Solve each system using the inverse of the coefficient matrix. 2 x+5 y+2 z=9 4 x-7 y-3 z=7 3 x-8

Solve each system using the inverse of the coefficient...

Key Concepts

Matrix Multiplication

Matrix multiplication is an operation that allows for the combination of two matrices to produce a new matrix. In the context of solving linear systems using the inverse of the coefficient matrix, multiplying the inverse with the constant terms vector yields the solution vector containing the values of the variables.

Matrix Inverse

The inverse of a matrix is a matrix that, when multiplied with the original matrix, yields the identity matrix. If a coefficient matrix is invertible (i.e., has a nonzero determinant), its inverse can be used to solve the system of linear equations by transforming the problem into a simple matrix multiplication.

Coefficient Matrix

In a system of linear equations, the coefficient matrix is formed by the coefficients of the variables. This matrix is used to compactly represent the system and is essential when applying matrix methods, such as inversion, to solve the equations.

Systems of Linear Equations

A system of linear equations consists of multiple linear equations that are solved simultaneously. The goal is to find the values of the variables that satisfy all equations at once. This concept is fundamental in linear algebra and is applicable in various fields such as engineering, physics, and economics.

Transcript

00:01 So this problem wants us to solve this set of equations, this system of equations, using the inverse of the coefficient matrix.

00:11 So this is a coefficient matrix a.

00:14 We just grab each coefficient.

00:16 So for the first equation, we put that in the first row, second equation, second row, third row.

00:23 And then this vector x will just be this vector x, y, z.

00:28 And the vector x, y, z.

00:29 And the vector b is just the right hand side.

00:32 So we will find x by multiplying the inverse of a and b.

00:39 So let's first find the inverse of a using gauss jordan elimination.

00:45 So first we make the augmented matrix with the identity matrix at the end of this a.

00:52 And then now we're going to eliminate with gauss jordan.

00:55 So first let's make the first pivot here, equal to one by dividing the whole row by 2 and then copy over the rest.

01:14 Now let's make the second entries of the first column equal to 0.

01:22 So we're illuminating the first column.

01:32 So row 2 will be subtracting 4 of row 1, and that gives us a 0 here.

01:44 This will be negative 17, and here we'll have a negative 7.

01:53 Negative 2, a 1, and a 0, and then row 3 will be itself minus 3 times row 1.

02:02 So that will be a 0 here, a negative 31 over 2, and then negative 5, then a negative 3 halves, a 0, and a 1.

02:15 So we eliminated the first column.

02:18 Now moving on to the second pivot, we want this to be a 1, so we divide the second row.

02:24 By negative 17.

02:47 Okay, now we're going to eliminate the second column by row one will be itself minus five halves of row two and row three will be itself plus 31 halves of row two.

03:23 Sorry, let me move this to be for the next matrix.

03:27 Okay.

03:30 So this will be be a 1 this will be a 0 this will be a negative 1 over 34 here we'll have a 7 over 34 and a 5 over 3 4 2nd 2nd row is unchanged here we'll have a 0 here we'll have 47 over 34 11 over 34 11 over 34 negative 31 over 34 and the one is unchanged.

04:17 Okay, finally for the last column, we want to divide the third row by its pivot.

04:33 So we're dividing by 47 over 34...

Please give Ace some feedback