Determinants. Recall that the 2 2 matrix a b c d. is invertible if

Transcription

1 Determinants Recall that the 2 2 matrix a b c d is invertible if and only if the quantity ad bc is nonzero. Since this quantity helps to determine the invertibility of the matrix, we call it the determinant. We will also write det a c b d = ad bc. We can find the determinant of a general 3 3 matrix A by recognizing that to invert A, we apply the row reduction process to the augmented matrix [ A I 3 ] to attempt to bring it to the form [ I 3 A 1 ]. For this to be possible, we must be able to reduce A to the identity matrix I 3. Let s study how this happens in the general case, assuming that the upper left entry is nonzero: a 11 a 12 a 13 a 11 a 12 a 13 a 21 a 22 a 23 scale rows 2, 3 a 11 a 21 a 11 a 22 a 11 a 23 a 31 a 32 a 33 a 11 a 31 a 11 a 32 a 11 a 33 a 11 a 12 a 13 0 a replace rows 2, 3 11 a 22 a 12 a 21 a 11 a 23 a 13 a 21 0 a 11 a 32 a 12 a 31 a 11 a 33 a 13 a 31

2 The entries in positions (2,2) and (3,2) in this last matrix cannot both be zero since A is invertible. In fact, swapping these two rows would yield the same matrix at this stage as if we had swapped rows 2 and 3 of A at the outset. So let s suppose that the (2,2) entry, a 11 a 22 a 12 a 21 is not 0. Then we can scale row 3 by this nonzero value and use it to eliminate the (3,2) entry below it with a row replacement. This brings A to row echelon form a 11 a 12 a 13 0 a 11 a 22 a 12 a 21 a 11 a 23 a 13 a a 11 Δ where Δ = a 11 a 22 a 33 + a 12 a 23 a 31 +a 13 a 21 a 32 a 11 a 23 a 32 a 12 a 21 a 33 a 13 a 22 a 31 We know that all the pivot entries (on the diagonal) must be nonzero for A to be invertible. As we have assumed that a 11 and a 11 a 22 a 12 a 21 are nonzero, this means that Δ must be nonzero as well. This leads us to define the determinant of A to be this quantity Δ and gives us the first case of the Theorem If A is a 3 3 matrix, then A is invertible if and only if its determinant Δ is nonzero.

3 Proof (cont d) Had we needed to swap rows 2 and 3 above, we would have arrived at an expression for Δ similar to the one above, but with every row index value 2 switched to a 3, and vice versa. If we do this in the above expression for Δ, we obtain precisely the expression Δ, implying that the definition for Δ works in this other case, too. Finally, if a 11 were zero, we could not have begun the reduction procedure above. But at least one of a 21 or a 31 would then have to be nonzero (else, having a zero column, A would not be invertible). We could then swap row 1 with row 2 or 3 to put a nonzero entry in the upper left corner and proceed as before. At the end we will obtain an expression similar to Δ except that every row index value of 1 is switched to either a 2 or a 3, and vice versa. However, making this switch in the expression we have for Δ above, we notice once again that this yields precisely the expression Δ. This proves that the expression for Δ defined above must be nonzero whenever A is invertible, independent of these particular cases. That is, the theorem above is proved. // Notice that

4 a 11 a 12 a 13 det a 21 a 22 a 23 = a 11 a 22 a 33 +a 12 a 23 a 31 + a 13 a 21 a 32 a 31 a 32 a 33 a 11 a 23 a 32 a 12 a 21 a 33 a 13 a 22 a 31 = a 11 (a 22 a 33 a 23 a 32 ) +a 12 (a 23 a 31 a 21 a 33 ) +a 13 (a 21 a 32 a 22 a 31 ) = a 11 det a 22 a 23 a 32 a 33 a 12 det a 21 a 23 a 31 a 33 +a 13 det a 21 a 22 a 31 a 32 That is, the determinant of the 3 3 matrix A can be computed as a linear combination of the entries of the first row with weights which are, up to sign, 2 2 determinants built from other entries of A. Further, the corresponding 2 2 determinant in each of the three terms is obtained from A by deleting the entries in the same row and column as the entry that appears as the weight.

5 The quantities C 11 = det a 22 a 23, a 32 a 33 C 12 = det a 21 a 23, a 31 a 33 C 13 = det a 21 a 22 a 31 a 32 are called cofactors of the corresponding entries a 11, a 12, a 13 of A. The (i, j) cofactor is the signed determinant of the 2 2 submatrix A ij of A found by deleting row i and column j of A; the sign has the value ( 1) i+ j (equals +1 when i and j have the same parity and 1 when they have different parity). With these definitions, we can express a formula for the determinant of A as det A = a 11 C 11 + a 12 C 12 +a 13 C 13. It turns out (and is straightforward to verify) that the same procedure can be used to compute a determinant in terms of a cofactor expansion when the entries are taken from any row or any column of A.

6 For instance, cofactor expansion along the second column of A gives the correct formula det A = a 12 C 12 + a 22 C 22 +a 32 C 32. Cofactor expansion also suggests that we can define the determinant of a 4 4 matrix recursively in terms of 3 3 cofactor determinants, the determinant of a 5 5 matrix recursively in terms of 4 4 cofactor determinants, and so on. That is, we define the determinant of the n n matrix A = a ij [ ] to be det A = a 11 C 11 + a 12 C 12 +L+a 1n C 1n, where C ij = ( 1) i+ j det A ij is the (i, j) cofactor, given in terms of the (n 1) (n 1) determinant of the submatrix A ij of A obtained by deleting row i and column j from A. It is possible to compute determinants by cofactor expansion along other rows and columns of A. We state this in a theorem whose proof is straightforward but very tedious.

7 Theorem If A = [ a ij ] is an n n matrix, then deta can be computed by cofactor expansion along any row (i = 1, 2,, n): det A = a i1 C i1 +a i2 C i2 +L+a in C in and along any column (j = 1, 2,, n): det A = a 1 j C 1 j +a 2 j C 2 j +L+a nj C nj. // Corollary If A is a triangular square matrix (either upper triangular or lower triangular), then det A is the product of the diagonal entries of A. // It turns out that the computation of determinants via cofactor expansions is not terribly efficient. Once again, we turn to row operations to simplify our methods.

8 Theorem Suppose A and B are n n matrices. 1. If B is obtained from A by a row operation that swaps a pair of rows, then detb = det A; 2. if B is obtained from A by a row operation that scales a row by the scalar factor k, then detb = k det A; and 3. if B is obtained from A by a row operation that replaces one row by its sum with some multiple of another row, then detb = det A. Proof We proceed by induction on n, the case n = 2 coming from the following computations: det c a d b = cb ad = det a b c d ; and det ka c kb d = det a kc = k det a b c d ; b kd = kad kbc det a + kc b + kd c d = (a + kc)d (b +kd)c = det a b c + ka d + kb = a(d + kb) b(c + ka) = ad bc = det a b c d

9 We now assume that n > 2 and that the theorem is true for all matrices of size less than n. The row operation is a swap. Cofactor expansion of B along any row that is not swapped (as n > 2, there must be some such row) is identical to cofactor expansion of A along the same row except that each of the submatrices in the corresponding cofactors for B has the same pair of rows swapped. By the induction hypothesis, these cofactors have opposite value. So detb = det A. The row operation is a scaling operation. Cofactor expansion of B along the row that is being scaled shows directly that detb = k det A. The row operation is a replacement. Suppose the operation replaces row i by its sum with k times row j. Cofactor expansion of B along row i shows that detb = det A +k det A where A is the matrix obtained from A by replacing row i with row j. In particular, since A has two identical rows, swapping them produces the same matrix, whence by our result above, det A = det A. This means that det A = 0, so detb = det A. // Recall that performing an elementary row operation on A is identical to multiplying A on the left by an elementary matrix E (which is the matrix obtained from the identity matrix I by

10 performing the same row operation). So the matrices denoted B in the previous theorem all have the form EA for some appropriate form of elementary matrix E. Since cofactor expansions allow us to conclude that deti = 1 (regardless of the size of I), if E is an elementary matrix which represents a row swap, then dete = 1, if E is an elementary matrix which represents scaling a row by the factor k, then dete = k, and if E is an elementary matrix which represents replacing one row by its sum with a multiple of another, then dete = 1, the theorem above can be restated as Theorem Suppose A and E are n n matrices with E elementary. Then det(ea) = dete det A. // Corollary Suppose A is an n n matrix that can be brought to row echelon form U = u ij [ ] by means of some number of row replacements, exactly r row swaps, and no row scalings. (This is always possible.) Then det A = ( 1) r u 11 u 22 Lu nn.

11 Proof Represent these row operations with elementary matrix multiplications: U = E p (L(E 2 (E 1 A ))L). Repeated application of the previous theorem shows that detu = ( 1) r det A, so det A = ( 1) r detu. But U is upper triangular, so detu can be computed by repeated cofactor expansions along the first column, showing that det A = ( 1) r u 11 u 22 Lu nn. // Corollary If A is an n n matrix, then A is invertible if and only if det A 0. // This shows us that our original definition of determinant for n n matrices with n > 2 accomplishes exactly what we want it to do: it determines when A is invertible! The last theorem can be extended much more generally: Theorem Suppose A and B are n n matrices. Then det( AB) = det A det B.

12 Proof If A is not invertible, then neither is AB (for if AB had inverse C, then A would have inverse BC since ABC = I ). So both det A and det( AB) are 0 and the identity det( AB) = det A det B is true. If A is invertible, then there is a sequence of row operations, represented by elementary matrices E 1,E 2,,E p, that brings A into the identity matrix I, whence E p LE 2 E 1 A = I. These elementary matrices are all invertible; let F 1, F 2,, F p be their respective inverses (all of which are themselves elementary matrices). Then A = F 1 F 2 LF p, so det( AB) = det( F 1 F 2 LF p B) = det F 1 (det F 2 (L(detF p det B)L)) = det F 1 (det F 2 (L(detF p )L)) detb = det( F 1 F 2 LF p ) det B = det A det B // It is important to note that it is not true in general that det( A + B) = det A +det B; that is, the determinant is not a linear function of matrices. However, the determinant is linear with respect to the columns of a matrix. That is, if we define a function

13 then Δ(x) = det [ a 1 L a j 1 x a j+1 L a n ], Δ(rx + sy ) = rδ(x )+ sδ(y), as can be seen by cofactor expansion of the matrix [ a 1 L a j 1 rx + sy a j+1 L a n ] along the jth column. Finally, there is a simple relation between the determinant of a matrix and of its transpose: Theorem If A is an n n matrix, then det A T = det A. Proof By induction on n. If n = 1, the result is obvious. If n > 1, then the cofactor expansion of det A T along the first row is identical to the cofactor expansion of det A along the first column, because the corresponding cofactors are equal by the induction hypothesis. //