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

step 1 Okay, let's solve this problem. So the first thing you do is you turn it into a matrix, and then

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

Key Concepts

Consistency of Systems

The consistency of a system of equations indicates whether there exists at least one solution that satisfies all equations simultaneously. By analyzing the row-reduced form of a matrix, one can determine if the system is consistent, meaning it has at least one solution, or inconsistent, meaning no such solution exists. This concept is crucial for validating the feasibility of a modeled situation.

Gaussian Elimination

Gaussian elimination is a method that uses row operations to reduce a matrix to row echelon form, from which back-substitution can be used to efficiently solve for the variables. This method systematically eliminates variables to simplify the system and determine if the system has a unique solution, infinitely many solutions, or no solution.

Row Operations

Row operations involve manipulating the rows of a matrix in order to simplify the system of equations. These operations, which include swapping rows, multiplying a row by a nonzero scalar, and adding or subtracting multiples of rows from each other, are essential in reducing a matrix to row echelon or reduced row echelon form, thus facilitating the solution process.

Matrix Representation

Representing systems of linear equations in matrix form allows for a compact and organized approach to solving the system. The coefficients of the variables can be organized into a coefficient matrix, while constants form a vector. This form is integral in using systematic techniques such as row operations and computer algorithms to solve large systems efficiently.

Systems of Linear Equations

A system of linear equations consists of multiple equations with several variables, where each equation represents a linear relationship among the variables. The goal is to find the set of variable values that satisfy all equations at once. This concept is foundational in algebra and applies to various fields requiring the modeling of real-world phenomena with linear relationships.

Transcript

00:01 Okay, let's solve this problem.

00:02 So the first thing you do is you turn it into a matrix, and then you use the role operation to solve it.

00:12 So first we turn it into a matrix.

00:15 1 -1 -9 -1 -3, negative 2, and 1.

00:23 1 -3 -negative 2.

00:27 Okay.

00:30 And here are 6, negative 5, 14.

00:43 Okay, so the next step we do is we label this is row 1, this is row 2, and this is row 3, okay? so the next step we do is to use the role operations to cancel some entries, okay? so the first step we do is we use the row 3 minus row 1, okay? and then we set this new value to row 3.

01:20 So we have 1.

01:37 So this is a result we get after the first row operations, right? so we use the third row to minus first row, and then we set the new value to the third row.

01:53 This is negative 1.

01:56 And the next step we do is we continue to label.

01:59 This is row 1, row 2, row 3.

02:03 And this time what we do is we use the we use row 2 or minus 3 times for 1 and we set this new value to row 2.

02:19 After this role operation, what we have is, okay, so this is what we have after this role operation.

02:43 So let's continue to label this row 1, this is row 2, and this is the row 3.

02:48 So next step we want to do.

02:50 Okay, so since now we have.

02:52 You know, successfully turn this two entries into zero.

02:57 So next step we want to do is we want to turn this entries into zero.

03:04 What do we do here? we use row 2 plus 2 and 0 2 times 2 plus row 3 times 5.

03:19 We add this together and we assign this to row 3.

03:24 Okay, so after this row operation, what we have is 1 -191 -90 -5 -4 and 0 -0 -903, negative 3, and 6, negative 23, and here we have 24.

03:48 So notice here, since we only have one entry in this row, okay, negative 3 times this variable, times this variable equals to 24, okay? we can divide negative 3 at this role, and we can change this term into 1, divided by negative 3, okay? so it's become 1, and so this entry, this term becomes negative 8, negative 8, okay? after this rule operation, the next step we want to do is we want to turn this entry, turn this entry into zero.

04:37 Okay, so we're getting closer, closer to the row at the floor, right? so how do we turn this term into zero? this entry into zero...

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