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Elementary Matrices of elementary row operations. In each case, we're looking for a square matrix E such that EA = B where A is the augmented matrix for the original system of equations and B is the augmented matrix Elementary transformations for the new system. In each case, we'll illustrate it and matrix inversion with a system of three equations in three unknowns. Math 130 Linear Algebra Let's start with this system of equations. D Joyce, Fall 2015 2y + 2z = 4 x + 2z = 3 Elementary row operations again. We used 3y + 3z = 6 the elementary row operations when we solved sys- It's almost in reduced echelon form, and only three tems of linear equations. We'll study them more steps are needed to put it in that form. Its aug- formally now, and associate each one with a partic- mented matrix is ular invertible matrix. 20 2 2 43 When you want to solve a system of linear equa- 1 0 2 3 tions Ax = b, form the augmented matrix by ap- 4 5 0 3 3 6 pending the column b to the right of the coeffi- cient matrix A. Then solve the system of equations For the first step, we'll exchange the first two by operating on the rows of the augmented matrix rows. The elementary matrix E1 to do that is al- rather than on the actual equations in the system. most the diagonal matrix. Only the two 1s in the There are three kinds of elementary row operations rows to be exchanged need to be moved to the op- are those operations on a matrix that don't change posite rows. the solution set of the corresponding system of lin- 20 1 03 20 2 2 43 21 0 2 33 ear equations. 41 0 05 41 0 2 35 = 40 2 2 45 0 0 1 0 3 3 6 0 3 3 6 1. Exchange two rows. Next, let's divide the second row by 2, that is, 2. Multiply or divide a row by a nonzero constant. multiply it by 0.5. Again, the elementary matrix E2 to do that is almost the diagonal matrix. Just 3. Add or subtract a multiple of one row from replace the 1 in the second row by 0.5. another. 21 0 03 21 0 2 33 21 0 2 33 With these three operations, we can convert the 40 0:5 05 40 2 2 45 = 40 1 1 25 augmented matrix to a particularly simple form| 0 0 1 0 3 3 6 0 3 3 6 the reduced echelon form|that allows us to read Finally, we'll subtract 3 times the second row off the solutions. from the third. Yet again, the elementary matrix Two martrices are said to be row equivalent if E3 to do that is almost the diagonal matrix. All one can be transformed to the other by means of a that's needed is to place a −3 in the second col- finite sequence of elementary row operations. umn of the third row. 21 0 03 21 0 2 33 21 0 2 33 Elementary matrices. Now to find the elemen- 40 1 05 40 1 1 25 = 40 1 1 25 tary matrices that correspond to these three kinds 0 −3 1 0 3 3 6 0 0 0 0 1 The matrix is now in reduced echelon form and In the first case where there's a 1 in each row of the general solution can be read off from it: the square matrix EA, because it's in reduced ech- elon form, therefore it has to be the identity. That (x; y; z) = (−2z + 3; −z + 2; z) says EA = I. Since E is the product Es ··· E2E1 of invertible elementary matrices, therefore E is also where z can be any number. invertible. Multiply the equation EA = I by E−1 Note that each kind of elementary matrix E , E , 1 2 on the left to get A = E−1, and therefore A−1 = E. and E is invertible, and its inverse is another (or 3 It is in this first case that A is invertible and you the same) elementary matrix of the same kind. The can find the inverse. inverse for E is itself. The inverse for E looks 1 2 In the second case, the last row of EA is all 0s. just like E except you use the reciprocal for the 2 We need to show that A cannot have an inverse special diagonal entry. And the inverse for E looks 3 in this case. We'll suppose A does have an inverse just like E except you use the negation for the off- 3 and derive a contradiction. Now, the matrix EA diagonal entry. has all 0s in its last row, so if you multiply it on the right by any matrix, then the product will also Proof that the inversion algorithm works. have all 0s in its last row. But A is invertible, so Recall the method used to find the inverse of a ma- multiply EA on the right by A−1E−1. The result trix. is EAA−1E−1I, which is the identity matrix I, and Theorem 1. To find the inverse of a square matrix I doesn't have 0s in its last row, a contradiction. A, first, adjoin the identity matrix to its right to get Therefore, A doesn't have an inverse in this second an n × 2n matrix [AjI]. Next, convert that matrix case. to reduced echelon form. If the result looks like That finishes the proof that this method will ei- [IjB], then B is the desired inverse A−1. But if the ther construct an inverse matrix or show that the square matrix in the left half of the reduced echelon inverse doesn't exist. q.e.d. form is not the identity, then A has no inverse. In the proof of the theorem in the case where −1 −1 Proof. Suppose the sequence of elementary row op- A is invertible, we had A = E , but E = −1 −1 −1 −1 erations performed on the matrix [AjI] to put it (Es ··· E2E1) = E1 E2 ··· Es . That gives us into reduced echelon form is described by the el- the following corollary. ementary matrices E1;E2;:::;Es. After the first Corollary 2. Every invertible matrix is the prod- elementary operation, the matrix [AjI] is con- uct of elementary matrices. verted to [E1AjE1I]. After the second, it becomes [E2E1AjE2E1I]. And after the last it looks like Math 130 Home Page at [Es ··· E2E1AjEs ··· E2E1I]: http://math.clarku.edu/~ma130/ Let's denote the product of the Eis as E, that is, E = Es ··· E2E1. Then we have [EAjE] as the reduced echelon form. Now, we have two cases to consider. One is where EA, the first half of the n × 2n matrix that's in reduced echelon form, has a 1 in each row. The other is where some row has no one in it but is all 0s. If any row is all 0s, then the last row will be all 0s since it's in reduced echelon form. 2.
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