k=1 ( 1)k+j M kj detm kj. detm = ad bc. = 1 ( ) 2 ( )+3 ( ) = = 0


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1 4 Determinants The determinant of a square matrix is a scalar (i.e. an element of the field from which the matrix entries are drawn which can be associated to it, and which contains a surprisingly large amount of information about the matrix. To define it we need some extra notation: Definition. If M is an n n matrix over a field, then M ij is the (n 1 (n 1 matrix obtained from M by throwing away the ith row and jth column. Example. If A then A 11 ( and A 23 ( Definition. If M is a 1 1 matrix then detm M 11. If M is an n n matrix with n 2 then. detm ( 1 k+1 M 1k detm 1k. Alternative Notation. Sometimes the determinant is written det(m or M. Remark/Exercise. If M which we saw in Section 1.6. ( a b c d is a 2 2 matrix then the formula simplifies to detm ad bc. Example. The determinant of the matrix A from the previous example is: deta A 11 deta 11 A 12 deta 12 +A 13 deta 13 ( ( ( det 2det +3det ( ( ( Warning. The determinant is only defined for square (n n matrices! 4.1 Expanding Along Rows and Columns The formula in the definition of the determinant is, for fairly obvious reasons, called the expansion along the first row. In fact, we can expand in a similar way along any row, or indeed down any column: Theorem 4.1. For any n n matrix M over a field and any 1 i n and 1 j n: (i detm n ( 1i+k M ik detm ik ; (ii detm n ( 1k+j M kj detm kj. Proof. Omitted. Computing the determinant using (i is called expanding along the ith row, while using (ii is expanding down the jth column. Notice that expanding along the first row just means using the definition of the determinant. But depending on the form of the matrix, choosing the row or column to expand along/down carefully can make it much easier to compute the determinant. 23
2 Example. Suppose we wish to compute det it makes sense to expand down this column, giving ( deta ( det ( +( det 1 6. Noticing the two 0 entries in the second column, ( +( det Using the definition (i.e. expanding along the first row would have given the same answer (check for yourself!, but with more calculation. Terminology. The value detm ij is called that (i,j minor of M, while the value ( 1 i+j detm ij is the (i,j cofactor of M. The method we have seen for computing the determinant is called cofactor expansion or Laplace 9 expansion. (Note that some people who write m ij for matrix entries use the notation M ij for the (i,j minor of M. 4.2 Upper Triangular Matrices A square matrix M is called upper triangular if M ij 0 for all i > j, that is, if all entries below the main diagonal are 0. Examples. ( and are upper triangular. Remark. A square (nonaugmented row echelon matrix is always upper triangular. Theorem 4.2. Let M be an n n upper triangular matrix over a field. Then detm is the product of the entries on the main diagonal: n detm M ii. Proof. See Exercise Sheet 3. Corollary 4.3. deti n 1 for all n. Proof. I n is clearly upper triangular, so deti n is the product of the entries on the main diagonal, which is Elementary Row Operations and the Determinant Cofactor expansion (even along a carefully chosen row or column is not a very efficient way to compute determinants. If the matrix is large it quickly becomes impractical. Just as with solving systems of equations and computing inverses, row operations ride to the rescue! Theorem 4.4. Let A and B be n n matrices over a field. Then (i if A r i αr i B then detb αdeta; (ii if A r i r i +λr j B then detb deta; (iii if A r i r j B then detb deta; Corollary 4.5. If A and B are square matrices over a field and are row equivalent, then i1 deta 0 detb 0. 9 After PierreSimon, Marquis de Laplace (pronounced laplass,
3 Proof. It follows from Theorem 4.4 that none of the types of elementary row operations ever change whether the determinant is 0. If A and B are row equivalent then there is a sequence of row operations transforming A into B, and none of these can ever change whether the determinant is 0. We will prove Theorem 4.4 shortly, but first let s see how it helps us calculate a determinant. The idea is to use row operations to transform the matrix into a form where the determinant is easy to calculate, and keep track of how the operations affect the determinant so that we can recover the determinant of the original matrix. Theorem 4.2 suggests it makes sense to aim for an upper triangular matrix; we know every square matrix is row equivalent to one of these because row echelon matrices are upper triangular, but usually it is not necessary to go as far as finding a row echelon matrix: Example. Let s find the determinant of the matrix Applying the operations r 1 6r 1, r 3 r 3 r 1, r 2 r 3 gives an upper triangular matrix: By Theorem 4.2 the determinant of this matrix is 1 ( 2 ( By Theorem 4.4 the operation r 1 6r 1 will have multiplied the determinant by 6, while the row swap will have changed the sign, so the determinant of our original matrix must be 1 6 (12 2. Another application of Theorem 4.4 is to obtain the determinants of elementary matrices; this will prove useful later. Corollary 4.6. (i dete ri αr i (ii dete ri r i +λr j (iii dete ri r j Proof. Each part follows by the corresponding part of Theorem 4.4 from the fact that E ρ is obtained from I n by the corresponding row operation ρ. For (iii, for example, E ri r j is obtained from I n by a row swap, so by Theorem 4.4(iii and Corollary 4.3 we have Exercise. Do the other two cases. dete ri r j deti n Proof of Theorem 4.4 (not directly examinable This section contains a proof of Theorem 4.4. It is not directly examinable and will not be covered in lectures. However, you are strongly encouraged to read it because (i it will increase your understanding of things which are examinable, (ii it will give you experience of learning mathematics by reading and (iii it will begin to show you how complex proofs are put together in stages using lemmas. Lemma 4.7. Let A be an n n matrix. If A has two different rows the same, then deta 0. Proof. Again, we use induction on n. The case n 1 is vacuously true: if there is only one row then there cannot be two different ( rows the same, so the statement cannot be false. 10 a b If n 2 then A for some a,b R, and now deta ab ab 0. a b Now let n 3 and suppose for induction that the result holds for smaller matrices. Pick two rows in A which are the same, and let r be the index of some other row. Expanding along row r, we have deta ( 1 k+r A rk deta rk. 10 Recall the flying pigs from Exercise Sheet 0! 25
4 For each k, notice that A rk is an (n 1 (n 1 matrix with two rows the same. Hence, by the inductive hypothesis, all of these matrices have determinant 0, so deta 0. Question/Exercise. Why, in the proof of Lemma 4.7, do we have to start the induction from n 3? In other words, why can t we just use n 1 as the base case? Lemma 4.8. Let X, Y and Z be n n matrices which are all the same except in row i, and suppose Z ij X ij +Y ij for all j. Then detz detx +dety Proof. Expand det Z along the ith row (exercise: write out the details!. We are now ready to prove Theorem 4.4. Proof of Theorem 4.4. (i Suppose A r i αr i B. Expanding along the ith row we have: detb α ( 1 k+i B ik detb ik ( 1 k+i (αa ik deta ik ( 1 k+i A ik deta ik αdeta. (ii Suppose A r i r i +λr j B. Let D be the matrix which is the same as A and B, except that the ith row is λ times the jth row of A. Then we are in the position of Lemma 4.8 with X A, Y D and Z B, so we have detb deta+detd. Now let C be the matrix which is the same as D, except that the ith row is exactly the jth row of A. Then D can be obtained from C by applying the transformation r i λr i, so by part (i, detd λdetc. But C has two rows (rows i and j the same, so by Lemma 4.7, detd λ0 0. Thus, detb deta+0 deta. (iii Suppose A r i r j B. We define matrices F, G, H, P and Q which are the same as A (and B except in rows i and j, where their entries are as follows: row i and row j of F are both the sum of rows i and j in A; row i of G is the sum of rows i and j in A; row j of G is row j of A; row i of H is the sum of rows i and j in A; row j of H is row i of A; rows i and j of P are row i of A; and rows i and j of Q are row j of A. Now we are in the position of Lemma 4.8 with X G, Y H and Z F, so we have detf detg+deth. Similarly, the conditions of Lemma 4.8 are satisfied with X A,Y Q,Z G and also with X B,Y P,Z H so we obtain detg deta+detq and deth detb +detp. Notice also that each of F, P and Q has two identical rows, so by Lemma 4.7 they all have determinant 0. Hence 0 detf detg+deth deta+detq+detp +detb deta+detb so that deta detb as required. 26
5 4.5 Determinants and Inverses We saw in Section 1.6 that a 2 2 matrix is invertible if and only its determinant is non0. We now know nearly enough to prove the corresponding statement for n n matrices: just one more lemma is needed. Lemma 4.9. Let M be an n n reduced row echelon matrix over a field. Then either M I n or M has a zero row and has determinant 0. Proof. See Exercise Sheet 3. Theorem An n n matrix A over a field is invertible if and only if deta 0. Proof. By Theorem 2.2, A is row equivalent to a reduced row echelon matrix M. By Theorem 3.4, A is invertible if and only if M I n. By Lemma 4.9, M I n if and only if detm 0. But M is row equivalent to A, so by Corollary 4.5, detm 0 if and only if deta 0. We also saw in Section 1.6 that for 2 2 matrices there is an explicit formula for the inverse in terms of determinants. The formula generalises to larger matrices, although it is often too complicated to be useful in practice: Theorem 4.11 (Cramer s Rule 11. If A is invertible then A 1 1 deta B where B is the matrix given by B ij ( 1 i+j deta ji. The matrix B in the statement of the theorem is called the adjugate of A. Exercise. Check that if A is a 2 2 matrix, Cramer s Rule simplifies to the formula from Section Multiplicativity of the Determinant In this section we will establish another important property of the determinant, namely that det(ab detadetb for all n n matrices A and B. This property is called multiplicativity of the determinant. To prove it we will use elementary matrices. Lemma Let E 1,...,E k be n n elementary matrices. Then for every n n matrix M, det(e k...e 1 M det(e k...e 1 detm. Proof. By strong induction on k. Suppose first k 1. Then E 1 E ρ for some elementary row operation ρ. Let M be an n n matrix. Now E ρ M is obtained from M by the operation ρ. For each of the three types of elementary row operation, we can check using the appropriate parts of Theorem 4.4 and Corollary 4.6 that det(e ρ M dete ρ detm. For example, if ρ is a row swap then by Theorem 4.4(i, det(e ρ M detm and by Corollary 4.6(i det(e ρ 1. Now let k 2 and suppose the statement is true for smaller k. Then applying the inductive hypothesis three times, with the role of M played first by E 1 M, then by M itself, and finally by E 1, we obtain: det(e k E k 1...E 1 M det(e k...e 2 det(e 1 M det(e k...e 2 dete 1 detm det(e k...e 1 detm. Theorem Let A and B be n n matrices over a field. Then 11 Gabriel Cramer ( det(ab (deta(detb 27
6 Proof. Write A E k...e 1 M where M is the reduced row echelon form of A and the E i s are elementary matrices. Then by Lemma 4.12: We now consider two cases: det(ab det(e k...e 1 MB det(e k...e 1 det(mb. Case 1: M I n. In this case we have MB B and E k...e 1 A so det(ab det(e k...e 1 det(mb detadetb. Case 2: M I n. In this case A is not invertible and so by Theorem 4.10, deta 0. Also, by Lemma 4.9 M has a row of zeros. It follows that MB has a row of zeros, and expanding along this row we see that det(mb 0. Now det(ab det(e k...e 1 det(mb 0 (deta(detb. 28
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