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Unformatted text preview: MA 265 LECTURE NOTES: WEDNESDAY, FEBRUARY 6 Properties of Determinants (cont’d) Type II: Multiplying Rows and Columns. Let A be an n × n matrix. We explain how det(A) changes if we apply an elementary operation of Type II. (1) (2) (3) (4) Let B be that matrix obtained from A by multiplying the ith row by k . Then det(B ) = k det(A). Let B be that matrix obtained from A by multiplying the ith column by k . Then det(B ) = k det(A). Say that A has a row of zeros. Then det(A) = 0. Say that A has a column of zeros. 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 multiplying the ith row by k . If we write the ith row of A as ri = ai1 ai2 ··· ain =⇒ k ri = k ai1 k ai2 ··· k ain ; We can express the determinant of B as det(B ) =
σ ∈Sn ε(σ ) a1 σ(1) · · · ai−1 σ(i−1) k ai σ(i) ai+1 σ(i+1) · · · an σ(n) ε(σ ) a1 σ(1) · · · ai−1 σ(i−1) ai σ(i) ai+1 σ(i+1) · · · an σ(n)
σ ∈Sn =k = k det(A). Now we prove the other statements. First we show (2). Let B be that matrix obtained from A by multiplying the ith column by k . Then B T is that matrix obtained from AT by multiplying the ith row by k . From (1), we see that det(B ) = det(B T ) = k det(AT ) = k det(A). Now we show (3). Say that A has a row of zeros. Let B be that matrix obtained from A by multiplying this row by k = 0; then B = A. But from (1), we see that det(A) = det(B ) = 0 · det(A), so that det(A) = 0. Finally we show (4). Say that A has a column of zeros. Let B be that matrix obtained from A by multiplying this column by k = 0; then B = A. But from (1), we see that det(A) = det(AT ) = det(B T ) = 0 · det(AT ), so that again det(A) = 0. Examples. Consider the following 3 × 3 matrix: We may compute the determinant via the following diagram: ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? 1? 2? 3 ?? ?? ? ? 1 5 3 ?? ?? 2 8 6
1 123 A = 1 5 3 . 286 ?? 1 ? ? ?? 1 ? ? ?2 ? ? ? ? 2 ?? 5 ?? ?8 ? ? ? ? ? ? ? ? ? 2 MA 265 LECTURE NOTES: WEDNESDAY, FEBRUARY 6 This gives the expansion det(A) = + (1 · 5 · 6) + (2 · 3 · 2) + (3 · 1 · 8) − (3 · 5 · 2) − (1 · 3 · 8) − (2 · 1 · 6) = 30 + 12 + 24 − 30 − 12 − 12 = 0. There is an easier way to arrive at this result. We can also factor out 2 from the last row, and 3 from the last column: det(A) = 1 1 2 23 53 86 123 = (2) 1 5 3 143 121 = (2) (3) 1 5 1 141 Now the ﬁrst and third columns are equal, so det(A) = 0. Type III: Adding Multiples of Rows and Columns. Let A be an n × n matrix. We explain how det(A) changes if we apply an elementary operation of Type III. (1) Let B be that matrix obtained from A by adding a multiple of the j th row to the ith row. Then det(B ) = det(A). (2) Let B be that matrix obtained from A by adding a multiple of the j th column to the ith column. Then det(B ) = det(A). 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 adding a multiple of the j th row by to the ith row If we write the ith row of A as ri = ai1 ai2 ··· ain =⇒ k rj + ri = ai1 + k aj 1 ai2 + k aj 2 ··· ain + k ajn ; We can express the determinant of B as det(B ) =
σ ∈Sn ε(σ ) a1 σ(1) · · · ai−1 σ(i−1) ai σ(i) + k aj σ(i) ai+1 σ(i+1) · · · an σ(n) ε(σ ) a1 σ(1) · · · ai−1 σ(i−1) ai σ(i) ai+1 σ(i+1) · · · an σ(n)
σ ∈Sn = +k
σ ∈Sn ε(σ ) a1 σ(1) · · · ai−1 σ(i−1) aj σ(i) ai+1 σ(i+1) · · · an σ(n) = det(A) + k det(C ) Here, C is that matrix obtained by A with the ith and j th rows equal. We know from before that det(C ) = 0, so we ﬁnd that det(B ) = det(A). Now we prove the other statements First we show (2). Let B be that matrix obtained from A by adding a multiple of the j th column the ith column. Then B T is that matrix obtained from AT by adding a multiple of the j th row to the the ith row. From (1), we see that det(B ) = det(B T ) = det(AT ) = det(A). Example. Consider the following 3 × 3 matrix: 4 32 A = 3 −2 5 . 2 46 MA 265 LECTURE NOTES: WEDNESDAY, FEBRUARY 6 3 We explain how to compute its determinant by performing a series of elementary row operations. Consider the following steps: (1/2) r3 → r3 as a Type II operation: 4 det(A) = (2) 3 1 1 det(A) = (−2) 3 4 1 det(A) = (−2) 0 0 1 det(A) = (+2) 0 0 32 −2 5 23 23 −2 5 32 2 3 −8 −4 −5 −10 2 3 −5 −10 −8 −4 r1 ↔ r3 (−3) r1 + r2 → r2 (−4) r1 + r3 → r3 r2 ↔ r3 as a Type I operation: as a Type III operation: as a Type I operation: (−1/5) r2 → r2 (−2) r2 + r1 → r1 8 r2 + r3 → r3 (1/12) r3 → r3 r1 + r1 → r1 (−2) r3 + r2 → r2 as a Type II operation: 1 2 3 1 2 det(A) = (−10) 0 0 −8 −4 10 det(A) = (−10) 0 1 00 10 det(A) = (−120) 0 1 00 −1 2 12 −1 2 1 as a Type III operation: as a Type II operation: as a Type III operation: 100 det(A) = (−120) 0 1 0 001 It is easy to check that det(I3 ) = 1, so that det(A) = −120. In general, we will see in the next lecture that det(In ) = 1. Relationship with Invertibility. Let A be an n × n matrix. Then A is nonsingular if and only if det(A) = 0. We explain why this is true. First assume that A is nonsingular. Then A is row equivalent to the n × n identity matrix In . That is, In = E A where E = E1 E2 · · · Ek is the product of elementary matrices. We have just shown that det(In ) = (det(E1 ) det(E2 ) · · · det(Ek )) det(A) in terms of −1 if Ei is of Type I; det(Ei ) = ki if Ei is of Type II; and +1 if Ei is of Type III. Since det(Ei ) = 0 and det(In ) = 1, we see that det(A) = 0 as well. In fact, we have just shown that det(A) = 1 1 1 ··· . det(E1 ) det(E2 ) det(Ek ) Now assume the converse, that det(A) = 0. We must show that A is nonsingular. Let B be the reduced row echelon form for A. That is, B = E A where E = E1 E2 · · · Ek is the product of elementary matrices. Then det(B ) = (det(E1 ) det(E2 ) · · · det(Ek )) det(A) = 0. If B had a row of zeros then det(B ) = 0, so B must be the identity matrix. This shows that In = E A, so that A−1 = E . That is, A must be invertible. Cramer’s Rule We have seen that the determinant gives us a quick way to determine when a matrix is invertible. We explain how the determinant can also help us to compute the inverse. 4 MA 265 LECTURE NOTES: WEDNESDAY, FEBRUARY 6 Cofactors. Recall that in the previous lecture we computed the following formula for a 3 × 3 determinant: a11 a21 a31 a12 a22 a32 a13 a23 a33 = a11 · a22 a32 a23 a33 − a12 · a21 a31 a23 a33 + a13 · a21 a31 a22 a32 . We wish to generalize this formula. To this end, we make the following deﬁnitions. Let A be an n × n matrix: a11 · · · a1j · · · a1n . . . . .. . . . . . . . . . A = ai1 · · · aij · · · ain . . . . . .. . . . . . . . . . an1 · · · anj · · · ann Deﬁne the minor Mij as the (n − 1) × (n − 1) matrix with the ith row and j th columns removed: a11 . . . ai1 . . . an1 ··· .. . a1j . . . ··· .. . a1n . . . ain . . . ann @ABC aij · · · GFED · · · .. . . . . anj .. . ··· ··· The cofactor of aij is the determinant Aij = (−1)i+j det(Mij ). As an example, when A is a 3 × 3 matrix, we have the cofactors A11 = + a22 a32 a23 a33 , A12 = − a21 a31 a23 a33 , and A13 = + a21 a31 a22 a32 . ...
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