JAWAHARLAL NEHRU TECHNOLOGICAL UNIVERSITY HYDERABAD Pharm. D I Year REMEDIAL MATHEMATICS Previous Question Paper (January/February - 2021 - R08)

To evaluate the determinant D = | a² b² c² | / | a b c | / | bc ca ab |, we perform row operations to simplify the matrix. Multiply the first row by abc, the second row by abc, and the third row by abc is not efficient; instead, perform R3 = aR3, R3 = bR3, R3 = cR3 or simply manipulate the columns. A more effective approach is to multiply R3 by abc/abc or observe the structure.

Multiply R3 by abc: (1/abc) * | a² b² c² | | a b c | | abc² ab²c a²bc |. This does not simplify easily. Instead, multiply C1 by a, C2 by b, and C3 by c:

Factor out abc from R3:

This is a standard circulant-like determinant. By applying column operations C1 = C1 - C2 and C2 = C2 - C3, we obtain factors (a-b)(b-c)(c-a)(ab+bc+ca). Specifically, the determinant evaluates to (a-b)(b-c)(c-a)(ab+bc+ca).

Given the system: 3x + 4y + 5z = 18; 2x - y + 8z = 13; 5x - 2y + 7z = 20.

First, calculate the determinant of the coefficient matrix Δ:

Next, calculate Δx, Δy, and Δz by replacing the respective columns with the constants (18, 13, 20):

Finally, find the variables: x = Δx/Δ = 408/136 = 3; y = Δy/Δ = 136/136 = 1; z = Δz/Δ = 136/136 = 1. The solution is (x, y, z) = (3, 1, 1).