lecture_13- Properties of Determinants Part 2 and Cramer's Rule Part 1

# Lecture_13- Properties of Determinants Part 2 and Cramer's Rule Part 1

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

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 . ...
View Full Document

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern