MAT1341-L16.Determinantes-VC

# Example 5 det parenleftbigg 1 3 2 7 paren 13 2

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Example 5. det parenleftbigg 1 3 2 7 parenrightbigg = ( 1)(7) (3)(2) = 13 . 2

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Example 6. det 1 3 3 3 5 2 4 4 6 + + = 1 det parenleftbigg 5 2 4 6 parenrightbigg 3 det parenleftbigg 3 2 4 6 parenrightbigg + ( 3) det parenleftbigg 3 5 4 4 parenrightbigg = 1( 30 8) 3(18 ( 8)) 3( 12 ( 20)) = 38 3(26) 3(8) = 140 Example 7. det 2 0 0 1 0 1 3 3 2 3 5 2 4 4 4 6 + + = 2 det 1 3 3 3 5 2 4 4 6 0 det 0 3 3 2 5 2 4 4 6 + 0 det 0 1 3 2 3 2 4 4 6 1 det 0 1 3 2 3 5 4 4 4 = 2(40) 0 + 0 1(48) = 32 Sometimes it is more convenient to compute the determinant of a ma- trix by moving, or expanding, along any row or column of the matrix, instead of moving along the first column. This is what the next theorem is about. 3
Theorem 8 (Cofactor Expansion) . Given an n × n A , the determinant can be computed as follows: (1) Let i be a fixed row number. Then det( A ) = ( 1) i +1 a i 1 det( A i 1 ) + ( 1) i +2 a i 2 det( A i 2 ) + · · · + ( 1) i + n a in det( A in ) (2) Let j be a fixed column number. Then det( A ) = ( 1) j +1 a 1 j det( A 1 j ) + ( 1) j +2 a 2 j det( A 2 j ) + · · · + ( 1) j + n a nj det( A nj ) Remark 9 . The signs ( 1) i + j alternate as we move one spot either horizontally or vertically. + + · · · + + · · · + + · · · . . . . . . Example 10. By expanding along the second row: det 1 3 3 3 5 2 4 4 6 + = ( 3) det parenleftbigg 3 3 4 6 parenrightbigg + (5) det parenleftbigg 1 3 4 6 parenrightbigg (2) det parenleftbigg 1 3 4 4 parenrightbigg = ( 3)( 18 + 12) + (5)( 6 12) (2)(4 + 12) = 18 90 32 = 140 4

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Remark 11 . The method of evaluating a determinant by expanding
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