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Review2 - T with respect to A is given by T A = T f 1 A T f...

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Determinants A = a 11 a 12 · · · a 1 n a 21 a 22 · · · a 2 n . . . . . . . . . . . . a n 1 a n 2 · · · a nn Recall that A i,j is the ( n - 1) × ( n - 1) matrix that is obtained from A by removing the i -th row and j -th column. To compute the determinant inductively, pick a row, say the i -th row, the determinant of A is given by 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 i,n det( A i,n ) . Can also pick a column, say the j -th column, 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 n,j det( A n,j ) . It is useful to think of the pattern + - + · · · - + - · · · + - + · · · . . . . . . . . . . . . to remember which sign to start with. Inverse of a matrix If A is an invertible n × n matrix, then A - 1 = 1 det( A ) + det ( A 11 ) - det( A 12 ) · · · ( - 1) 1+ n det( A 1 n ) - det( A 21 ) + det( A 22 ) · · · ( - 1) 2+ n det( A 2 n ) . . . . . . . . . . . . ( - 1) n +1 det( A n 1 ) ( - 1) n +2 det( A n 2 ) · · · ( - 1) n + n det( A nn ) T Matrix of a Linear Transformation and change of basis. Let V be an n -dimensional linear space, and let A = ( f 1 , f 2 , . . . , f n ) be a basis for
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Unformatted text preview: T with respect to A is given by [ T ] A = [ [ T ( f 1 )] A [ T ( f 2 )] A ··· [ T ( f n )] A ] . If B = ( g 1 ,g 2 ,...,g n ) is another basis for V , then we have the change of basis matrix S A → B = ± [ f 1 ] B [ f 2 ] B ··· [ f n ] B ² . Notice that ( S A → B )-1 = S B → A , and [ T ] B = S A → B [ T ] A S B → A . Review Problems (Optional, but recommended) • From the textbook: p. 184: 1, 3, 4, 8, 9, 10, 15, 20, 37, 42, 50, 51, 64; p. 209: 3, 6, 35; p. 273: 3, 5, 6, 8 (compute the determinants); p. 5: 3, 6, 10 (solve the systems of equations using Cramer’s rule); p. 292: 1, 2, 5, 6, 19, 20, 40; p. 163: 9, 25, 26, 47; p. 170: 12, 17, 51, 73; p. 181: 7, 13, 18, 41, 57....
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