Sol4 - SOLUTION TO ASSIGNMENT 4 - MATH 251, WINTER 2007 1....

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1. Deduce from the theorems on determinants the following: (1) If a column is zero, the determinant is zero. (2) det( A ) = det( A t ), where A t is the transposed matrix. (3) If a row is zero, the determinant is zero. (4) Let A be a matrix in “block form”: A = A 1 ? ? 0 A 2 . . . 0 0 A k . Here each A i is a square matrix say of size r i , and A 2 starts at the r 1 + 1 column and r 1 + 1 row, etc. Prove that det( A ) = det( A 1 ) det( A 2 ) ··· det( A k ) . Conclude that the determinant of a triangular matrix is given by det a 11 ? 0 a 22 . . . 0 0 a kk = a 11 a 22 ··· a kk . (Here each a ii is a scalar). Proof. (1) Let A be a matrix such that its i -th column v i is zero. Then det( A ) = det( v 1 | . . . | v i | . . . | v n ) = det( v 1 | . . . | v i + v i | . . . | v n ) = det( v 1 | . . . | v i | . . . | v n ) + det( v 1 | . . . | v i | . . . | v n ) = 2 det( A ) . The claim follows. (2) Let A = ( a ij ) be a square n × n matrix. Note that for every two permutations σ, τ we have a σ (1)1 ··· a σ ( n ) n = a σ ( τ (1)) τ (1) ··· a σ ( τ ( n )) τ ( n ) . We apply that for τ = σ - 1 and obtain det( A ) = X σ S n sgn ( σ ) a σ (1)1 ··· a σ ( n ) n = X σ S n sgn ( σ ) a 1 σ - 1 (1) ··· a - 1 ( n ) = X σ S n sgn ( σ - 1 ) a 1 σ - 1 (1) ··· a - 1 ( n ) = X σ S n sgn ( σ ) a 1 σ (1) ··· a ( n ) . (We used sgn ( σ ) = sgn ( σ - 1 ) and that summing over all σ S n is the same as summing over all σ - 1 S n ). It remains to note that if we let A t = ( b ij ) (with b ij = a ji ) then X σ S n sgn ( σ ) a 1 σ (1) ··· a ( n ) = X σ S n sgn ( σ ) b σ (1)1 ··· b σ ( n ) n = det( A t ) . (3) Apply (1) and (2)!
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Sol4 - SOLUTION TO ASSIGNMENT 4 - MATH 251, WINTER 2007 1....

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