Solve each system of equations using matrices (row operations). If the system has no solution, s

Solve each system of equations using matrices (row...

Key Concepts

Row Echelon Form

Row echelon form is a structured form of an augmented matrix where each non-zero row is positioned above rows of all zero entries and the leading coefficient (pivot) of each non-zero row is to the right of the pivot in the row above it. Achieving this form is a key step in solving linear systems via elimination techniques.

Consistency of Linear Systems

The concept of consistency addresses whether a system of linear equations has at least one solution. A system is consistent if a solution exists and inconsistent if no solutions exist. Row operations can reveal inconsistency by leading to contradictory equations, such as a row that translates to an impossible statement like 0 = 1.

Gaussian Elimination

Gaussian elimination is a method used to solve systems of linear equations. It involves applying row operations to convert the augmented matrix into an upper triangular or row echelon form, which simplifies the process of back-substitution to find the solution.

System of Linear Equations

This refers to a set of equations that involve the same set of variables, where each equation is linear. The goal is to determine the values of these variables that satisfy all equations simultaneously, which can require finding unique solutions, infinitely many solutions, or determining that no solution exists.

Augmented Matrix

An augmented matrix is a compact representation of a system of linear equations that includes both the coefficients of the variables and the constant terms. This form is particularly useful for applying systematic procedures such as row operations to simplify the system.

Row Operations

Row operations are techniques applied to the rows of an augmented matrix to simplify the system of equations. These operations include swapping rows, multiplying a row by a non-zero constant, and replacing a row by the sum of itself and a multiple of another row.

Transcript

00:01 In this problem you will solve a system of linear equations by using matrices and row operations.

00:09 You are given a system that has four equations and four variables.

00:15 We will write an augmented matrix for that system and then we will go through a series of row operations to get that augmented matrix into reduced row echelon form.

00:33 And notice on the reduced row echelon form that it has the ones down the diagonal and zero is everywhere else on the left side of the vertical bar.

00:44 And then some constants on the right side of the vertical bar.

00:48 Now if your variables are x, y, z, and w, then when you have your equation in reduced row, x long form, then on the first row of that matrix, you know the x is whatever a is, y is the b value, z is the c value, and then w is the d value.

01:17 So let's take the problem that you are given and let's put it in an augmented matrix.

01:25 Looking at the first equation the coefficients of the variables our first row we'll have all one to the coefficients and then on the other side of the vertical bar we have the constant 4th okay the second equation will give us this row the third equation from the coefficients and the constant we have this third row and then for the fourth row from the fourth equation the co -efficient the co -efficient efficiency and the constant so there's our augmented matrix and our task is to go through some various row operations and transform it into this reduced row its wrong form okay i'm going to go through a series of various row operations there's not any one way to do this problem but what i like to do is if i have a one in this first slot for this column first row i want to keep it and try to make everything boil it into zeros and that's how i start the problem then i will go through and try to produce as many zeros as i can and when opportunities arise to get the ones in the slots that they need to be for this down diagonal guess i'm going to go for the zeros and then as opportunity arises i'll go for the ones all right so my first step on this problem is i'm going to produce a new row two by taking row one and adding it to row two.

03:19 Since i'm going to use row one, i'm just going to copy it over again, not changing it.

03:27 I'll take several row operations on this step and they're all going to be based off of row one.

03:34 So i'm just going to copy row one down.

03:36 Okay so if i take and just add row 1 and 2 together here's my new row 2 i'll have 0 3 2 1 and 4 keep in mind what i'm trying to do is get some zeros down this first column okay then i'm going to get a new row 3 by again taking row 1 but multiplying it times a negative 2 and add it to the present row 3 guess i'm taking row 1 times a negative 2 and i'll add it to the row 3 and this is what i will produce 0 1 negative 1 negative 3 and negative 2 all right and then i will produce a new row 4 by again taking row 1 this time time this time times a positive 2 and add it to the current row 4 so now i'm going to multiply my positive 2 and add it to row 4 and this is what it will produce 0 3 0 4 and 7 so i've got my zeros down that first column which is heading in the right direction right so the next thing i'm going to do i'm going to try to produce some more zeros so the next thing i'm going to do is i'm going to get a new row 1 and i'm going to do that by taking row 3 row 3 and multiply by a negative 1 and add that to my current row 1 since i am using row 3 i'm going to copy it over again i'm not going to change it the row that i'm using the row that i'll be multiplied by i don't change so i like to just go ahead and copy it down to remind myself i'm not changing and i'm going to use it and work with them.

05:46 Okay, so if i take the current row 3 times negative 1, and i'm going to add that to row 1.

05:53 This is what my new row 1 will look like.

05:59 1, 0, 2, 4, and 6 produced another 0.

06:09 Then i'm going to get a new row 2 by again using row 3, but this time multiplied it by negative 2 and adding that to the current row 2 okay so now i'm taking row 3 now times a negative 2 and adding it to row 2 so here's my new row 2 be 0 1 negative 1 negative back up on that back up back up me back up on that one okay that'll give me a 4 here here and a 7 and then an 8.

07:02 So i've got my 1 on that diagonal like i needed.

07:06 Okay, and then i'm going to produce a new row 4.

07:12 And what i'm going to do is i'm going to again use row 3.

07:15 This time i'm going to multiply by a negative 3 and add it to the current row 4.

07:21 So now this is going to be times a negative 3 and add the result to row 4.

07:28 So this will give me a zero so go back to that color still have a zero i will produce a zero here this will be three 13 and 13 so see i'm producing zeros as i'm moving along that's what i need to do produce as many zeros as i can okay so now the next thing i'm going to do is let's see i've got the zero in the row two i've got the zero where i wanted i've got to got the one where i wanted.

08:04 So i think what i'm going to do is i'm going to produce a new row three by taking row two times a negative one and add back to the current row three.

08:16 And that's the only thing i'm going to do in this step.

08:19 It's the only operation i will do in this step.

08:22 So i'm going to leave row one like it is.

08:25 Like i said, this is tedious and long, but have a game plan in mind and go for it.

08:31 Get those zeros.

08:32 Get the one.

08:33 Where they need to be okay now i'm going to produce a new row 3 by taking the current row 2 times a negative 1 and add 2 row 3 so that's going to produce 0 0 negative 5 negative 10 negative 10 and i will keep my current row 4 okay can you see we're getting somewhere we'll take a few more matrices but we will we'll get there.

09:11 All right, so when i look at this now, i've got all the zeros and ones where i need them to be in column one and column two.

09:20 So i'm going to have to do some work to columns three and four...

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