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

step 1 So here we have a system of equations. 4x or x plus 4y minus 3 z equals negative 8. 3x minus y p

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

Key Concepts

System of Linear Equations

A system of linear equations is a collection of equations involving the same set of variables. The goal is to find values for the variables that satisfy all the equations simultaneously. This concept forms the basis for many problems in linear algebra and allows for both analytical and matrix-based approaches to finding solutions.

Augmented Matrix

An augmented matrix is a compact representation of a system of linear equations where each row represents an equation and the columns represent the coefficients of the variables, with an additional column for the constant terms. This matrix form facilitates the use of systematic techniques such as row operations for solving the system.

Row Operations

Row operations are methods applied to the rows of an augmented matrix in order to simplify the matrix. These operations include swapping rows, multiplying a row by a nonzero scalar, and adding a multiple of one row to another row. They are essential for reducing the matrix to a form where the solutions to the system can be easily identified.

Reduced Row Echelon Form

Reducing a matrix to row echelon form or reduced row echelon form (RREF) simplifies the process of solving the system by making the relationships between the variables clear. In RREF, every leading entry is 1 and is the only nonzero entry in its column, facilitating straightforward back-substitution to find the solutions.

Inconsistency

Inconsistency in a system is identified when the reduction process reveals a row that represents a false statement, such as 0 = k, where k is a nonzero constant. This indicates that the system has no solution, as there exists at least one equation that cannot be satisfied by any set of variable values.

Transcript

00:01 So here we have a system of equations.

00:03 4x or x plus 4y minus 3 z equals negative 8.

00:08 3x minus y plus 3z equals 12.

00:12 X plus y plus 6 z equals 1.

00:15 We want to solve for this system of equations.

00:17 And the way we're going to do that is by using matrices and raw operations on those matrices.

00:23 Now we're going to have two matrices.

00:26 One, that's going to be a 3 by 3.

00:28 We know that's a 3 by 3 because there's 3 variables and there's 3 equations and then we're going to have another matrixe for the constants that the variables are equal to so let's plug in the coefficients for the variables we have a 1 3 and 1 1 1 3 1 1 4 negative 1 1 and negative 3 and 6 and we can do a dividing line to indicate that we're drawing a different matrixy, even though we're drawing them very close to each other, and that's going to be for the constants of negative 8, 12, and 1.

01:11 So i can close up this matrixe, and we do want to get this in row echelon form, where everything on the diagonal is equal to 1, everything beneath the diagonal is equal to 0.

01:24 We can start off.

01:26 You want to get this 3 here equal to a 0, and the easiest way to do that is going to be by taking three copies of row one and subtracting or and subtracting that from row two.

01:42 Remember this is row one, this is row two, and this will be row three.

01:48 So i can go row two is equal to the old row two minus three row one.

02:01 So now you can write out row two, which is three negative one.

02:05 3 and 12, then we can subtract 3 times row 1.

02:10 Remember that negative 3 times a number, instead of adding, instead of subtracting a 3, we can add a negative 3.

02:20 So that just means that we can take a negative 3, multiply everything here times negative 3, and then we just have to add them.

02:26 So negative 3 times 1 is going to be negative 3.

02:30 Negative 3 times 4 is going to be negative 12.

02:33 Negative 3 times negative 3 is going to be 9 and negative 3 times negative 8 is going to be 24.

02:48 So i can add these values together.

02:51 We get 3 minus 3 is 0.

02:54 Negative 1 minus 12 is negative 13.

02:57 9 plus 3 is 12 and 12 plus 24 is going to be 36.

03:04 So now we can erase our old r2 because remember, this is our new r2 right here.

03:11 We can erase our old r2 and replace it with our new r2.

03:15 Okay, so we can go 0, negative 13, 12, and 36.

03:25 So now we've got that.

03:27 We can clean up our work.

03:30 And then what we want to do is remember we're getting everything beneath the diagonal equal to 0.

03:35 So both of the numbers left are in row 3, 1 in 1.

03:42 1.

03:44 So we can go row 3, we want to get this value here first to be equal to 1 or to be equal to 0.

03:54 And you kind of notice that since row 3, this position of this position are the same, we can simply take our new row 3 and set it equal to the old row 3 minus the old row 1.

04:07 Right, so now we can write that out.

04:10 You got a 1, 1, 6, 1.

04:13 Now we're subtracting row 1.

04:14 So it's a negative 1 minus 4.

04:20 Now negative 1 times negative 3 is going to be positive 3.

04:24 Negative 1 times negative 8 is going to be positive 8.

04:29 Now we can simply do the addition.

04:32 1 minus 1 is 0.

04:34 1 minus 4 is negative 3.

04:36 6 plus 3 is 9.

04:39 And 1 plus 8 is 9.

04:43 Now we can write this out.

04:46 Remember, since this is our new row 3, we can race our old row.

04:49 Rule 3, right? we'll place it with our new row 3.

04:55 9 and 9.

04:57 Okay.

04:59 Now we can erase our work, just clear up a little space.

05:04 Now we notice that we do want this number here, this fraction r3, or we want this value here, row 3, this negative 3 to be 0.

05:20 And we can't do that with row 1, because if we add subtract row 1, this 1's gonna change.

05:25 Is zero.

05:27 So we want to look for something that already has a zero in the first position, which is row two.

05:32 So now we can go and three isn't going to go into 13 evenly.

05:40 13 is not going to go into three evenly.

05:43 The best way to do this is going to, we're going to take 13 copies of row three, right? so then this value here will become 39.

05:52 So take 13 row three, and we're going to subtract three copies of row two.

06:01 Three row two.

06:02 What we want to do that is because we've got negative three here, right? let's say that's negative 39 minus negative 39, minus negative 39.

06:16 This negative and this negative cancel each other out.

06:19 So we get negative 39 plus 39.

06:22 That's equal to zero, which is what we want.

06:25 When you get more than a couple negative signs, it's always good to write it out just so you don't get confused.

06:32 Okay, so now we got that.

06:35 And i did write that wrong.

06:37 We do want three row twos, minus three row two.

06:41 Okay, so now i can write this out.

06:45 13 times three, it's 39.

06:49 So it's 13 times negative 3.

06:51 It's going to be negative 39.

06:54 9 times 13 is 117.

06:59 Oh, don't forget the 0.

07:01 0, negative 39.

07:05 So we'll clean that up a little bit.

07:07 0, negative 39, 117, and since this is also 9, this will be 117 also, 117 and 117.

07:22 Okay, so now we're going to take negative 3, multiplied times 02, negative 3 times 0 is 0, negative 3 times negative 13 is going to be positive 39, 12 times 13, or 12 times 3, sorry, 12 times 3 is going to be 36, but remember this is negative 3, so it's going to be negative 36, and 36 times negative 3 is going to be negative 108.

07:57 So now we can add these up.

07:59 Looks a little confusing, but we can do some addition.

08:04 Sorry, and it's a little shaky right now.

08:08 So you got a zero.

08:09 Plus 0 ,0, 9039 plus 39 is going to be 0.

08:15 117 minus 36 is going to be 81.

08:19 Place multiple 9 there.

08:21 And then 117 minus 108 is going to be 9.

08:30 Now remember this is our new row 3, right? so you can erase our old row 3.

08:38 And we'll place it with our new row 3, which is 0 -0 -801.

08:46 And 9.

08:48 Now remember, we've got everything beneath the diagonal equal to 0.