Solve each system using the inverse of the coefficient matrix. 2 x+3 y+3 z =1 x+4 y+3 z

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

Key Concepts

System of Linear Equations

A system of linear equations consists of multiple linear equations involving the same set of variables. The objective is to find values for these variables that satisfy all of the equations simultaneously. In the context of linear algebra, these systems can be efficiently handled using matrix techniques.

Coefficient Matrix

The coefficient matrix is formed by the coefficients of the variables in the system of linear equations. It encapsulates all the linear relationships in a compact form, allowing for systematic manipulation and analysis of the system using matrix operations.

Inverse of a Matrix

The inverse of a matrix is a key concept in linear algebra. For a given square matrix, if an inverse exists, it is the unique matrix that when multiplied by the original matrix yields the identity matrix. The existence of an inverse indicates that the matrix (and hence the system it represents) is non-singular, meaning it has a unique solution.

Solving Systems Using Matrix Inversion

When the coefficient matrix of a system of linear equations is invertible, the unique solution of the system can be found by multiplying the inverse of the coefficient matrix by the constant terms vector. This method leverages the property that the product of a matrix and its inverse is the identity matrix, effectively isolating the variables.

Transcript

00:01 In this problem, we're asked to find the solution to this system of equation using the inverse of the coefficient matrix.

00:12 So in order to do that, the first thing we need to do is rewrite this system of equations in the matrix form.

00:19 So in matrix a, it's going to contain all the coefficients.

00:23 So my coefficients are 2, 3, 3, 1, 4, 5, 1, 3, 4.

00:33 And then in matrix x, it's going to contain all the variables, x, y, and z.

00:45 And then in matrix b, it's going to be the answer, 1 ,0, negative 1.

00:56 So we already know that if we multiply matrix a times matrix x, that would give me the matrix b.

01:11 And to solve for x, we know, to solve for matrix x, we know that x is going to equal to inverse of a times matrix b.

01:27 So for this problem, you have already found out what matrix, what inverse of matrix is from problem 25.

01:41 So if you don't know how to find the matrix inverse of matrix a, look at problem 25 video.

01:48 So we know that, so therefore we have inverse of matrix a times b will be equal to.

01:59 So the inverse of matrix a is this.

02:02 Positive 7 .8, negative 3 eighth, negative 3 eighth, negative 1 8, 5 eighths, negative 3 eighth, negative 1 eighth, negative 3 eighth, and 5 eighth.

02:23 And then you're going to multiply the inverse of the matrix a times matrix b, which is 1 ,0, negative 1.

02:36 And we know that to multiply the matrix, what we do is that we take each row in the first matrix, multiplied by the first column in the second matrix.

02:51 So therefore, what this means is that i'm going to take the first row from inverse of 8.

02:56 So that's going to be 7 eighth times 1 minus 3 eighth times 0 minus 3 eighth times negative 1...

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