a 62 d negative 62d and negative 196 added to negative 68 negative 264 okay remember i said we were going to come back to this equation at some point well this is the point and so i tried to find something that they both had in common easily fortunately it didn't work very well so what i did was is i looked at the d's why did i look at the d's because they were they just seemed simpler and i wrote down right 16 plus 16 plus 16 plus 16 you know and i did the same thing with 62 and 62 in hopes to find something i also did it with the a's and didn't find anything so what i needed to do is i'm going to start with this one down here is i'm going to multiply this by four and you're thinking wait is this going to work i'll show you so when i multiplied this by four i did four times 78 and got 1 ,300, oh, sorry, 1, or 312a.

10:32 And i did 4 times 62, i got negative 4, 248 d.

10:39 And when i did 4 times negative 264, i got negative 156.

10:50 Okay...

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