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

step 1 So here we're given the following system of equations. 2x plus y minus 3 z equals 0, negative 2x

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

Key Concepts

Inconsistency in Linear Systems

Inconsistency occurs when a system of linear equations has no solution. In the context of matrices, a system is identified as inconsistent when the augmented matrix contains a row with all zero coefficients and a non-zero constant, which represents a contradictory equation.

Gaussian Elimination

Gaussian elimination is an algorithm that uses row operations to systematically reduce an augmented matrix to an upper triangular (row-echelon) form. This process simplifies the solving of systems of equations by allowing back-substitution to find the values of the variables.

Row Operations

Row operations consist of interchanging rows, multiplying a row by a nonzero scalar, and adding a multiple of one row to another row. These operations are fundamental for transforming an augmented matrix into a simpler form, such as row-echelon or reduced row-echelon form, to reveal the solutions of the system.

Augmented Matrix

An augmented matrix combines both the coefficient matrix and the constants from the equations into one larger matrix. This streamlined representation is crucial for applying systematic row operations to simplify and solve the system of equations efficiently.

Systems of Linear Equations

This concept involves a collection of linear equations with multiple variables where the goal is to find the set of values that satisfy all equations simultaneously. In algebra, understanding such systems is essential as they model numerous real-world situations and provide a foundation for further study in linear algebra and applied mathematics.

Matrices

Matrices are rectangular arrays of numbers or symbols arranged in rows and columns. They serve as a compact way to represent systems of linear equations, facilitating operations, computations, and transformations that are pivotal in solving these systems.

Transcript

00:01 So here we're given the following system of equations.

00:03 2x plus y minus 3 z equals 0, negative 2x plus 2y, plus z equals negative 7, and 3x minus 4.

00:14 Whoops, that should be a y, minus 3 z equals 7.

00:22 So we're going to solve this system of equation by using matrices and raw operations.

00:28 We know we have a 3x3 matrix because there are three variables, x, y, and z, and there are three rows because there are three equations.

00:39 We'll plug the coefficients in, 2, negative 2, 3, 1, 2, negative 4, negative 3, 1, and negative 3.

00:53 We'll do the dividing line to indicate this is a different matrix.

00:56 We're just drawing them close to each other.

00:58 This is where we'll put the early consonants of 0, negative.

01:03 7 and 7.

01:07 So now we want to get this into row echelon form.

01:11 If you don't remember everything, row echelon form is where everything beneath the diagonal is equal to 0.

01:17 Everything on the diagonal is equal to 1.

01:23 So we want to start out and we can label these rows first.

01:28 This will be row 1.

01:29 This will be row 2 and this will be row 3.

01:32 We notice that we want to get this to be 0.

01:36 We can get that just by adding row 1 and row 2 together.

01:40 So we'll take our new row 2 indicated by capital r is equal to row 2 plus row 1.

01:50 So we can do the math.

01:54 Negative 2, 2, 1, negative 7, and 2, 1, negative 3, 0.

02:05 Add these together, and we get 03, and we get 0, 3 ,000, and we get 0 ,000, and negative 7.

02:14 Since this is our new r2, we can erase our old r2 and replace it with our new r2 of 0.

02:24 Let's draw that a little bit near, so we don't think it's a 6.

02:27 0, 3, negative 2, and negative 7.

02:33 Now we can erase the work right here just to give us a little more room.

02:37 And now we want to get our 3, then 3 and the negative 4 equal to 0.

02:43 So that we have everything underneath the diagonal equal to zero.

02:48 First step we're going to do is we can take r1 and manipulate it with r3 to get this value here, this 3, equal to zero.

03:01 We notice that if we take new r3 and set that equal to 2 times r3, because this will be 6, 3 times r1, and this will be 6.

03:13 We subtract those, and this value here will be zero.

03:17 So we can take 2r3 minus 3r1.

03:26 Now we can take this r3, multiply it times 2.

03:30 So we get 6, negative 8, negative 6, and 14.

03:37 And then we are subtracting negative 3.

03:40 Remember that adding negative 3 is the same thing as subtracting 3.

03:45 So what we're going to do is we're going to multiply each value times negative 3 and then add them.

03:52 So negative 3 times 2 is negative 6.

03:56 Negative 3 times 1 is negative 3.

03:59 Negative 3 times negative 3 is positive 9 and negative 3 times 0 is 0.

04:05 So now we can add these values together.

04:09 6 plus negative 6 is 0.

04:12 Negative 8 plus negative 3 is negative 11.

04:15 Negative 6 plus 9 is positive 3.

04:22 14 plus 0 is 14.

04:26 Now we know that this is our new r3.

04:31 We can erase our old r3 here and replace it with our new r3, which is 0, negative 11, 3, 14.

04:49 Now we can erase all the work we did here, and we're going to try and get this negative 11 to be equal to 0.

04:59 Now you may be thinking we can use row 1 again.

05:02 That would be wrong because any operations we do to roll 1, we're going to have to add or subtract them to row 3.

05:10 And this 2 is going to change this 0 into being a 0.

05:15 So what we're going to do is we're going to take row 2 because this here is a 0.

05:19 And that way these 0s, no matter what we do to them, add and subtract them multiply them times 100, they're still going to be 0.

05:27 So now we can take r3.

05:30 Take three copies of r3 so this value here becomes 33 11 copies of r2 so this value becomes 33 we can add those values together and this negative 11 will become 0 we take 3 r3 plus 11 r2 so we can add these together 3 r3 is going to be 0 negative 30 3 9 and 14 times 3 is 42.

06:07 Now we can take 11 times r2, 11 times 0 is 0, 11 times 3 is 33, negative 2 times 11 is negative 22, negative 7 times 11 is negative 77.

06:26 Then we can add these values like so.

06:31 0 plus 0 is 0, negative 33 plus 33 is 33 is 30.

06:37 There is 0, 9 minus 22 is 13, is 13, negative 13.

06:50 Because the 9 is smaller than the 22, so the negative comes like that.

06:54 The 42 minus 77 is negative 35.

07:00 Now this is our new r3.

07:05 We can erase, we have here, we can put our old, our new r3 in here.

07:16 3, negative 13, negative 35.

07:24 Now we've got this, r3, and we've got these values here, equal to 0, all the values underneath the diagonal.

07:33 So now what we want to do is set the values on the diagonal equal to 1.

07:38 The way we're going to do that is by taking the value on the diagonal like 2, 3, or negative 13, multiplying the entire row by its reciprocal.

07:49 So let's do it like this.

07:51 This.

07:53 You can go and r1, the new r1 is going to be equal to 1 half r1.

08:04 This 2 will become a 1 and we don't really care about whatever else is left.

08:10 We just want everything on the diagonal to be equal to 1.

08:14 So 2 times 1 half is 1.

08:19 1 times 1 half is 1 half and negative 3 times 1 half, our net is negative 3 halves.

08:33 And since this is 0, we can just leave it as a 0.

08:38 So now we can take r2 and we want to reduce that...

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