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lecture_12- Determinants Part 2 and Properties of Determinants Part 1

# Lecture_12- Determinants Part 2 and Properties of Determinants Part 1

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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 right-ward arrows correspond to “+” and the left-ward 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 right-ward arrows correspond to “+” and the left-ward 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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