Mathametics 03 – Matrices and determinants

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        ]                 [-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)/3              (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)             (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

ABOUT US

NERAM classroom is Tamil nadu’s one of the top NATA and JEE coaching center which made great change in students especially from small towns. Started in 2017,and successfully we have alumni who have been placed under top institutes such as NIT ,SPA ,AU and some other good institutes under Anna universities. NERAM classroom offers highly personalised and effective learning programs for aspirants of competitive exams like JEE, NATA etc . NERAM classroom has become one of the most preferred education platforms across India.

 

WHY WE ARE INCOMPATIBLE ?

Our niche is creating personalised learning experiences for every type of learner. NERAM breaks away from a typical classroom teaching structure and uses technology and customisation to enable students to interact with their tutors.In India, students often feel too shy to clarify their doubts in a crowded coaching class. Many students travel long hours to get to class. NERAM breaks these shackles by liberating learning and teaching from the limits of location.Students are given full freedom to interact with teachers,experts and alumni as per their convenient time.

 

WHY OPT US ?

Traditionally, students rely on brick-and-mortar coaching centers, but companies like NERAM classroom are springing up and more students are supplementing offline courses with online tutoring.
NERAM classroom is an incredibly personalized tutoring platform for you, while you are staying at your home. We have grown leaps and bounds to be the best Online Tuition Website in India with immensely talented Master Teachers, from the most reputed institutions. We provide you year-long structured coaching classes for JEE and NATA entrance exam preparation at affordable tuition fees, with an exclusive session for clearing doubts, ensuring that neither you nor the topics remain unattended. Revision notes and formula sheets are shared with you, for grasping the toughest concepts. Assignments, Regular Homeworks, Subjective & Objective Tests promote your regular practice of the topics.

 

PERSONALISED LEARNING AND TUTORING

Traditionally, students rely on brick-and-mortar coaching centers, but companies like NERAM classroom are springing up and more students are supplementing offline courses with online tutoring.
Personalised Learning is the foundation of our unique programs for students. Using knowledge graphs, the program adapts and creates personalised learning journeys for students.Based on the student's progress, personal learning profiles are created which help analyse strengths and areas of improvement. Each concept gets tagged at different levels of difficulty to create a smooth learning curve which keeps students motivated through any new learning challenge. Whenever errors are made, motivating remedial methods are generated to strengthen and improve the student's learning experience.

 

TECHNOLOGY ENABLED LEARNING

At NERAM, we leverage technology to merge best practices like use of well verified study materials, videos, engaging content and quizzes with the best teachers so that every student across India has access to the best learning and guidance for their future .

KEY FEATURES OF NERAM CLASSROOM

  • INTERACTIVE LEARNING SESSIONS
  • NUMBER OF MOCK TESTS
  • FRIENDLY GUIDE
 

BEST TEACHERS & ENGAGING CONTENT

NERAM classroom Learning Programs provide students a holistic learning experience. Students across regions can access the best teachers.Very few teams like BYJU’S, VEDANTA in India have a great sense for the education sector as well as how technology can be used for effective scaling, NERAM is one such team.
Hours and Hours of Study with no fun, is a bad idea for you, foreseeing the long run. To ensure that motivation is stirred in the best proportion for your clear understanding, a good number of quizzes and Objective tests are organized to impart knowledge. FREE Sample Papers and Important questions are extracted, solved and discussed, ensuring that you are 100% prepared before any exam.
Hours and Hours of Study with no fun, is a bad idea for you, foreseeing the long run. To ensure that motivation is stirred in the best proportion for your clear understanding, a good number of quizzes and Objective tests are organized to impart knowledge. FREE Sample Papers and Important questions are extracted, solved and discussed, ensuring that you are 100% prepared before any exam.

  Website Contents © NeramClasses. All rights reserved.
× Whatsapp!