equal. It is left as an exercise to the reader to verify that the values of |A| by expanding along R , C and C are equal to the value of |A| obtained in ( ), ( ) or ( ). Hence, expanding a determinant along any row or column gives same value. Remarks (i) For easier calculations, we shall expand the determinant along that row or column which contains maximum number of zeros.
(ii) While expanding, instead of multiplying by (– ) i + j , we can multiply by + or – according as ( i + j ) is even or odd. (iii) Let A = and B = . Then, it is easy to verify that A = 2B. Also |A| = – = – and |B| = – = – .
Observe that, |A| = (– ) = |B| or |A| = n |B|, where n = is the order of square matrices A and B. In general, if A = k B where A and B are square matrices of order n , then | A| = k n | B |, where n = , , Example Evaluate the determinant ∆ = . Solution Note that in the third column, two entries are zero. So expanding along third column (C ), we get ∆ = – = (– – ) – + = – Example Evaluate ∆ = –cos –sin –sin .
DETERMINANTS Solution Expanding along R , we get ∆ = –sin –sin – sin – cos –sin –sin = – sin α ( – sin β cos α ) – cos α (sin α sin β – ) = sin α sin β cos α – cos α sin α sin β = Example Find values of x for which . Solution We have i.e. – x = – i.e. x = Hence x = ± EXERCISE .