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

step 1 So here we're given a system of equations. X minus Y plus Z equals 4. 2x minus 3Y plus 4 Z equal

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

Key Concepts

Gaussian Elimination

Gaussian elimination is a systematic method of using row operations to reduce an augmented matrix to row echelon form. Once in row echelon form, the system can be solved by back substitution. This process is the backbone of determining whether a system has a unique solution, infinitely many solutions, or no solution.

Inconsistent Systems

An inconsistent system is one where no set of values satisfies all equations simultaneously. In the context of matrices, inconsistency is often revealed by a row that implies a false statement (such as 0 = a non-zero number) after performing row operations.

Row Operations

Row operations are the fundamental tools used in matrix methods to simplify an augmented matrix. They include swapping rows, multiplying a row by a nonzero scalar, and adding a multiple of one row to another row, all of which help in reducing the matrix to a simpler form.

Matrix Representation

The matrix representation is a compact way to express a system of linear equations. The coefficients of the variables and the constants are arranged into matrices, allowing the use of matrix operations to solve for the unknowns.

System of Linear Equations

A system of linear equations consists of two or more linear equations involving the same set of variables. Solving the system means finding the values of these variables that satisfy all the equations simultaneously.

Augmented Matrix

An augmented matrix is formed by appending the constants from the right-hand side of each equation to the coefficient matrix. This format is particularly useful for applying row operations to solve the system.

Transcript

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

00:03 X minus y plus z equals 4.

00:07 2x minus 3y plus 4 z equals negative 15, and 5x plus y minus 2z equals 12.

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

00:20 We're going to start off by filling in the matrix.

00:22 We know it's a 3x3 matrix because there's three variables and three equations.

00:27 So we can put in the coefficients of the variables.

00:30 2 and 5 for the x's, negative 1, negative 3 and 1 for the y's, and 1, 4, and 2 for the zs.

00:41 Draw the dividing line to indicate that we're drawing a different matrix, even though we're drawing them side by side.

00:49 And then we'll put the constants, negative 4, negative 15, and 12.

00:56 We close up this matrix, and now what we want to do is get this in row echelon form.

01:04 Everything on the diagonal is equal to 1, sorry, everything beneath the diagonal is equal to 0.

01:15 Okay, so we can do that.

01:18 And the first thing we want to do in order to do that is make this 2 here equal to 0.

01:27 So since that's row 2, we can write down r2, capital r2 means that that's going to be our new row 2.

01:35 And we notice that if we take row 1, multiply it times 2, and subtract it from row 2, we get 0 here.

01:42 So we can go old row 2 is equal to row 2 minus 2, row 1.

01:49 So now we can write this out.

01:51 So we don't write row 2 first.

01:53 We get 2, negative 3, 4, and negative 15.

02:00 Then we want to take 2 times row 1 and subtract it from that.

02:06 So we can go negative 2 times 1.

02:10 Is going to be negative 2.

02:12 Negative 2 times negative 1 is going to be positive 2.

02:16 Negative 2 times 1 is negative 2.

02:19 And negative 2 times negative 4 is going to be positive 8.

02:24 So now we can do the addition and subtraction.

02:28 2 minus 2 is going to be 0.

02:31 Negative 3 plus 2 is negative 1.

02:34 4 minus 2 is 2.

02:37 And negative 15 plus 8 is going to be negative 7.

02:43 Now this is our new row 2.

02:47 So we can erase our old row 2 here and replace it with our new row 2 of 0, negative 1, 2, and negative 7.

03:03 Okay, that's one step closer to getting this in row echelon form.

03:09 So now let's go through and we want to get, remember everything beneath the diagonal equal to 0.

03:17 The next step to doing that is going to be by taking this value here, this 5, and making that equal to 0.

03:26 Now just like row 1 or row 2, we notice that 1 evenly goes into 5, 5 times, so we can make our new row 3 equal to our old row 3 minus 5 times row 1.

03:43 5 row 1 3.

03:44 So row 3 is 5 -1.

03:51 Negative 2, this should be a negative 2 right here.

03:56 So remember this is what we're going off of is this equation.

04:00 Negative 2, 12, and then we have minus 5 row 1.

04:05 So negative 5 times 1 is negative 5.

04:09 Negative 5 times negative 1 is positive 5.

04:13 Negative 2 times, or 1, sorry, times negative 5 is negative 5.

04:18 And negative 5 times negative 4 is positive 20.

04:25 So now we can add these values together, and we get 5 minus 5 is 0, 1 plus 5 is 6, negative 2 minus 5 is negative 7, and 10 plus 12 plus 20 is 32.

04:52 So now we can go, this is our new r3.

04:55 We can replace our old r3 with our new r3.

05:01 We get 0, 6, negative 7, and 32.

05:09 So now we can just clear up our work just to give some room.

05:14 So now we know we've got one thing left underneath the diagonal to make equal to 0.

05:20 And we can work with row 1 again.

05:24 But if we did that, we made this 6, right? we multiply this row times six, make this negative six, we're adding a six here.

05:33 And that's going to become six.

05:35 We don't want that...