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Consider the set A of all determinants of order 3 with entries 0 or 1 only. Let B be the subset of A consisting of all determinants with value 1. Let C be the subset of A consisting of all determinants with value –1. Then
For every ‘det, with value 1’ (∈ B) we can find a det. with value – 1 by changing the sign of one entry of ‘1’.
Hence there are equal number of elements in B and C.
∴ (b) is the correct option
If ω (≠1) is a cube root of unity, then
Operating R_{1}→ R_{1} – R_{2} + R_{3}
Let a, b, c be the real numbers. Then following system of equations in x, y and z
Then the given system of equations becomes
X + Y – Z = 1
X – Y + Z= 1, – X + Y + Z = 1
This is the new system of equations For new system, we have
∴ New system of equations has unique solution.
If A and B are square matrices of equal degree, then which one is correct among the followings?
If A and B are square matrices of same degree then matrices A and B can be added or subtracted or multiplied. By algebra of matrices the only correct option is A + B = B + A
The parameter, on which the value of the determinant does not depend upon is
Expanding along C_{1}, we get
Δ = (1 + a^{2} – 2a cos dx) [sin (p + d) x cos px – sin px cos (p + d)x]
⇒ Δ = (1 + a^{2} – 2a cos dx) [sin {(p + d)x – px}]
⇒ Δ = (1 + a^{2} – 2a cos dx) [sin dx]
which is independent of p.
f (100) is equal to
= 0 [∵ C_{1} and C_{2} are identical] which is free of x, so the function is true for all values of x. Therefore, at x = 100, f (x) = 0, i.e., f (100) = 0
If the system of equations x – ky – z = 0, kx – y – z = 0, x + y – z = 0 has a nonzero solution, then the possible values of k are
For the given homogeneous system to have non zero solution determinant of coefficient matrix should be zero; i.e.,
Then the value of the determinant
Given that
Also 1 + ω + ω^{2} = 0 and ω^{3} = 1
Now given det. is
[Using ω = –1 – ω^{2} and ω^{3} = 1]
Operating C_{1} → C_{1} + C_{2} + C3
(as 1 + ω + ω2 = 0)
Expanding along C_{1}, we get
3 (ω2 – ω4) =3 (ω2 – ω) = 3ω (ω– 1)
The number of values of k for which the system of equations (k + 1)x + 8y = 4k; kx + (k + 3) y = 3k – 1 has infinitely many solutions is
For infinitely many solutions the two equations become identical
hen value of α for which A^{2} = B, is
Given that
and A^{2} = B
∵ There is no common value
∴ There is no real value of a for which A^{2} = B
If the system of equations x + ay = 0, az + y = 0 and ax + z = 0 has infinite solutions, then the value of a is
The given system is, x + ay = 0
az + y = 0
ax + z = 0
It is system of homogeneous equations therefore, it will have infinite many solutions if determinant of coefficient matrix is zero. i.e.,
Given 2x – y + 2z = 2, x – 2y + z = – 4, x + y + λz = 4 then the value of λ such that the given system of equation has NO solution, is
Since the system has no solution, Δ = 0 and any one amongst D_{x}, D_{y}, D_{z} is nonzero.
then the value of α is
Now,  A  = α^{2} – 4
⇒ (α^{2} – 4)^{3} = 125 = 5^{3}
⇒ α^{2} – 4 = 5
⇒ α = + 3
then the value of c and d are
∴ Characteristic eqn of above matrix A is given by
Also by Cayley Hamilton thm (every square matrix satisfies its characteristic equation) we obtain
A^{3} – 6A^{2} + 11A – 6I = 0
Multiplying by A^{–1}, we get
Comparing it with given relation,
we get c = – 6 and d = 11
and Q = PAP^{T} and x = P^{T}Q^{2005}P then x is equal to
We observe that Q = P A PT
⇒ Q^{2} = (P A P^{T}) (P A P^{T})
= P A (P^{T} P) A P^{T} = PA (I A) P^{T}
∴ P A^{2} P^{T}
Proceeding in the same way, we get Q^{2005} = P A^{2005} PT
and proceeding in the same way
Consider three points
P = ( sin(β  α),  cosβ), Q = (cos(βα), sin β) and
R = (cos(β α+θ), sin(β θ)) , where
Then,
The given points are P ( sin(β α), – cosβ) , Q(cos(β  a), sinβ)
R(cos(β α+θ), sin(β θ))
Operating C_{3}  C_{1} sin_{θ} C_{2} cosθ , we get
= (1  sinθ  cosθ)[cosβ cos(β α)  sin β sin(βα)]
⇒ Δ = [1  (sinθ+ cosθ)] cos(2βα)
∴ Δ≠ 0 ⇒ the three points are non collinear..
The number of 3 × 3 matrices A whose entries are either 0 or1 and for which the system has exactly twodistinct solutions, is
Then the given system is equivalent to
Which represents three distinct planes. But three planes can not intersect at two distinct points, therefore no such system exists.
Let ω ≠ 1 be a cube root of unity and S be the set of all nonsingular matrices of the form where each of a, b and c is either ω or ω^{2}. Then the number of distinct matrices in the set S is
For the given matrix to be nonsingular
⇒ 1 – (a + c) ω + a cω^{2} ≠ 0 ⇒ (1 – aω) (1 – cω) ≠ 0
⇒ a ≠ ω^{2} and c ≠ ω^{2} where ω is complex cube root of unity.
As a, b and c are complex cube roots of unity
∴ a and c can take only one value i.e. w while b can take 2 values i.e. ω and ω^{2}. ∴ Total number of distinct matrices = 1 × 1 × 2 = 2
Let P = [a_{ij}] be a 3 x 3 matrix and let Q = [b_{ij}], where b_{ij} = 2^{i + j } a_{ij} for 1 < i , j < 3. If the determinant of P is 2, then the determinant of the matrix Q is
We have
= 2^{12} x P = 2^{12} x 2 = 2^{13}
If P is a 3 x 3 matrix such that P^{T} = 2P + I, where P^{T} is the transpose of P and I is the 3 x 3 identity matrix, then there
exists a column matrix
We have P^{T} = 2P + I
⇒ P = 2P^{T} + I ⇒ P = 2(2P + I) + I
⇒ P = 4P + 3I ⇒ P + I = 0
⇒ PX + X = 0 ⇒ PX = – X
Let and I be the identity matrix of order 3. If Q = [q_{ij}] is a m atrix such that P^{50} – Q = I, then equals
Now P^{50} = (I + A)^{50} = ^{50}C_{0} I^{50} + ^{50}C_{1} I^{49} A + ^{50}C_{2} I^{48} A^{2} + O
= I + 50A + 25 × 49 A^{2}.
∴ Q = P^{50} – I = 50A + 25 × 49 A^{2}.
⇒ q_{21} = 50 × 4 = 200
⇒ q_{31} = 50 × 16 + 25 × 49 × 16 = 20400
⇒ q_{32} = 50 × 4 = 200
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