Use Cramer’s Rule to solve each system of equations. a+9 b-2 c=2 -a-3 b+4 c=1 2 a+3 b-6 c=-5

Name: Problem 6, Chapter 4: Algebra 2
Uploaded: 2020-05-22T16:03:30-08:00
Duration: 7 min 7 s
Channel: Jessica Waltersdorff
Description: Use Cramer’s Rule to solve each system of equations. a+9 b-2 c=2 -a-3 b+4 c=1 2 a+3 b-6 c=-5

step 1 We're using Kramer's rule to solve a system with three variables. So the first thing we need to

Use Cramer’s Rule to solve each system of equations. a+9 b-2...

Key Concepts

Systems of Linear Equations

A set of equations involving the same set of variables. In such a system, a solution is a set of values that satisfy all equations simultaneously. The system is often represented in a matrix form to utilize various techniques for finding the solution.

Coefficient Matrix

This is the matrix formed by the coefficients of the variables in the system of equations. The structure of the coefficient matrix is crucial in applying methods like Cramer’s Rule, as its determinant is used to check for the uniqueness of the solution.

Determinants

A scalar value computed from a square matrix that provides important properties of the matrix, such as whether the matrix is invertible. In the context of solving systems of equations, determinants are key in assessing the existence and uniqueness of solutions.

Cramer’s Rule

A method for solving systems of linear equations where the number of equations equals the number of unknowns. It uses the determinant of the coefficient matrix and replaces individual columns with the constant terms to compute the values of the variables. It is particularly useful when the system has a unique solution indicated by a non-zero determinant.

Transcript

00:01 We're using kramer's rule to solve a system with three variables.

00:06 So the first thing we need to do is we need to figure out the determinant for the system.

00:12 So we are going to create our matrix using our coefficients.

00:16 So i have 1, negative 1, and 2 for my a's.

00:21 I have 9, negative 3 and 3 for my b's, and i have negative 2, 4, and negative 6 for my c's.

00:31 And in order to solve my matrix, i'm going to move my first two rows out here so that i can cross multiply.

00:42 All right, and i'm going to start from the bottom right corner and go up, and then i'm going to add these values and subtract them from the values that i get from the bottom left corner.

01:02 Okay, so my red values are going to be 18 plus 17.

01:09 72 plus 6, so that's going to give me 96.

01:14 And then my green values are going to be 12 plus 12 plus 54.

01:20 So that's going to give me 78.

01:23 All right, and i'm going to subtract those two values, 96 minus 78.

01:28 And i'm going to get 18.

01:30 So that's going to be my determinant for the system or my denominator when i'm solving for each variable.

01:37 All right.

01:37 So let's look at solving for a.

01:40 In order to solve for a, i have to substitute the solutions in the a column.

01:46 So when i set this back up, i'm not going to put 1, negative 1, 2 in the a column.

01:51 I'm going to put my solutions of 2, 1, and negative 5.

02:00 And then i'm going to just fill in the rest of my b coefficients and my c coefficients...

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