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# lecture_14 - MA 265 LECTURE NOTES FRIDAY FEBRUARY 8...

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MA 265 LECTURE NOTES: FRIDAY, FEBRUARY 8 Computing Determinants Determinants of Special Matrices. Let A be an n × n matrix. (1) Assume that A is upper triangular: A = a 11 a 12 · · · a 1 n a 22 · · · a 2 n . . . . . . a nn . Then det( A ) = a 11 a 22 · · · a nn is the product of the elements on the main diagonal. (2) Assume that A is lower triangular: A = a 11 a 21 a 22 . . . . . . . . . a n 1 a n 2 · · · a nn . Then det( A ) = a 11 a 22 · · · a nn is the product of the elements on the main diagonal. (3) Assume that A is diagonal: A = a 11 a 22 . . . a nn . Then det( A ) = a 11 a 22 · · · a nn is the product of the elements on the main diagonal. In particular, the n × n identity matrix I n is a diagonal matrix with a ii = 1, so it has determinant det( I n ) = 1. We show why these are true. First we show (1). Let A = [ a ij ] be an upper triangular matrix. Then a ij = 0 whenever i > j . The determinant is the sum det( A ) = X σ S n ε ( σ ) a 1 σ (1) a 2 σ (2) · · · a n σ ( n ) . Since a ij = 0 whenever i > j , we see that the product in the sum is zero whenever i > σ ( i ). This means the only term than remains corresponds to the permutation σ = 1 2 · · · n , so that the formula det( A ) = a 11 a 22 · · · a nn follows. Now we show the other statements. We show (2). Let A be a lower triangular matrix. Then A T is an upper triangular matrix with the same elements on the mail diagonal. The result follows from (1) because det( A ) = det( A T ). We show (3). Let A be a diagonal matrix. The results follows from (1) because a diagonal matrix is a special case of an upper triangular matrix. Product Rule. Let A and B be n × n matrices. Then det ( A B ) = det( A ) det( B ) . We prove this statement. Let C be the reduced row echelon form for A ; then C = E A for some matrix E = E 1 E 2 · · · E k that is the product of elementary matrices. Recall that the determinant of A changes depending on the Type of elementary row operation of the elementary matrix E i : det( C ) = det( E 1 ) det( E 2 ) · · · det( E k ) det( A ) where det( E i ) =

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