Multiple Choice QuestionsIf D = 
for x ≠ 0, y ≠ 0 then D is
divisible by neither x nor y
divisible by both x and y
divisible by x but not y
divisible by y but not x
B.
divisible by both x and y
The system of equations
x + y + z = 0
2x + 3y + z = 0
and x + 2y = 0
has
a unique solution; x = 0, y = 0, z = 0
infinite solutions
no solution
finite number of non-zero solutions
B.
infinite solutions
The given system of equations are
x + y + z = 0,
2x + 3y + z = 0,
and x + 2y = 0
Let A be a 2 × 2 matrix with real entries. Let I be the 2 × 2 identity matrix. Denote by tr (A), the sum of diagonal entries of A. Assume that A2= I.
Statement −1: If A ≠ I and A ≠ − I, then det A = − 1.
Statement −2: If A ≠ I and A ≠ − I, then tr (A) ≠ 0.
Statement −1 is false, Statement −2 is true
Statement −1 is true, Statement −2 is true, Statement −2 is a correct explanation for Statement −1
Statement −1 is true, Statement −2 is true; Statement −2 is not a correct explanation for Statement −1.
Statement − 1 is true, Statement − 2 is false.
D.
Statement − 1 is true, Statement − 2 is false.
Let 
a , b ∈ N. Then
there cannot exist any B such that AB = BA
there exist more than one but finite number of B’s such that AB = BA
there exists exactly one B such that AB = BA
there exist infinitely many B’s such that AB = BA
D.
there exist infinitely many B’s such that AB = BA
Let A be a square matrix all of whose entries are integers. Then which one of the following is true?
If det A = ± 1, then A–1 exists but all its entries are not necessarily integers
If detA ≠ ± 1, then A–1 exists and all its entries are non-integers
If detA = ± 1, then A–1 exists and all its entries are integers
If detA = ± 1, then A–1 need not exist
C.
If detA = ± 1, then A–1 exists and all its entries are integers
Each entry of A is integer, so the cofactor of every entry is an integer and hence each entry in the adjoint of matrix A is integer. Now detA = ± 1 and A–1 =(1/ det(A)) (adj A)
⇒ all entries in A–1 are integers
The simultaneous equations Kx + 2y - z = 1, (K - I)y - 2z = 2 and (K + 2)z = 3 have only one solution when :
K = - 2
K = - 1
K = 0
K = 1
B.
K = - 1
The system of given equations are
Kx + 2y - z = 1 ...(i)
(K - 1)y - 2z = 2 ...(ii)
and (K + 2)z = 3 ...(iii)
This system of equations has a unique solution, if
If a1, a2, a3,…, an,… are in G.P., then the determinant
1
0
4
2
B.
0
C1 – C2, C2 – C3 two rows becomes identical Answer: 0
Let a, b, c be any real numbers. Suppose that there are real numbers x, y, z not all zero such that x = cy + bz, y = az + cx and z = bx + ay. Then a2 + b2 + c2 + 2abc is equal to
2
-1
0
1
D.
1
The system of equations x – cy – bz = 0, cx – y + az = 0 and bx + ay – z = 0 have non-trivial solution if
If |A2| = 25, then |α| equals
