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

Matrix Representation

This concept refers to writing a system of linear equations in the form of a matrix, where the coefficients of the variables are organized in rows and columns. This representation simplifies the process of applying systematic operations, and it serves as the foundation for techniques like row reduction and Gaussian elimination.

Augmented Matrix

An augmented matrix is created by appending the constants from the right-hand side of the equations to the coefficient matrix. This form combines both the coefficients and the constant terms into a single matrix, which is then used to apply row operations and eliminate variables systematically.

Elementary Row Operations

Elementary row operations are the basic manipulations allowed on the rows of a matrix, including row swapping, scaling a row by a nonzero scalar, and adding a multiple of one row to another row. These operations are key to transforming the matrix into a simpler form, such as row echelon form or reduced row echelon form, without changing the solution set of the system.

Gaussian Elimination

Gaussian elimination is a systematic method of applying elementary row operations to convert an augmented matrix into an upper triangular or row echelon form, where it becomes easier to solve for the variables through back-substitution. This method is central to solving systems of linear equations and understanding the structure of the solutions.

System Consistency Analysis

System consistency analysis involves examining the reduced form of the augmented matrix to determine whether the system has a unique solution, infinitely many solutions, or no solution at all. This often includes checking for rows that represent impossible equations (such as 0 = nonzero constant), which indicate that the system is inconsistent.

Transcript

00:01 So we're given the following system of equations, 2x minus 3y minus c and equals 0, and 3x plus 2y plus 2z equals 2, and x plus 5y plus 3z equals 2.

00:14 We want to solve this system of equations by using matrices and raw operations on those matrices.

00:21 So the first thing we're going to do is obviously make a matrix.

00:26 We know it's going to be a 3x by 3 matrix because there are three variables, x, y, and z, and there are 3.

00:31 Three equations, so there'd be three rows.

00:35 Now i can fill in the coefficients.

00:37 Two, three, one, negative three, two, and five, and negative one, positive two, or just two, and three.

00:50 So a dividing line to indicate that we're drawing another matrix.

00:55 We're just drawing them close to each other so we can work with them.

00:58 This could be the matrix of constants, 0, 2, and 2.

01:05 What we want to do is get this matrix into reduced, or just simply row echelon form.

01:13 Roe echelon form is everything on the diagonal is equal to 1, and everything beneath the diagonal is equal to 0.

01:24 So you can do that.

01:25 Let's start off by reminding ourselves.

01:30 This is row 1, this is row 2, and this is row 3.

01:35 Now let's solve to get this equal to 0.

01:40 We can take two copies of row 2 and three copies of row 3 to make both of these values 6, then we can subtract them to make this 3 right here equal to 0.

01:55 Let's do that.

01:57 Capital r indicating that once we do this row calculation, we'll have our whatever the values are, and that will be equal to the capital r2, and then that capital r2 will replace this old r2.

02:13 R2, new r2, is going to be equal to 2 times r2 minus 3 times r1.

02:26 So now i can do the math.

02:29 2 times 3 is 6, 2 times 2 is 4, 2 times 2 is 4, 2 times 2 is 4, and 2 times 2 times 2 is 4, and 2 times 2 times 2.

02:40 2 is 4.

02:42 Now remember that subtracting 3 is the same thing as adding a negative 3.

02:46 So we're going to multiply everything in the first row times negative 3 and add those values.

02:51 So 2 times negative 3 is negative 6.

02:59 Negative 3 times negative 3 is positive 9.

03:03 Negative 3 times negative 1 is 3 and 3 times 0 is 3.

03:14 Times zero is going to be zero.

03:16 So now we can add these values up and then we'll get our r2 which we can replace our old r2 with our new r2.

03:23 So 6 minus 6 is 0, 4 plus 9 is 13, 4 plus 3 is 7 and 4 plus 0 is of course 4.

03:35 Now this is our new r2 so we can replace our old r2 with our new r2.

03:48 So that's 0, 13, 7, and 4.

03:56 Now we want to get every other value underneath the diagonal equal to 0.

04:05 And both of the values we have left are in row 3.

04:09 So now we're just going to work with row 3 for a little bit.

04:14 Raise that.

04:16 Remember we're working with new r3.

04:19 So we can double r3, right? so this value becomes 2.

04:23 We don't care about this.

04:24 And we know that we can subtract row 1, and then this will be a 0.

04:29 So we can do 2r3 minus r1.

04:34 So let's do the math.

04:37 1 times 2 is 2.

04:39 1 times 5 is 10.

04:42 1 or 2 times 3 is 6.

04:44 2 times 2 is 4.

04:46 And we're simply subtracting r1.

04:49 So this becomes minus 2.

04:52 This becomes adding 3.

04:54 This comes adding one.

04:57 This comes zero.

04:59 So now we can add these values together.

05:01 2 minus 2 is 0.

05:03 10 plus 3 is 13.

05:07 6 plus 1, 7.

05:08 4 plus 0 is 4.

05:10 So remember this is our new r3.

05:14 So now we can erase our old r3 and replace it with our new r3, which is 0, 13, 7, and 4.

05:27 So now we're going to notice something that these two rows right here, row 2 and row 3, they are the same.

05:42 When we have 0x plus 13y plus 7, that's equal to 4.

05:47 The same information is repeated.

05:50 So we don't really need to worry or use this row anymore.

05:54 That's just taking up time and space.

05:57 So since these are exactly the same, we can kind of just erase.

06:01 Row 3.

06:02 We don't really need it because it's the same.

06:06 Now if this was 0, 14, 8, and 5, we would still need it even though it's one -off.

06:14 Since these are exactly the same, there's no new information gain from this row.

06:20 So, i'll use this and clean it up.

06:25 So now we're going to be making a rush line form with something that's not square.

06:30 It's really simple.

06:31 It's still a diagonal here.

06:34 We want to equal to 1 .5.

06:35 And beneath the diagonal here is equal to 0.

06:39 Then we'll work with this later.

06:42 So now in order to get these values here equal to 1, we can multiply the entire row by the reciprocal...

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