Chapter 6 Matrices, Gaussian Elimination, and Matrix Inverses Chapter 6 of the textbook covers a number of topics, but we shall only cover those topics related to matrices. We shall present matrices within the context of linear equations. The discussion will be limited to 3-by-3 matrices, as smaller matrices are boring and larger matrices are a bit too messy for our discussions. Consider the following system of linear equations in three variables. 2 X Y + Z = 1 4 X + Y Z = 5 X + Y + Z = 0 As we have learned in high-school algebra, the variables X, Y, and Z are just &quot;place holders&quot; in that the real structure of the problem is contained in the numbers. One use of matrices is to focus on this structure without attending to the variables. There are two matrices associated with this problem, the 3-by-3 matrix of coefficients and the 3-by-4 augmented matrix. We show these two matrices just below. Coefficient Matrix 2 -1 1 4 1 -1 1 1 1 Augmented Matrix 2 -1 1 4 1 -1 1 1 1 1 5 0 There are a number of ways to solve this set of equations, but first it is important to discuss what the mathematicians call <a href="/keyword/elementary-row-operations/" >elementary row operations</a> . The student should note that there are also elementary column operations, but that these are of less use in our studies. There are three <a href="/keyword/elementary-row-operations/" >elementary row operations</a> . 1) Exchanging two rows in a matrix, and 2) Multiplying a row by a non-zero constant, and 3) Replacing a row by the sum of that row and a non-zero multiple of another row. Note that the third operation specifies the non-zero product of a row, as adding a zero product of one row to another row does not change the matrix. The insight of the method of <a href="/keyword/elementary-row-operations/" >elementary row operations</a> is that in some sense they do not change the matrix. For example, the solution to a set of linear equations does not depend on what equation is written first; thus swapping rows should not change the solution. Before working with this sample matrix and its associated set of equations, let us consider two related sets of equations. Each set of equations is related to the example above. The first set of equations is shown below. 2 X Y + Z = 1 2 X + 2 Y + 2 Z = 8 X + Y + Z = 4 Examine the augmented matrix for the system. 2 -1 1 1 2 2 2 8 1 1 1 4 Perform the elementary row operation: (Row 2) 2 (Row 3) to get the following. 2 -1 1 1 0 0 0 0 1 1 1 4 This merely emphasizes the fact that the second equation is just a multiple of the third equation and does not convey any additional information. Thus, all we really have is the following set of two equations in three variables the system is &quot;under-determined&quot;. 2 X Y + X + Y + Z = 1 Z = 4 All that can be said of the solution to this set of equations is that it lies on the line specified by the equation 3 X + 2 Z = 5. For some problems, this may be an adequate solution. Consider now the next set of equations. 2 X Y + Z = 1 2 X + 2 Y + 2 Z = 8 X + Y + Z = 3 Examine the augmented matrix for the system. 2 -1 1 1 2 2 2 8 1 1 1 3 Perform the elementary row operation: (Row 2) 2 (Row 3) to get the following. 2 -1 1 1 0 0 0 2 1 1 1 3 This merely emphasizes the fact that this set of equations is not consistent there is no possible solution. Mathematicians use the rank of a matrix as a tool to investigate these problems in the solutions of systems of linear equations; we shall just ignore the problems. Gaussian Elimination We now consider the solution of the original set of equations by the use of Gaussian elimination. This involves the use of <a href="/keyword/elementary-row-operations/" >elementary row operations</a> to turn the matrix of coefficients into an upper triangular form, in which all entries below the diagonal are 0. We actually work on the augmented matrix. The set of equations is 2 X Y + Z = 1 4 X + Y Z = 5 X + Y + Z = 0 The augmented matrix for this system is given by. 2 - 1 1 1 4 1 - 1 5 1 1 1 0 Add 2 (Row 1) to (Row 2) to get 2 - 1 1 1 0 3 - 3 3 1 1 1 0 Add (Row 1) to (Row 3) to get 1 2 - 1 1 0 3 - 3 3 0 1.5 0.5 - 0.5 Add (Row 2) to (Row 3) to get 1 2 - 1 1 0 3 - 3 3 0 0 2 - 2 At this point, we can use the method to solve the equations, as follows. The third equation is 2 Z = 2, so Z = 1 With the substitution of the value for Z, the second equation becomes 3 Y + 3 = 3, or Y = 0. With the substitution of Y = 0 and Z = 1, the first equation becomes 2 X + 0 1 = 1, or 2 X = 2 and X = 1. The solution is (1, 0, 1). I want to present another solution to this set of equations. We left the augmented matrix as 1 2 - 1 1 0 3 - 3 3 0 0 2 - 2 Divide (Row 3) by 2, (Row 2) by 3, and (Row 1) by 2 to get the following 1 - 0.5 0.5 0.5 0 1 -1 1 0 0 1 - 1 Now we see that the solution to the third equation is obviously Z = 1, and we can proceed with the solution to the other two equations as we did just above. I want to proceed with <a href="/keyword/elementary-row-operations/" >elementary row operations</a> , just to make the students completely tired of them. First add (Row 2) to (Row 1) to get. 1 0 0 1 0 1 - 1 1 0 0 1 - 1 Now add (Row 3) to (Row 2) to get 1 0 0 1 0 1 0 0 0 0 1 -1 This corresponds to the following set of equations X=1 Y=0 Z = 1, and we obviously have another solution to the set of equations. We next consider a method of solutions based on the inverse of the matrix of coefficients. Matrix multiplication is a well defined operation, and the definition of a matrix inverse follows from this operation. Suppose that matrix A has an inverse, denoted A-1. Then we 1 0 0 have A A-1 = A-1 A = I, where I is the identity matrix. For a 3-by-3 matrix I = 0 1 0 . 0 0 1 Matrix Inverse We have examined the set of equations given by 2 X Y + Z = 1 4 X + Y Z = 5 X + Y + Z = 0 This can be written in matrix form as A W = B, where 2 - 1 1 A = 4 1 - 1 , W = 1 1 1 X Y , and B = Z 1 5 . 0 We can use the fact that A-1 A = I to convert the equation to I W = A-1 B, or W = A-1 B. There are a number of methods by which the inverse of a matrix can be computed. We focus on the method of <a href="/keyword/elementary-row-operations/" >elementary row operations</a> . In order to compute the inverse of a matrix, we first augment the matrix with the identity, 2 - 1 1 1 0 0 creating A|I = 4 1 - 1 0 1 0 and use <a href="/keyword/elementary-row-operations/" >elementary row operations</a> to produce I|A-1. 1 1 1 0 0 1 First divide (Row 1) by 2, to get 1 - 1 / 2 1 / 2 1 / 2 0 0 4 1 - 1 0 1 0 1 1 1 0 0 1 Now add 4 (Row 1) to (Row 2) and 1 (Row 1) to (Row 3) to get 1 - 1 / 2 1 / 2 1 / 2 0 0 0 3 - 3 - 2 1 0 0 3 / 2 1 / 2 - 1 / 2 0 1 Divide (Row 2) by 3 to get 0 0 1 - 1 / 2 1 / 2 1 / 2 0 1 - 1 - 2 / 3 1 / 3 0 0 3 / 2 1 / 2 - 1 / 2 0 1 Add (Row 2) to (Row 1) and 3/2 (Row 2) to (Row 3) to get 1/ 6 1 / 6 0 1 0 0 0 1 - 1 - 2 / 3 1 / 3 0 0 0 2 1 / 2 - 1 / 2 1 Divide (Row 3) by 2 to get 1/ 6 1/ 6 0 1 0 0 0 1 - 1 - 2 / 3 1 / 3 0 0 0 1 1 / 4 - 1 / 4 1 / 2 Add (Row 3) to (Row 2) to get 1/ 6 0 1 0 0 1 / 6 0 1 0 - 5 / 12 1 / 12 1 / 2 0 0 1 1 / 4 - 1 / 4 1 / 2 1/ 6 0 2 0 1/ 6 2 1 - 5 / 12 1 / 12 1 / 2 According to this A = = 12 - 5 1 6 . 1/ 4 3 - 3 6 - 1 / 4 1 / 2 -1 And, W 2 0 1 X 1 2 1 Y 5 - 5 1 6 5 = = A-1 = 12 Z 0 3 - 3 6 0 2 1 + 2 5 + 0 0 12 1 1 1 - 5 1 + 1 5 + 6 0 = 0 = 0 = 12 12 3 1 - 3 5 + 6 0 - 12 -1