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Example text
Let is and js , 1 ≤ s ≤ r, r ≤ n, denote r integers such that 1 ≤ i1 < i2 < · · · < ir ≤ n, 1 ≤ j1 < j2 < · · · < jr ≤ n and let n xi = aij ek , 1 ≤ i ≤ n, k=1 r yi = aijt ejt , 1 ≤ i ≤ n, t=1 zi = xi − yi . Then, any vector product is which the number of y’s is greater than r or the number of z’s is greater than (n − r) is zero. ir where the vector product on the right is obtained from (z1 · · · zn ) by replacing zis by yis , 1 ≤ s ≤ r, and the sum extends over all nr combinations of the numbers 1, 2, .
B2n ... . . bnn . 15) 2n Reduce all the elements in the first n rows and the first n columns, at present occupied by the aij , to zero by means of the row operations n Ri = Ri + aij Rn+j , j=1 1 ≤ i ≤ n. 16) 34 3. Intermediate Determinant Theory The result is: c11 c21 ... cn1 An Bn = −1 b11 −1 b21 ... −1 bn1 c12 c22 ... cn2 b12 b22 ... bn2 . . c1n . . c2n ... . . cnn . . b1n . . b2n ... . . bnn . 17) 2n The product formula follows by means of a Laplace expansion. cij is most easily remembered as a scalar product: b1j b cij = ai1 ai2 · · · ain • 2j .
B1n . . b2n ... . . bnn . 17) 2n The product formula follows by means of a Laplace expansion. cij is most easily remembered as a scalar product: b1j b cij = ai1 ai2 · · · ain • 2j . 18) ··· bnj Let Ri denote the ith row of An and let Cj denote the jth column of Bn . Then, cij = Ri • Cj . Hence An Bn = |Ri • Cj |n R1 • C1 R1 • C2 R2 • C1 R2 • C2 = ······ ······ Rn • C1 Rn • C2 · · · R1 • Cn · · · R2 • Cn ··· ······ · · · Rn • Cn . 19) n Exercise. If An = |aij |n , Bn = |bij |n , and Cn = |cij |n , prove that An Bn Cn = |dij |n , where n n dij = air brs csj .