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

Solve each system of equations using matrices (row operations). If the system has no solution, say that it is inconsistent.
$$
\left\{\begin{array}{r}
x+2 y-z=-3 \\
2 x-4 y+z=-7 \\
-2 x+2 y-3 z=4
\end{array}\right.
$$

Key Concepts

Inconsistent System

An inconsistent system is one that has no solution. When performing matrix row operations, an inconsistent system becomes evident when a row reduces to a form where all coefficients are zero but the constant term is nonzero, indicating a contradiction.

Row Operations

Row operations are the allowed manipulations on the rows of an augmented matrix—such as swapping rows, scaling a row by a nonzero constant, and adding a multiple of one row to another. These operations are fundamental to the process of Gaussian elimination and preserve the solution set of the system.

Augmented Matrix

An augmented matrix is a compact representation of a system of linear equations where the coefficients of the variables are arranged in matrix form alongside the constants from the right-hand side of the equations. This structure facilitates systematic application of algebraic methods like row operations to solve the system.

Gaussian Elimination

Gaussian elimination is a method used to solve systems of linear equations by transforming the augmented matrix into a row-echelon form using a series of row operations. Once in this form, back-substitution can be applied to efficiently solve for the variables.

Transcript

00:01 Let's look at this problem.

00:03 So how do we solve this problem? first step we do is we turn it into a matrix.

00:10 1, 2, negative 1, 2, negative 4, 1, negative 2, negative 3, and negative 7, and 4.

00:33 So let's label this is row 1, this is row 2, and this is row 3.

00:40 So what we do here is, the first step we want to do is we want to turn this entry into 0.

00:51 Okay, so we use 0 .3 at row 2 and we assign the value, assign the new sum to row 3.

01:01 Okay.

01:02 So what we have here after the role operation is 1 to negative 1, 2, 94, 1 ,000, 1.

01:15 And the 0, negative 2, negative 2, negative 3, negative 7, negative 3.

01:26 Okay, so now we successfully turn this entry into 0.

01:33 So what we can do here, we can further simplify this role.

01:37 Okay, we can turn this entry and this entry into 1.

01:42 So it is there for our further role operation.

01:47 Operation.

01:49 So here i write 1, 1.

01:52 Okay, so this negative 3 divided by negative 2 will become 3 over 2.

02:03 Okay, so the next step we do, okay, let's label this row 1, row 2, row 3.

02:09 I want to turn this negative, this 2 into 0.

02:14 Okay, so how do we do that? we use row 2, subtract track row 1 times 2 and we assign this new sum to row 2.

02:25 So after the role operation, what do we have is, okay? so this is what we get after this role operation.

02:48 Okay, so notice here we can see that this entry is turned into 0.

02:56 So next step, let's continue.

02:59 Okay, i want to turn this entry into 0.

03:04 How do we do that? so i use row 3 and times 8 plus row 2 and i assign this new sum to row 3...

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