lesson 6 - introduction to determinants (slides+notes)

Lesson 6Introduction to Determinants (Section 13.1–2)

Math 20

October 1, 2007

Announcements

I Thomas Schelling at IOP (79 JFK Street), Wednesday 6pm

I Problem Set 2 is on the course web site. Due October 3

I Sign up for conference times on course website

I Problem Sessions: Sundays 6–7 (SC 221), Tuesdays 1–2 (SC116)

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G], A[, and the Euclidean algorithm9:15pm, Tuesday, October 2, at the SOCH (Student Organization Center at Hilles)

Free coffee, tea, and refreshments. No special mathematics ormusic knowledge required! Contact shlo@fas with any questions.

Consider the system of two equations in two variables:

a11x1+a12x2 =b1

a21x1+a22x2 =b2

Can you find the solutions for x1 and x2 in terms of thecoefficients?

Math 20 - October 01, 2007.GWBTuesday, Oct 2, 2007

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Solutions

The solutions are

x1 =a22b1 − a12b2

a11a22 − a21a12

x2 =a11b2 − a21b1

a11a22 − a21a12

Observations

I If a11a22 − a21a12 6= 0, the system has a unique solution.

I If a11a22 − a21a12 = 0 and a22b1 − a12b2 6= 0, the system hasno solution

I If a11a22 − a21a12 = 0 and a22b1 − a12b2 = 0, the system hasinfinitely many solutions

Math 20 - October 01, 2007.GWBTuesday, Oct 2, 2007

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Solutions

The solutions are

x1 =a22b1 − a12b2

a11a22 − a21a12

x2 =a11b2 − a21b1

a11a22 − a21a12

Observations

I If a11a22 − a21a12 6= 0, the system has a unique solution.

I If a11a22 − a21a12 = 0 and a22b1 − a12b2 6= 0, the system hasno solution

I If a11a22 − a21a12 = 0 and a22b1 − a12b2 = 0, the system hasinfinitely many solutions

The determinant

Definition

The determinant of a 2× 2 matrix A =

(a11 a12

a21 a22

)is the number

∣∣∣∣a11 a12

a21 a22

∣∣∣∣ = a11a22 − a21a12

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Math 20 - October 01, 2007.GWBTuesday, Oct 2, 2007

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Theorem (Cramer’s rule)

The solution to the system of linear equations

a11x1+a12x2 =b1

a21x1+a22x2 =b2

is

x1 =a22b1 − a12b2

a11a22 − a21a12=

∣∣∣∣b1 a12

b2 a22

∣∣∣∣∣∣∣∣a11 a12

a21 a22

∣∣∣∣x2 =

a11b2 − a21b1

a11a22 − a21a12=

∣∣∣∣a11 b1

a21 b2

∣∣∣∣∣∣∣∣a11 a12

a21 a22

∣∣∣∣

Leontief

Example

On the first day we had to solve the system

0.6x1 − 0.3x2 = 75000

−0.2x1 + 0.7x2 = 50000

Solution

x1 =67, 500

0.36= 187, 500

x2 =45, 000

0.36= 125, 000

Math 20 - October 01, 2007.GWBTuesday, Oct 2, 2007

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Leontief

Example

On the first day we had to solve the system

0.6x1 − 0.3x2 = 75000

−0.2x1 + 0.7x2 = 50000

Solution

x1 =67, 500

0.36= 187, 500

x2 =45, 000

0.36= 125, 000

Math 20 - October 01, 2007.GWBTuesday, Oct 2, 2007

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Math 20 - October 01, 2007.GWBTuesday, Oct 2, 2007

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The 3× 3 case

Does anybody want to do:

a11x1+a12x2+a13x3 =b1

a21x1+a22x2+a23x3 =b2

a31x1+a32x2+a33x3 =b3

We get

x1 =

b1a22a33 − b1a23a32 − b2a12a33

+ b2a13a32 + b3a12a23 − b3a22a13

a11a22a33 − a11a23a32 − a21a12a33

+ a21a13a32 + a31a12a23 − a31a22a13

The 3× 3 case

Does anybody want to do:

a11x1+a12x2+a13x3 =b1

a21x1+a22x2+a23x3 =b2

a31x1+a32x2+a33x3 =b3

We get

x1 =

b1a22a33 − b1a23a32 − b2a12a33

+ b2a13a32 + b3a12a23 − b3a22a13

a11a22a33 − a11a23a32 − a21a12a33

+ a21a13a32 + a31a12a23 − a31a22a13

DefinitionThe determinant of a 3× 3 matrix is∣∣∣∣∣∣

a11 a12 a13

a21 a22 a23

a31 a32 a33

∣∣∣∣∣∣ = a11a22a33 − a11a23a32 − a21a12a33

+ a21a13a32 + a31a12a23 − a31a22a13

Cofactors

We can compute a 3× 3 determinant in terms of smallerdeterminants:∣∣∣∣∣∣

a11 a12 a13

a21 a22 a23

a31 a32 a33

∣∣∣∣∣∣ = a11

∣∣∣∣a22 a23

a32 a33

∣∣∣∣− a12

∣∣∣∣a21 a23

a31 a33

∣∣∣∣ + a13

∣∣∣∣a21 a22

a31 a32

∣∣∣∣

Theorem (Cramer’s rule)

The solution to the 3× 3 system

a11x1+a12x2+a13x3 =b1

a21x1+a22x2+a23x3 =b2

a31x1+a32x2+a33x3 =b3

is

x1 =

∣∣∣∣∣∣b1 a12 a13

b2 a22 a23

b3 a32 a33

∣∣∣∣∣∣∣∣∣∣∣∣a11 a12 a13

a21 a22 a23

a31 a32 a33

∣∣∣∣∣∣, x2 =

∣∣∣∣∣∣a11 b1 a13

a21 b2 a23

a31 b3 a33

∣∣∣∣∣∣∣∣∣∣∣∣a11 a12 a13

a21 a22 a23

a31 a32 a33

∣∣∣∣∣∣, x3 =

∣∣∣∣∣∣a11 a12 b1

a21 a22 b2

a31 a32 b3

∣∣∣∣∣∣∣∣∣∣∣∣a11 a12 a13

a21 a22 a23

a31 a32 a33

∣∣∣∣∣∣

Example

Solve

x1+2x2+3x3 = 6

2x1−3x2+2x3 = 14

3x1+ x2− x3 =−2

Solutionx1 = 50

50 = 1, x2 = −10050 = −2, x3 = 150

50 = 3

Example

Solve

x1+2x2+3x3 = 6

2x1−3x2+2x3 = 14

3x1+ x2− x3 =−2

Solutionx1 = 50

50 = 1, x2 = −10050 = −2, x3 = 150

50 = 3

A geometric interpretation