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

step 1 Here we're given a system of equations that we want to solve. The way we're going to solve it is

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

Key Concepts

System of Linear Equations

A system of linear equations consists of multiple linear equations that are solved simultaneously to find common values for the variables involved. Each equation represents a linear relationship, and the solution is the set of values that satisfies all equations concurrently.

Matrix Representation

Systems of linear equations can be represented in matrix form by identifying a coefficient matrix and an augmented matrix. This representation organizes the coefficients and constants systematically, allowing for the application of matrix operations to solve the system.

Augmented Matrix

An augmented matrix is formed by appending the constants from the right-hand side of the equations to the coefficient matrix. This provides a compact form of the system, which can be manipulated using row operations to reveal the solutions.

Row Operations

Row operations are used in the process of matrix reduction to solve systems of linear equations. These operations, which include swapping rows, scaling rows, and combining rows, simplify the augmented matrix into a form where the solutions can be easily determined.

Consistent and Inconsistent Systems

A consistent system is one that has at least one solution, either unique or infinitely many, while an inconsistent system has no solutions. The application of row operations to the augmented matrix can help reveal whether a system is consistent or inconsistent.

Transcript

00:01 Here we're given a system of equations that we want to solve.

00:03 The way we're going to solve it is by transforming these equations into a matrix and then performing a row operations on that matrix.

00:12 First, let's transform this system of equations into a matrix.

00:17 Since there are two variables, x and y, there are going to be two columns in the matrix.

00:22 And since there are two equations, there are going to be two rows.

00:26 So we take the coefficient of the first x here, put it here.

00:31 Then the coefficient of the y, there.

00:35 Same with here, the coefficient of x, coefficient of the y.

00:39 Do this line, which indicates that we're having two matrices basically, and then we put in the constants of five and three.

00:51 Now in order to solve this using row operations, we want to get into reduced echelon form.

00:58 Reduced row echelon form is when all the numbers on the diagonal are one.

01:03 Which they already are, and all the numbers below the diagonal are 0.

01:09 So we just have to get this equal to 0, then we'll have it in reduced row echelon form.

01:15 Now let's see, in order to row 2 into reduced row echelon form, we want to get this into a 0.

01:24 So we can do that by taking r2 minus r1.

01:30 This will cancel this out and we'll get a zero into this flat.

01:34 So we can do r2 minus r1 equals capital r2.

01:40 That just reminds us that when we're done with this equation, we plug this value back into r2.

01:46 Now you can do the subtraction.

01:49 R2 minus r1.

01:50 We get 1, 1, 3, minus 1, 2.

02:00 So now you can do the subtraction.

02:02 1 minus 1 is 0...

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