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Instructor: Irvin Roy Hentzel Office 432 Carver Phone 515-294-8141
Math 307 Spring, 2003 Hentzel Time: 1:10-2:00 MWF Room: 1324 Howe Hall Instructor: Irvin Roy Hentzel Office 432 Carver Phone Text: Linear Algebra With Applications, Second Edition Otto Bretscher
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Friday, April 4 Chapter 6.2 Page 266 Problem 4,14,46,48
Main Idea: Expand along any row or column. Key Words: Adjoint, Aij Det[A] = SUM sgn(p) a 1 p(1) a 2 p(2) a n p(n) all p Goal: Learn how to expand a determinant.
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Theorem: If A n x n is any square matrix, then
Det[A] = a11 Det[A11] - a12 Det[A12]+...+(-1)n+1 Det[A1n] Where Aij is the n-1 x n-1 matrix obtained by deleting row i and column j from A.
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Proof Det [A] = SUM sgn(p) a 1 p(1) a 2 p(2) ... a n p(n) all p There are n possible choices for a 1 p(1) . One can choose any element from the top row. The possibilities are a11, a12, ... a1n.
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We split up the sum into n parts depending on which element was chosen from the top row.
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SUM sgn(p) a 1 p(1) a 2 p(2) …a n p(n)
Det[A] = SUM sgn(p) a 1 p(1) a 2 p(2) …a n p(n) p (1)=1 + SUM sgn(p) a 1 p(1) a 2 p(2) .... a n p(n) p(1)=2 + SUM sgn(p) a 1 p(1) a 2 p(2) … a n p(n) p(1)= : p(1)=n
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Det[A] = SUM sgn(p) a 11 a 2 p(2) ... a n p(n) p(1) = 1 + SUM sgn(p) a 12 a 2 p(2) ... a n p(n) p(1) = 2 + SUM sgn(p) a 13 a 2 p(2) ... a n p(n) p(1) = 3 : + SUM sgn(p) a 1n a 2 p(2) ... a n p(n) p(1) = n
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Det[A] = a11 SUM sgn(p) a 2 p(2) ... a n p(n) p(1) = 1 + a12 SUM sgn(p) a 2 p(2) ... a n p(n) p(1) = 2 + a13 SUM sgn(p) a 2 p(2)... a n p(n) p(1) = : + a1n SUM sgn(p) a 2 p(2) ... a n p(n) p(1) = n
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These SUMS are close to Det[Aij].
The only difference is that the sgn(p) is based on all n positions of p(1) p(2) p(3) ... p(n). In Det [A ij] the sgn is based on n-1 positions of p(2) p(3) ... p(n).
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Det[A] = a11 Det[A11] - a12 Det[A12] + a13 Det[A13] . (-1)n+1 a1n Det[A1n]
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n Theorem: Det[A] = Sum (-1) i+j aij Det[Aij] j=1 Proof: This is expansion by the i th row.
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We first interchange rows i-1 times to move the ith row to the first row. Then we expand along the first row. This gives | ai1 Det[Ai1] | | ai2 Det[Ai2] | Det[A]=(-1)i-1 | ai3 Det[Ai3] | | : | | (-1)n+1 ain Det[Ain] |
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This can be written as Det[A] = SUM (-1) i+j-2 aij Det[Aij] Which can be simplified since (-1)2 = 1 to Det[A] = SUM (-1)i+j aij Det[Aij]
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Definition: Given a matrix A, then
|+Det [A11] - Det [A21] +Det [A31] ....| |-Det [A12] + Det [A22] -Det [A32] ....| Adj(A) =|+Det [A13] - Det [A23] +Det [A33] ....| |-Det [A14] + Det [A24] -Det [A34] ....| | |
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Theorem: A Adj(A) = Adj(A) A = Det[A] I
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Find the inverse of | | | | | |
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check |3 1 7| | -7 (-5) -1| T | | | | Adj|1 2 3| =|(-20)-11 (7)| = | | | | |2 3 1| | -11 (2) 5 | | | | |
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| | | | | | = -1/23 | | | | | |
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Theorem: Det[A] = Det[AT]
Proof: Since (A-1)T = (AT) -1 If A is invertible, then A is a product of Elementary Row Operation Matrices. For Elementary Row Operation Matrices Det [ET] = Det [E], we have Det [A] = Det[AT]. If A is not invertible, then neither is AT and Det [A] = Det[AT] = 0.
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Theorem: The determinant of a linear transformation is invariant under change of basis.
Proof: Det [ P -1A P] = Det[P-1] Det [A] Det [P] = Det[ P-1 P] Det [A] = Det [A].
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Find the Determinant of differentiation on the space with basis Sinh[x] Cosh[x].
Det[ derivation ] = -1.
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Find the Determinant of differentiation on the space with basis e x, e -x.
Det [ derivation ] = -1.
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Page 263 Example 8. A = | | | | | | | | Find the determinant of A.
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Add -2 Row 2 to row 3 | | | | | | | |
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Expand along second column
| | +1 | | | | | |
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Add 4 Row 1 to Row 2 | | +1 | | | | | |
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Expand along column 2 | | (-1) | | | | | | Evaluate the final 2 x 2 determinant. -( ) = 55.
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Page 265 Problem 5. | | | | A = | | | | | | Find the determinant of A.
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Eliminate the first column
| | | | A = | | | | | | The determinant of a triangular matrix is the product of the elements on the diagonal. Det [A] = 24.
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