Consider a system of linear equations aix + biy + ciz = di (where ai, bi, ci ≠ 0 and i = 1,2,3 ) & (α,β,γ) is its unique solution, then match list-I with list-IIList-I(I) If ai = di = k2, (k ≠ 0) and α + β + γ = 2, then k is(II) If ai = di = k ≠ 0, then α + β + γ is(III) If ai = k > 0, di = k + 1, then α + β + γ can be(IV) If ai = k di = k + 1, then α + β + γ can beList-II(P) 1(Q) 2(R) 0(S) 3(T) -1
Let a1, a2, a3, ..., a10 be in G.P. with ai > 0 for i = 1,2,..., 10 and S be the set of pairs (r, k), r, k ∈ N (the set of natural numbers) for which . Then the number of elements in S, is :
If \([\cdot]\) denotes the greatest integer less than or equal to the real number under consideration and \(x \in [0,1), y \in [1,2), z \in [2,3)\), the value of the determinant \(\begin{vmatrix} [x]+1 & [y] & [z] \\ [x] & [y]+1 & [z] \\ [x] & [y] & [z]+1 \end{vmatrix}\) is
Let α, β and γ be real numbers. consider the following system of linear equationsx + 2y + z = 7x + αz = 112x - 3y + βz = γMatch each entry in List-I to the correct entries in List-IIList-IList-II(P) If β = 1/2(7α - 3) and γ = 28, then the system has(1) a unique solution(Q) If β = 1/2(7α - 3) and γ ≠ 28, then the system has(2) no solution(R) If β ≠ 1/2(7α - 3) where α = 1 and γ ≠ 28, then the system has(3) infinitely many solutions(S) If β ≠ 1/2(7α - 3) where α = 1 and γ = 28, then the system has(4) x = 11, y = -2 and z = 0 as a solution(5) x = -15, y = 4 and z = 0 as a solution
Let S = \left\{ A = \begin{pmatrix} 0 & 1 & c \\ 1 & a & d \\ 1 & b & e \end{pmatrix} : a, b, c, d, e \in \{0, 1\} \text{ and } |A| \in \{-1, 1\} \right\}, \text{ where } |A| \text{ denotes the determinant of } A. \text{ Then the number of elements in } S \text{ is } \_\_\_\_.