Operations on Determinants
Multiplication of two Determinants
Two determinants can be multiplied together only if they are of same order. The rule of multiplication is as under:
Take the first row of determinant and multiply it successively with 1st, 2nd & 3rd rows of other determinant. The three expressions thus obtained will be elements of 1st row of resultant determinant. In a similar manner the element of 2nd & 3rd rows of determinant are obtained.
where, R1 → first row of first determinant
Ri → first row of second determinant
Find the product of the determinants
Differentiation of a Determinant
Yes, let f(x) = be a given function.
=> f(x) = a(x) d(x) - c(x) b(x)
=> f'(x) = a'(x) d(x) + a(x) d'(x) - c'(x) b(x) - c(x) b'(x)
=> a'(x) d(x) - c'(x) b(x) + a(x) d'(x) - c(x) b'(x)
Thus, the differential coefficient of a determinant is obtained by differentiating a single row (or column) at a time and finally adding the determinants so obtained.
Thus, for a determinant of order 'n'.
Summation of Determinants
Let Δr=, where a,b,c,l,m and n are constants independents of r,
then =∑nr=1 ∆r =
Here functions of r can be the elements of only one row or column. None of the elements other of than that row or column should be dependent on r.
Let Δa=. Showthat ∑na=1 ∆a = 0
By adding all the corresponding elements of C1 of all determinants Δa we have
By taking (n-1)n/2 as common factor from C1 and 6 as common factor from C3, we get
Since C1 and C3 are identical ∑na=1 ∆a =0.
1. Symmetric determinant
The elements situated at equal distance from the diagonal are equal both in magnitude and sign.
= abc + 2fgh - af2 - bg2 - ch2.
2. Skew symmetric determinant
All the diagonal elements are zero and the elements situated at equal distance form the diagonal are equal in magnitude but opposite in sign. The value of a skew symmetric determinant of odd order is zero.
3. Circulant determinant:
The elements of the rows (or columns) are in cyclic arrangement.
= -(a3 + b3 + c3 - 3abc).
4. = (a-b)(b-c)(c-a).
5. = (a-b)(b-c)(c-a)(a+b+c).
6. = (a-b)(b-c)(c-a)(ab+bc+ca).
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