Unformatted text preview: MA 265 LECTURE NOTES: MONDAY, FEBRUARY 4 Determinants Review. In the previous lecture, we deﬁned the determinant of an n × n matrix as the following number: a11 a21 . . . an1 a12 a22 . . . an2 ··· ··· .. . ··· a1n a2n . . . ann =
σ ∈Sn ε(σ ) a1 σ(1) a2 σ(2) · · · an σ(n) as a sum over the n! permutations σ , where ε(σ ) = ±1 depending on whether σ is even or odd. As speciﬁc examples, we showed that ab det [a] = a and = a d − b c. cd (Be careful not to confuse this with absolute value!) Example. We compute the determinant of a 2 × 2 matrix: 2 4 −3 5 = (2) (5) − (−3) (4) = 22. Visually, we compute the determinant using the following arrows: ?? ?? ? a? b ?? ? c d? ? ? ? ? The arrows mean “multiply all of these numbers together”, where rightward arrows correspond to “+” and the leftward arrows correspond to “−”. This diagram says “+(a · d) − (b · c)”. 3 × 3 Determinants. We show how to compute the determinant of a 3 × 3 matrix. Recall that we have 3! = 6 permutations to consider: 1 → 1 1 → 2 1 → 3 2→2 2→3 2→1 σ2 : σ3 : σ1 : 3 → 3 3 → 1 3 → 2 σ4 : 1 → 1 2→3 3 → 2 σ5 : 1 → 2 2→1 3 → 3 σ6 : We showed in the previous lecture that the ﬁrst three have signs ε(σ1 ) = ε(σ2 ) = ε(σ3 ) = +1, and the last three have signs ε(σ4 ) = ε(σ5 ) = ε(σ6 ) = −1. Using the deﬁnition of the determinant, the 3 × 3 matrix a11 a12 a13 A = a21 a22 a23 a31 a32 a33
1 1 → 3 2→2 3 → 1 2 MA 265 LECTURE NOTES: MONDAY, FEBRUARY 4 has determinant det(A) = ε(σ1 ) a11 a22 a33 + ε(σ2 ) a12 a23 a31 + ε(σ3 ) a13 a21 a32 + ε(σ4 ) a11 a23 a32 + ε(σ5 ) a12 a21 a33 + ε(σ6 ) a13 a22 a31 = (a11 a22 a33 + a12 a23 a31 + a13 a21 a32 ) − (a11 a23 a32 + a12 a21 a33 + a13 a22 a31 ) . This deﬁnition is a little awkward to use, so we give a couple of diﬀerent ways to view this formula. Consider the matrix above, but say we list the ﬁrst two columns again on the right: ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? a11? a12? a13? a11 a12 ? ? ? ? ? ? ? ? ? a21 a22? a23? a21? a22 ? ? ? ? ? ? ? ? ? a31 a32 a33? a31? a32? ? ? ? ? ? ? ? ? ? ? ? ? Remember that the arrows mean “multiply all of these numbers together”, where rightward arrows correspond to “+” and the leftward arrows correspond to “−”. This diagram says A = +a11 a22 a33 + a12 a23 a31 + a13 a21 a32 − a11 a23 a32 − a12 a21 a33 − a13 a22 a31 . Unfortunately, this “trick” using the arrows to keep track of the products only works for n × n matrices when n = 2 and n = 3. There is no similar trick when n ≥ 4. This formula can also be written as a sum of 2 × 2 determinants: A = (a11 a22 a33 + a12 a23 a31 + a13 a21 a32 ) − (a11 a23 a32 + a12 a21 a33 + a13 a22 a31 ) = a11 (a22 a33 − a23 a32 ) − a12 (a21 a33 − a23 a31 ) + a13 (a21 a32 − a22 a31 ) = a11 · a22 a32 a23 a33 − a12 · a21 a31 a23 a33 + a13 · a21 a31 a22 a32 . You can remember these matrices by crossing out the row and column where a11 , a12 and a13 appear: @ABC GFED a12 a11 a21 a31 a22 a32 a13 a23 a33 @ABC a11 GFED a13 a12 a21 a31 a22 a32 a23 a33 a11 a21 a31 @ABC a12 GFED a13 a22 a32 a23 a33 This diagram says “a11 · a22 a23 a21 a23 a21 ”, “a12 · ”, and “a13 · a32 a33 a31 a33 a31 remember to take the alternating sum of these three in order to compute A. a22 ”, respectively. Just a32 Example. We will compute the determinant of the following 3 × 3 matrix two diﬀerent ways: 123 A = 2 1 3 . 312 First we consider this 3 × 3 array, where we add the ﬁrst two columns again to the right: 123 213 312 We then draw arrows to consider how to multiply: 1 2 3 2 1 1 MA 265 LECTURE NOTES: MONDAY, FEBRUARY 4 3 ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? 1? 2? 3 ?? ?? ? ? 2 1 3 ?? ?? 3 1 2 Hence we ﬁnd that ?? 1 ? ? ?? 2 ? ? ?3 ? ? ? ? 2 ?? 1 ?? ?1 ? ? ? ? ? ? ? ? ? A = + (1 · 1 · 2) + (2 · 3 · 3) + (3 · 2 · 1) − (3 · 1 · 3) − (1 · 3 · 1) − (2 · 2 · 2) = 2 + 18 + 6 − 9 − 3 − 8 = 6. For another method, we expand the 3 × 3 determinant as a sum of 2 × 2 determinants. Consider the following three diagrams: 89:; ?>=< 1 2 3 2 1 1 3 3 2 1 2 3 89:; ?>=< 2 1 1 3 3 2 1 2 3 2 1 1 89:; ?>=< 3 3 2 We see that we have the expansion A = 1 · 1 1 3 2 −2· 23 32 +3· 21 31 = 1 · (−1) − 2 · (−5) + 3 · (−1) = 6. Properties of Determinants We have seen that in order to compute the determinant we must study each of the permutations σ ∈ Sn . Since this set has n! elements, this would be a tedious task. Indeed, to compute even a 5 × 5 determinant we must study 5! = 120 permutations. We seek a simpler way to compute determinants by consider the reduced row echelon form for a matrix. Transpose. We begin by discussing several basic properties of matrices. Let A be an n × n matrix. We will show that det(A) = det(AT ). To see why, write A = [aij ], so that AT = [aji ]. (Remember that rows become columns.) Then by deﬁnition, det(AT ) =
τ ∈Sn ε(τ ) aτ (1) 1 aτ (2) 2 · · · aτ (n) n . Now each permutation τ just gives a rearrangement of S = {1, 2, . . . , n} to {τ (1), τ (2), . . . , τ (n)}. Let σ be that permutation that “undoes” this rearrangement; that is, σ = τ −1 . In particular, if τ (j ) = i, then j = σ (i); which is the same as saying aτ (j ) j = ai σ(i) . Recall that ε(σ ◦ τ ) = ε(σ ) · ε(τ ). Since σ ◦ τ = 1 is just the trivial permutation, it has no inversions so that ε(σ ◦ τ ) = +1. That means ε(τ ) = ε(σ ) have the same sign! That means we have det(AT ) =
τ ∈Sn ε(τ ) aτ (1) 1 aτ (2) 2 · · · aτ (n) n =
τ ∈Sn ε(σ ) a1 σ(1) a2 σ(2) · · · an σ(n) = det(A). 4 MA 265 LECTURE NOTES: MONDAY, FEBRUARY 4 We compute the determinant of its transpose Example. We found before that the 1 A= 2 3 3 × 3 matrix 23 1 3 has determinant 12 123 AT = 2 1 1 . 332 A = 6. Consider the following diagram: ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? 1? 2? 3 ? ?? ? ?? 2 1? 1 ??? 3 3 2 This gives the expansion ?? 1 ? ? ?? 2 ? ? ?3 ? ? ? ? 2 ?? 1 ?? ?3 ? ? ? ? ? ? ? ? ? AT = + (1 · 1 · 2) + (2 · 1 · 3) + (3 · 2 · 3) − (3 · 1 · 3) − (1 · 1 · 3) − (2 · 2 · 2) = 2 + 6 + 18 − 9 − 3 − 8 = 6. Hence det(A) = det(AT ). Type I: Interchanging Rows and Columns. Let A be an n × n matrix. We explain how det(A) changes if we apply an elementary operation of Type I. (1) Let B be that matrix obtained from A by interchanging the ith and j th rows. Then det(B ) = − det(A). (2) Let B be that matrix obtained from A by interchanging the ith and j th columns. Then det(B ) = − det(A). (3) Say that A has two equal rows. Then det(A) = 0. (4) Say that A has two equal columns. Then det(A) = 0. We explain why these statements are true. We begin by showing (1). We write the determinant of A as det(A) =
σ ∈Sn ε(σ ) a1 σ(1) a2 σ(2) · · · an σ(n) . Let B be that matrix obtained from A by interchanging the ith and j th rows. If we write the ith and j th rows of A as ri = ai1 ai2 ··· ain and rj = aj 1 aj 2 ··· ajn ; we see that the elements ai σ(i) and aj σ(j ) in the determinant above for A interchange to ai σ(j ) and aj σ(i) in the determinant below for B . Let τ be that permutation that interchanges i ↔ j and leaves all other integers ﬁxed. Then σ (j ) = (σ ◦ τ ) (i) and σ (i) = (σ ◦ τ ) (j ) with σ (k ) = (σ ◦ τ ) (k ) for all other k = i, j ; so that we can express the determinant of B as det(B ) =
σ ∈Sn ε(σ ) a1 (σ◦τ )(1) a2 (σ◦τ )(2) · · · an (σ◦τ )(n) . MA 265 LECTURE NOTES: MONDAY, FEBRUARY 4 5 Recall that ε(σ ◦ τ ) = ε(σ ) · ε(τ ). It is easy to see that τ is an inversion, so that ε(τ ) = −1. This gives det(B ) =
σ ∈Sn (−1) · ε(σ ◦ τ ) a1 (σ◦τ )(1) a2 (σ◦τ )(2) · · · an (σ◦τ )(n) ε(σ ◦ τ ) a1 (σ◦τ )(1) a2 (σ◦τ )(2) · · · an (σ◦τ )(n)
σ ∈Sn = (−1) = − det(A). Now we prove the other statements. First we show (2). Let B be that matrix obtained from A by interchanging the ith and j th columns. Then B T is that matrix obtained from AT by interchanging the ith and j th rows. From (1), we see that det(B ) = det(B T ) = − det(AT ) = − det(A). Now we show (3). Say that A has two equal rows. Let B be that matrix obtained from A by interchanging these rows; then B = A. But from (1), we see that det(A) = det(B ) = − det(A), so that det(A) = 0. Finally we show (4). Say that A has two equal columns. Let B be that matrix obtained from A by interchanging these columns; then B = A. But from (1), we see that det(A) = det(AT ) = det(B T ) = − det(AT ) = − det(A), so that again det(A) = 0. Examples. Consider the following 2 × 2 matrices: A= 2 −1 3 2 and B= 3 2 2 −1 . Note that B can be obtained from A by interchanging the ﬁrst and second rows. We compute the determinants det(A) = (2) (2) − (3) (−1) = 7 =⇒ det(B ) = − det(A). det(B ) = (3) (−1) − (2) (2) = −7 Now consider the following 3 × 3 matrix: 123 A = −1 0 7 . 123 Note that its ﬁrst and third rows are equal. We may compute the determinant via the following diagram: ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? 1? 2? 3? 1 2 ? ? ?? ? ? ? ?? −1 −1 0 7 0 ?? ??? ? ? ?? ?? ? 1 2 3 1 2 ? ? ? ?? ?? ?? ? ? ? ? ? ? This gives the expansion det(A) = + (1 · 0 · 3) + (2 · 7 · 1) + (3 · (−1) · 2) − (3 · 0 · 1) − (1 · 7 · 2) − (2 · (−1) · 3) = 0 + 14 + (−6) − 0 − 14 − (−6) = 0. ...
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