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The Matrix Inverse, and the Determinant

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The Matrix Inverse, and the Determinant We now use the interpretation of Gauss reduction given above to find the inverse of a matrix (if it has one). Theorem 2.70 (Row operations and the matrix inverse). If row operations are applied to a n×n matrix A to bring it to reduced row echelon form R, then either (i) R = In (the identity matrix) and A is invertible, or (ii) R has at least one row of zeros and A is not invertible. Proof. Since R is a square matrix, either: (i) R has a leading 1 in every column of the matrix or; (ii) the leading 1s skip (at least) one column so there is (at least) one row of zeros at the end. That is, either 21 0 ∗ 0 ::: 03 21 0 ::: 0 03 0 1 ∗ 0 ::: 0 0 1 ::: 0 0 6 7 6 7 60 0 0 1 ::: 07 6. .7 6 7 (i) R = 6. .. .7 = I or (ii) R = 6. .7 : 6 7 6. .. .7 60 0 ::: 1 07 6 7 4 5 60 0 0 0 ::: 17 0 0 ::: 0 1 4 5 0 0 0 0 ::: 0 Now suppose E1;E2;:::;Ek are the elementary matrices such that R = Ek :::E2E1A and consider the two cases. (i) Here R = I, so I = EA, where E = Ek :::E2E1 is the product of the elementary matrices used to reduce A to the reduced row echelon form R. We find E is the inverse of A. Certainly EA = I, so E works on the left of A. But also E is invertible (as it is a product of elementary matrices which are invertible), and hence −1 −1 −1 −1 In = E E = E IE = E (EA)E = (E E)AE = IAE = AE; so E also works on the right. Thus −1 (2.11) A = E = Ek :::E2E1: (ii) Here R has a row of zeros, so any product RC has a row of zeros for any matrix C. Therefore RC 6= I for any matrix C. Thus R is not invertible, so neither is A (if A were invertible, then R = En :::E2E1A would be invertible too, since it would be a product of invertible matrices, which is a contradiction). The proof of Theorem 2.70 gives an algorithm for finding the inverse of a matrix (when it has one). In fact the argument shows that if A is invertible then it is row equivalent to the identity matrix I. Further if E1;E2;:::;Ek are the elementary matrices such that R = I = Ek :::E2E1A; then the inverse is given by the formula (2.11)! So to compute the inverse A−1 of an invertible matrix A, we simply apply row operations to A to reduce A to the identity matrix I and keep a record of the elementary matrices E1;E2;:::;Ek involved. We then multiply these together in reverse order to find −1 A = Ek :::E2E1. 21 1 33 Example 2.71. Given A = 40 1 15 , find A−1. 0 0 1 56 Solution: 4.1. The supermatrix method. A compact way of presenting the working above is to use the supermatrix method. Write down the n × n matrix A and the identity matrix In in one large supermatrix A∗ = [AjI] of n rows and 2n columns. Now apply the row operations to A∗ that reduce A to reduced row echelon form R. As A is reduced to R = I, the same operations applied to I give A−1. If A is not row equivalent to I then A is not invertible. That is, if row operations applied to the supermatrix A∗ = [AjI] produce [RjC], where R is in reduced row echelon form. If R = I, then C = A−1. On the other hand, if R has a row of zeros then A is not invertible. Example 2.72. Use the supermatrix method to find A−1, where A is the matrix of Example 2.71. Solution: The algorithm given in the supermatrix method is quite mechanical and produces the inverse in a rea- sonably straightforward way. The arithmetic may become a bit tedious for matrices bigger than 3 × 3, but there are computer packages available than do these sorts of matrix calculations. 21 1 −13 Example 2.73. Given A = 43 1 −25, use the supermatrix method to find A−1. 1 −2 1 57 Solution: 5. Determinants 5.1. The 2 × 2 case. a b Recall that, if A = is a 2 × 2 matrix, then by Lemma 2.53 the matrix A is invertible if and c d only if ab − cd 6= 0. In particular, the value of ad − bc determines whether or not A is invertible. This observation leads to the following definition. a b Definition 2.74 (Determinant). The determinant of a 2 × 2 matrix A = is the number ad − bc. c d We denote the determinant of A by det(A) or jAj. In the latter case we use vertical lines instead of brackets to border the array, so we write a b det(A) = jAj = = ad − bc: c d 2 −5 1 0 3 −2 Example 2.75. Find the determinants of A = , I = , and B = . 3 7 0 1 −3 2 Solution: In the next two results we list some properties of determinants, which we justify when A and B are 2 × 2 matrices and I = I2, the 2 × 2 identity matrix. However, later on we will define the determinant of a general n × n matrix, and these properties remain true (though we shall not prove them) when A and B are n × n matrices and I = In is the n × n identity matrix. Lemma 2.76 (Properties of determinants). If A and B are square matrices (of the same size) and I is the identity matrix then: (i) det(I) = 1. (ii) det(AT ) = det(A). 58 (iii) det(AB) = det(A) det(B). −1 1 (iv) If det(A) 6= 0, then A is invertible and det(A ) = det(A) . Conversely, if A is invertible, then det(A) 6= 0. Proof. As mention above, we only consider the 2 × 2 case. The first property (i) is contained in Example 2.75. For the remaining properties, we let a b e f ae + bg af + bh A = and B = ; and so AB = : c d g h ce + dg cf + dh We now state how the determinant of a matrix is affected by the elementary row operations of the last section. Again we only give the proof in the 2×2 case, but it also holds for general square n×n matrices. Lemma 2.77 (Row operations on determinants). If the square matrix B is obtained from A by: • multiplying a row of A by a scalar λ, then det(B) = λ det(A); • adding a multiple of one row of A to another, then det(B) = det(A); • interchanging two rows of A, then det(B) = − det(A): Further, each of these remains true if we replace row by column. Proof. We leave (i), (ii) and (iii) as exercises, and just consider the last remark on columns. Remember the rows of AT are the columns of A, so any column operation on A can be achieved by first taking the transpose of A, then performing the corresponding row operation on AT , and then taking the transpose of the resulting matrix. Since by Lemma 2.76 a matrix and its transpose have the same determinant, the result follows. 1 2 Example 2.78. Compare det A and det B, for A = and B which is obtained from A by: −2 3 (i) multiplying the first row of A by 2; (ii) adding twice the first row of A to the second; (iii) interchanging the two rows of A: 59 Solution: Beware: If B = 2A, and A is a 2 × 2 matrix, then both rows of A get multiplied by 2, so det(B) = 2 × 2 det(A) = 22 det(A). 5.2. The 3 × 3 case. Before defining the determinant of a 3 × 3 matrix A, we define some 2 × 2 subdeterminants of A called cofactors. Definition 2.79. Let A = [aij] be a 3 × 3 matrix. The cofactor of the entry aij of A, denoted Aij, is given by determinant of the 2 × 2 matrix obtained from A by deleting row i and A = (−1)i+j × ij column j of A. 21 2 33 Example 2.80. If A = 44 5 65 find A11;A12;A13 and A21. 7 8 0 Solution: Remark: Since (−1)i+j is 1 if i + j is even and −1 if i + j is odd, the signs of the cofactors from a 2+ − +3 "chessboard" type pattern. For instance in the 3 × 3 case, the ± signs take the form 4− + −5. + − + 60 Definition 2.81. The determinant of the 3 × 3 matrix A = [aij] is given by det(A) = a11A11 + a12A12 + a13A13: That is, det(A) is the sum of the products of each entry in the first row of A times its corresponding cofactor. For this reason we say we have found det(A) by expanding about its first row. In terms of the elements of A a11 a12 a13 a22 a23 a21 a23 a21 a22 det(A) = a21 a22 a23 = a11 − a12 + a13 : a32 a33 a31 a33 a31 a32 a31 a32 a33 Example 2.82. Find det(A), where A is the matrix of Example 2.80. Solution: 5.3. Expansion of det(A) by any row or column. We have defined the determinant of a 3 × 3 matrix by expanding along the first row.
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