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# X 4 3 2 7 1 1 5 3 1 4 x y z ax b example 3 contd

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x 4 3 2 7 1 1 0 5 3 0 1 4 x y z                     Ax b

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Example 3 contd Method Find: 1. | A | 2. all the cofactors 3. adj( A ) 4. A -1 5. Solve for x, y & z. 13
Example 3 - determinant Det( A ) 14 2 1 2 2 2 3 3 2 4 2 4 3 1 ( 1) 1 ( 1) 0 ( 1) 0 1 3 1 3 0 1 1 0 4 ( 6) 3 0 7              A 3 1 3 2 3 3 3 2 4 2 4 3 3 ( 1) 0 ( 1) 1 ( 1) 1 0 1 0 1 1 3 0 1 0 ( 2) 4 3 7            A 1 1 n n ij ij ij ij i j a a C C A Expanding along Row 2: Expanding along Row 3:

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Example 3 - cofactors 15 1 1 11 1 0 1 1 0 1 C 3 12 1 0 1 1 3 1 C   4 13 1 1 3 1 3 0 C   5 23 4 3 9 1 3 0 C 4 22 4 2 10 1 3 1 C 3 21 3 2 3 1 0 1 C   6 33 4 3 1 1 1 1 C 5 32 4 2 2 1 1 0 C   4 31 3 2 2 1 1 0 C
Example 3 adjoint and inverse Matrix of cofactors Adjoint of A 16 1 1 3 3 10 9 2 2 1 C 1 3 2 Adj ( ) 1 10 2 3 9 1 A C -1 1 3 2 Adj 1 = 1 10 2 7 3 9 1 A A A

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Example 3 - solution 17 -1 1 3 2 7 1 1 10 2 5 7 3 9 1 4 x y z                x A b 1 0 1 7 ( 3) 5 2 4 7 x   1 5 1 7 10 5 ( 2) 4 7 y     1 4 3 7 9 5 1 4 7 z    
Expansion by Alien Cofactors What happened if we multiply cofactors by the wrong (alien) row? If A = ( a ij ) 3 3 then expanding along row 2 gives Replacing the elements from row 2 with another row, say row 1, is equivalent to having a matrix with two equal rows. 18 21 21 22 22 23 23 a C a C a C A 11 11 12 13 31 32 3 12 3 13 a a a a a a a a a

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Expansion by Alien Cofactors We know: Theorem 16.5.1 Let A = ( a ij ) n n with cofactors C ij Then: (i) (ii) and similarly for expanding along columns. 19 11 21 12 22 13 23 0 a C a C a C 1 1 2 2 i i i i in in a C a C a C A 1 1 2 2 if , 0 i k i k in kn k i a C a C a C
Deriving formula: Let A = ( a ij ) n n with cofactors C ij Equivalently: 20 1 1 ( ) | | adj A Α A 1 1 2 2 , if 0, if i k i k in kn i k a C a C a C i k   A 21 1 21 22 2 22 2 1 2 11 12 1 2 1 11 2 1 0 0 0 0 0 0 n n n n n nn n nn n n C C a a a C C a a a a C C a a C C C             A A A

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