1) If A and B are symmetric matrices of same order and X = AB + BA and Y = AB – BA, then (XY)T is equal
(a) XY (b) YX (c) -YX (d) none of these
(c) X = AB + BA ⇒ XT = X
and Y = AB – BA ⇒ YT = -Y
Now, (XT)T = YT × XT = -YX.
2) If A is a skew–symmetric matrix of order 3, then the matrix A4 is
(a) skew symmetric (b) symmetric (c) diagonal (d) none of these
(b) We have AT = –A
(A4)T = (A.A.A.A.)T = AT AT AT AT
⇒ [–(A)][–(A)][–(A)][–(A)]
= (–1)4 A4 = A4
Q3.If A = i -i and B = 1 -1
[ -i i ] [-1 1 ] , then A8 equals
(a) 4B (b) 128B (c) -128B (d) -64B
(B) We have A = iB
⇒ A2 = [i(B)]2 = i2B2 = -B2 = – 2 -2 = -2B
[-2 2]
⇒ A4 = [-2(B)]2 = 4B2 = 4[2(B)] = 8B
⇒ (A4)2 = [8(B)]2
⇒ A8 = 64B2 = 128B
4) A square matrix A is said to be nilpotent of index m. If Am = 0, now, if for this A [I – (A)]n = I + A + A2 + … + Am-1, then n is equal to
(a) 0 (b) m (c) -m (d) -1
(d) Let B = I + A + A2 + … + Am-1
⇒ B (I – (A) = (I + A + A2 + … + Am-1) [I – (A)] = I – Am = I
⇒ B = [I – (A)]-1
⇒ n = -1.
5) If A and B are symmetric matrices then AB – BA is a
(a) symmetric matrix (b) skew symmetric matrix
(c) diagonal matrix (d) null matrix.
(b) A’ = A , B’ = B
(AB – BA)’ = (AB)’- (BA)’ = B’A’- A’B’
= BA – AB
= – (AB – BA) = skew symmetric
6) If A and B are square matrices of same order & A is nonsingular then for a +ve integer n, [A–1 B(A)]n is equal to
(a) A–n Bn An (b) An Bn A–n (c) A–1 Bn A (d) n[A–1 B(A)]
(c) [A–1 B(A)]2 = [A–1B(A)] = [A–1B(A)]
= A–1B2A
Sum [A–1 B(A)]n = A–1BnA.
7) Let A and B be two 2 × 2 matrices. Consider the statements
(i) AB = O ⇒ A = O or B = O
(ii) AB = I2 ⇒ A = B-1
(iii) (A + (B)2 = A2 + 2AB + B2 then
(a) (i) is false, (ii) and (iii) are true (b) (i) and (iii) are false, (ii) is true
(c) (i) and (ii) are false, (iii) is true (d) (ii) and (iii) are false, (i) is true
(C) [A–1 B(A)]2 = [A–1B(A)] = [A–1B(A)]
= A–1B2A
Sum [A–1 B(A)]n = A–1BnA.
Q8.If A = [aij]3 × 3, such that aij = when i = j ,
{when i ≠ j , then 1 + log1/2det[Adj(Adj (A)] =
(A) -12 (B) -10 (C) -13 (D) -11
(d) A = [aij]
aij = 2, i = j = 0, i ≠ j
|A| = 8
Now, det [Adj(Adj(A)] = (8)4 = 212
9) A is a 3×3 matrix with entries from the set { -1,0,1} . The probability that A is neither symmetric nor skew symmetric is
(a) (39-36-33+1)/39 (b) (39-36-33)/39 (c) (39-1)/310 (d) (39-33+1)/39
(a) Total number of matrices that can be formed is 39 .
Let A = [aij]3×3 where aij ∈ { 1,0,1}
If A is symmetric then aij = aji ∀ i , j
If A is skew-symmetric then aij = -aji ∀ i , j
Q10.A is a 3×3 matrix with entries from the set { -1,0,1} . The probability that A is neither symmetric nor skew symmetric is
(A) 1/abc(ab+bc+ca) (B) ab+bc+ca (C) 0 (D) a+b+c
(C) Given determinant = b2c2(ca2 – a2b)+c2a2(ab2 – b2c)+a2b2(bc2 – c2a)
Q11.A is a 3×3 matrix with entries from the set { -1,0,1} . The probability that A is neither symmetric nor skew symmetric is
(A) 0 (B) 1 (C) 100 (D) -100
(C) Given determinant = b2c2(ca2 – a2b)+c2a2(ab2 – b2c)+a2b2(bc2 – c2a)
Q12.A is a 3×3 matrix with entries from the set { -1,0,1} . The probability that A is neither symmetric nor skew symmetric is
(A) 31/15 (B) 41/18 (C) 50/21 (D) 61/27
(C) Given determinant = b2c2(ca2 – a2b)+c2a2(ab2 – b2c)+a2b2(bc2 – c2a)
Q13.A is a 3×3 matrix with entries from the set { -1,0,1} . The probability that A is neither symmetric nor skew symmetric is
(A) -5 (B) -4 (C) -3 (D) -2
(C) Given determinant = b2c2(ca2 – a2b)+c2a2(ab2 – b2c)+a2b2(bc2 – c2a)
Q14.A is a 3×3 matrix with entries from the set { -1,0,1} . The probability that A is neither symmetric nor skew symmetric is
(A) 5π/4 (B) -3π/4 (C) π/4 (D) -π/4
(C) Given determinant = b2c2(ca2 – a2b)+c2a2(ab2 – b2c)+a2b2(bc2 – c2a)
Q15.A is a 3×3 matrix with entries from the set { -1,0,1} . The probability that A is neither symmetric nor skew symmetric is
(A) 5π/4 (B) -3π/4 (C) π/4 (D) -π/4
(d) pqr(a3+b3+c3) – abc(p3+q3+r3)
⇒pqr(a3b3c3 – 3abc) – abc(p3+q3+r3-3pqr)
⇒pqr(a3+b3+c3 – 3abc) – abc(p+q+r)(p2+q2+r2 – pq – qr – rp)
= pqr(a3+b3+c3 – abc).
Q16.A is a 3×3 matrix with entries from the set { -1,0,1} . The probability that A is neither symmetric nor skew symmetric is
(A) (3n – r) (B) 2(3n – r) (C) 3(3n – r) (D) 3n + r
(b) LHS = (24mx+6n) – (12mx2+nx+2r)
x = 0 ⇒ 6n – 2r ⇒ 2(3n – r)
17) The number of values of k for which the system of equations kx+(k+3)y = 3k – 1; (k+1)x+8y = 4k has infinitely many solutions is
(a) 0 (b) 1 (c) 2 (d) infinite
(b) Solving the equations we get
(k – 1)(k – 3)y = -(k -1)2
If k = 1 , infinite sol.
If k ≠ 1 then inconsistent if k = 3
If k ≠ 1,3, then kx = [4k(k -2)]/(k -3)
If k = 0 unique sol.
If k ≠ 0 unique sol.
18) 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
(a) 3 (b) 1 (c) 0 (d) -3
(B) Solving the equations we get
(k – 1)(k – 3)y = -(k -1)2
If k = 1 , infinite sol.
If k ≠ 1 then inconsistent if k = 3
If k ≠ 1,3, then kx = [4k(k -2)]/(k -3)
If k = 0 unique sol.
If k ≠ 0 unique sol.
19) If the system of linear equations x+y+z = 6 , x+2y+3z = 14 and 2x+5y+λz = μ , (λ , μ ∈ R) has no solution , then
(a) λ ≠ 8 (b) λ = 8, μ ≠ 36 (c) λ = 8, μ = 36 (d) None of these
(B) Solving the equations we get
(k – 1)(k – 3)y = -(k -1)2
If k = 1 , infinite sol.
If k ≠ 1 then inconsistent if k = 3
If k ≠ 1,3, then kx = [4k(k -2)]/(k -3)
If k = 0 unique sol.
If k ≠ 0 unique sol.
Q20.If the system of linear equations x+y+z = 6 , x+2y+3z = 14 and 2x+5y+λz = μ , (λ , μ ∈ R) has no solution , then
(A) 3ω (B) 3ω(ω – 1) (C) 3ω2 (D) 3ω(1 – ω)
(B) Solving the equations we get
(k – 1)(k – 3)y = -(k -1)2
If k = 1 , infinite sol.
If k ≠ 1 then inconsistent if k = 3
If k ≠ 1,3, then kx = [4k(k -2)]/(k -3)
If k = 0 unique sol.
If k ≠ 0 unique sol.
Q21.If the system of linear equations x+y+z = 6 , x+2y+3z = 14 and 2x+5y+λz = μ , (λ , μ ∈ R) has no solution , then
(A) -11,6 (B) -6,11 (C) 6,11 (D) -6,-11
(B) Solving the equations we get
(k – 1)(k – 3)y = -(k -1)2
If k = 1 , infinite sol.
If k ≠ 1 then inconsistent if k = 3
If k ≠ 1,3, then kx = [4k(k -2)]/(k -3)
If k = 0 unique sol.
If k ≠ 0 unique sol.
Q22.If the system of linear equations x+y+z = 6 , x+2y+3z = 14 and 2x+5y+λz = μ , (λ , μ ∈ R) has no solution , then
(A) -11,6 (B) -6,11 (C) 6,11 (D) -6,-11
(a) If Q = PAPT
PTQ = APT (1)(as PPT = 1)
PTQ2005P = APTQ2004P
= A2PTQ2003P
= A3PTQ2002P
= A2004PT(QP)
= A2004PT(PA) (Q = PAPT ⇒ QP = PA)
= A2005
= A2005
