# lecture_12 - MA 265 LECTURE NOTES MONDAY FEBRUARY 4...

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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: a 11 a 12 ··· a 1 n a 21 a 22 ··· a 2 n . . . . . . . . . . . . a n 1 a n 2 ··· a nn = X σ ∈ S n ε ( σ ) a 1 σ (1) a 2 σ (2) ··· a nσ ( 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 det[ a ] = a and a b c d = ad- bc. (Be careful not to confuse this with absolute value!) Example. We compute the determinant of a 2 × 2 matrix: 2- 3 4 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: σ 1 : 1 7→ 1 2 7→ 2 3 7→ 3 σ 2 : 1 7→ 2 2 7→ 3 3 7→ 1 σ 3 : 1 7→ 3 2 7→ 1 3 7→ 2 σ 4 : 1 7→ 1 2 7→ 3 3 7→ 2 σ 5 : 1 7→ 2 2 7→ 1 3 7→ 3 σ 6 : 1 7→ 3 2 7→ 2 3 7→ 1 We showed in the previous lecture that the first three have signs ε ( σ 1 ) = ε ( σ 2 ) = ε ( σ 3 ) = +1, and the last three have signs ε ( σ 4 ) = ε ( σ 5 ) = ε ( σ 6 ) =- 1. Using the definition of the determinant, the 3 × 3 matrix A = a 11 a 12 a 13 a 21 a 22 a 23 a 31 a 32 a 33 1 2 MA 265 LECTURE NOTES: MONDAY, FEBRUARY 4 has determinant det( A ) = ε ( σ 1 ) a 11 a 22 a 33 + ε ( σ 2 ) a 12 a 23 a 31 + ε ( σ 3 ) a 13 a 21 a 32 + ε ( σ 4 ) a 11 a 23 a 32 + ε ( σ 5 ) a 12 a 21 a 33 + ε ( σ 6 ) a 13 a 22 a 31 = ( a 11 a 22 a 33 + a 12 a 23 a 31 + a 13 a 21 a 32 )- ( a 11 a 23 a 32 + a 12 a 21 a 33 + a 13 a 22 a 31 ) ....
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lecture_12 - MA 265 LECTURE NOTES MONDAY FEBRUARY 4...

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