Lesson 6 - Introduction To Determinants (Slides+Notes)
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Transcript of Lesson 6 - Introduction To Determinants (Slides+Notes)
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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)
I My office hours: Mondays 1–2, Tuesdays 3–4, Wednesdays1–3 (SC 323)
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Math 20 - October 01, 2007.GWBTuesday, Oct 2, 2007
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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.
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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?
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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
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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
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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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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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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
∣∣∣∣
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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
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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
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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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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
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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
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DefinitionThe determinant of a 3× 3 matrix is∣∣∣∣∣∣
a11 a12 a13
a21 a22 a23
a31 a32 a33
∣∣∣∣∣∣ = a11a22a33 − a11a23a32 − a21a12a33
+ a21a13a32 + a31a12a23 − a31a22a13
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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
∣∣∣∣
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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
∣∣∣∣∣∣
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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
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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
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A geometric interpretation