Q 11 :

Let  A=[20050102]  and  B=[504023], then |AB| is equal to:

  • 460

     

  • 2000

     

  • 3000

     

  • - 7000

     

(4)

Ans.   -7000

Explanation:

|AB|=|A||B|=(400-500)(150-80)

         =(-100)(70)=-7000



Q 12 :

If A is a square matrix of order 3, such that A(adjA)=10I, then |adjA| is equal to:

  • 1

     

  • 10

     

  • 100

     

  • 101

     

(3)

Ans.    100

Explanation:

Given:             A(adjA)=10I

We know that,  A.adjA=|A|.I

∴                            |A|=10

∵                   |adjA|=|A| n-1

∴                   |adjA|=|A|3-1=102=100



Q 13 :

If A is a 3×3 matrix such that |A|=8, then |3A| equals:

  • 8

     

  • 24

     

  • 72

     

  • 216

     

(4)

Ans.     216

Explanation:

Given:                |A|=8

We know that,  |kA|=kn|A|, where k is a constant 

∴                    |3A|=33|A|=27×8=216



Q 14 :

If A is a skew-symmetric matrix of order 3, then the value of |A| is:

  • 3

     

  • 0

     

  • 9

     

  • 27

     

(2)

Ans.    0

Explanation:

Since, A is a skew-symmetric matrix,

∴                  AT=-A

⇒              |AT|=|-A|

Since |AT|=|A| and for a 3×3 matrix,

|-A|=(-1)3|A|,

we get  |A|=(-1)3|A|

Hence, |A|=-|A|,

⇒    2|A|=0

⇒       |A|=0



Q 15 :

If  |x218x|=|62186|,  then the value of x is:

  • ±2

     

  • 0

     

  • ±3

     

  • ±6

     

(4)

Ans.   ±6

Explanation:

∵            |x218x|=|62186|

⇒           x2-36=36-36

⇒           x2-36=0

⇒                      x=±6



Q 16 :

The determinant  |xsinθcosθ-sinθ-x1cosθ1x|  is:

  • Independent of θ only

     

  • Independent of x only

     

  • Independent of both θ and x

     

  • None of the above

     

(1)

Ans.     Independent of θ only

Explanation:

Let   Δ=|xsinθcosθ-sinθ-x1cosθ1x|

          =x(-x2-1)-sinθ(-xsinθ-cosθ)+cosθ(-sinθ+xcosθ)

         =-x3-x+xsin2θ+sinθcosθ-sinθcosθ+xcos2θ

         =-x3-x+x(sin2θ+cos2θ)

         =-x3-x+x                           [∵ sin2θ+cos2θ=1]

         =-x3,  which is independent of θ.



Q 17 :

If M11=-40, M12=-10 and M13=35 of the determinant  Δ=|13-24-56352|, then the value of Δ is:

  • -80

     

  • 60

     

  • 70

     

  • 100

     

(1)

Ans.    -80

Explanation:

Δ=a11A11+a12A12+a13A13

   =a11M11-a12M12+a13M13

   =1(-40)-3(-10)+(-2)(35)

   =-40+30-70=-80



Q 18 :

Let A be a square matrix of order 3×3, then |kA| is equal to:

  • k|A|

     

  • k2|A|

     

  • k3|A|

     

  • 3k|A|

     

(3)

Ans.   k3|A|

Explanation:

We know that for n×n matrix,

                 |(λA)|=λn|A|

Hence,        |kA|=k3|A|



Q 19 :

Which of the following is correct?

  • Determinant is a square matrix

     

  • Determinant is a number associated to a matrix

     

  • Determinant is a number associated to a square matrix

     

  • None of the above

     

(3)

Ans.    Determinant is a number associated to a square matrix.

Explanation:

We know that to every square matrix, A=[aij] of order n, we can associate a number called the determinant of the square matrix A, where aij=(i,j)th element of A. Thus, the determinant is a number associated to a square matrix.



Q 20 :

Value of  |cos15°sin15°sin15°cos15°|  is:

  • 1

     

  • 12

     

  • 32

     

  • None of these

     

(3)

Ans.    32

Explanation:

                      Δ=|cos15°sin15°sin15°cos15°|

                          =cos215°-sin215°

                          =cos(2×15°)=cos30°

                          =32



Q 21 :

If A is a square matrix x, such that A2=I, then A-1 is equal to:

  • 2A

     

  • 0

     

  • A

     

  • A + 1

     

(3)

Ans.     A

Explanation:

                        A2=I

then              A·A=I

or       A-1(A·A)=A-1I

⇒     (A-1A)A=A-1

⇒                  IA=A-1

⇒              A-1=A



Q 22 :

If a, b and c are all different from zero and  |1+a1111+b1111+c|=0,  then the value of 1a+1b+1c  is:

  • 0

     

  • abc

     

  • -1

     

  • 1abc

     

(3)

Ans.     -1

Explanation:

Given,          |1+a1111+b1111+c|=0

Expanding the determinant,

(1+a)[(1+b)(1+c)-1]-1(1+c-b)+1(1-1-b)=0

                                                 (1+a)(b+c+bc)-c-b=0

                                  b+c+bc+ab+ac+abc-b-c=0

                                                                          ab+bc+ca=-abc

Dividing both sides by abc,

          1a+1b+1c=-1



Q 23 :

Let A be a 3×3 matrix of non-negative real numbers such that A[222]=4[111] then (detA)max is

  • 12

     

  • 8

     

  • 16

     

  • 32

     

(2)

a1+a2+a3=2,  b1+b2+b3=2,  c1+c2+c3=2

(detA) is maximum when a1=2, b2=2, c3=2 and all others are 0.



Q 24 :

If a,b,c are real numbers satisfying |(a2+1)2(ab+1)2(ac+1)2(ab+1)2(b2+1)2(bc+1)2(ac+1)2(bc+1)2(c2+1)2|=k(a-b)2(b-c)2(c-a)2 then k=

  • 0

     

  • 1

     

  • 2

     

  • 4

     

(3)

|1aa21bb21cc2|      |1112a2b2ca2b2c2|



Q 25 :

Let S={A=[abcd]:a,b,c,d∈{0,1,2,3,4} and A2–4A+3I=0} be a set of 2 x 2 matrices. Then the number of matrices in S, for which the sum of the diagonal elements is equal to 4, is :          [2026]

  • 20

     

  • 17

     

  • 21

     

  • 19

     

(4)

We have, Tr = (A) = 4

⇒  a + d = 4          ... (i)

and |A| = 3

[∵ A2–4A+3I=0 is characteristic equation of matrix A]

  ⇒ad – bc = 3

Case 1: If a = 0, d = 4, then bc = –3 (Reject)

Case 2: If a = 1, d = 3, then 3 – bc = 3

              ⇒ bc = 0

              If b = 0, then 'c' has 5 choices and if c = 0, 'b' has 4 choices, so total 9 cases.

Case 3: If a = 2, d = 2, then 4 – bc = 3

             ⇒ bc = 1 ⇒ b = 1, c = 1 (only 1 case)

Case 4: If a = 3, d = 1, then 3 – bc = 3

              ⇒ bc = 0 ⇒ 9 cases

Case 5: If a = 4, d = 0, then bc = -3 (Reject)

∴  Total 19 cases.



Q 26 :

Let A=[1274–2838–7] and det(A–αI)=0, where α is a real number. If the largest possible value of α is p, then the circle (x–p)2+(y–2p)2=320, intersects the co-ordinate axes at          [2026]

  • 1 point

     

  • 2 points

     

  • 3 points

     

  • 4 points

     

(3)

A=[1274–2838–7]

Now, |A–αI|=0

⇒ [1–α274–2–α838–7–α]=0

⇒ (1–α){α2+9α+14–64}–2{–4α–28–24}+7{32+6+3α}=0

⇒ α2+9α–50–α3–9α2+50α+8α+104+266+21α=0

⇒ α3+8α2–88α–320=0

⇒ α2(α-8)+16α(α–8)+40(α–8)=0

⇒ (α-8)+(α2+16α+40)=0

⇒ α=8,α=–16±962

⇒ α=–8±24

Largest possible value of α is 8.

∴  p=8

⇒  (x–8)2=(y–16)2=320

For y = 0

      (x–8)2=320-256=64

⇒ x–8=±8 ⇒ x=16,0

For x = 0

      (y–16)2=320–64

⇒ (y–16)2=256 ⇒ y–16=±16

       y = 32, 0

So, points are (0, 32), (0, 0), (16, 0).



Q 27 :

If f:N→Z is defined by f(n)=|n–1–5–2n23(2k+1)2k+1–3n33k(2k+1)3k(k+2)+1|, k∈N and ∑n=1kf(n)=98, then k is equal to :          [2026]

  • 3

     

  • 4

     

  • 5

     

  • 6

     

(1)

We have, 

∑f(n)=|∑n–1–5–2∑n23(2k+1)2k+1–3∑n33k(2k+1)3k(k+2)+1|, k∈N

Now, ∑n=k(k+1)2, ∑n2=k(k+1)(2k+1)6

and  ∑n3=k2(k+1)24

 ∑n=1kf(n)=|k(k+1)2–1–5–2·k·(k+1)(2k+1)63(2k+1)2k+1–3·k2(k+1)243k(2k+1)3k(k+2)+1|

⇒ 7k(k+1)(2k+1)6=98

⇒ k(k+1)(2k+1)=84

∴ k=3



Q 28 :

Match the following columns:

  Column I (A,B,C are matrices)   Column-II
a. If |A3×3|=2, then |2A-1|= p. 1
b. If |A3×3|=18, then |adj(adj(2A))|= q. 4
c. If (A+B)2=A2+B2, and |A2×2|=2, then |B2×2|= r. 8
d. |A2×2|=2,|B3×3|=3 and |C4×4|=4, then |ABC| is equal to s. 0
    t. Does not exist

 

  • a-q; b-p; c-s; d-t

     

  • a-p; b-r; c-q; d-s

     

  • a-p; b-q; c-r; d-s

     

  • a-s; b-q; c-r; d-p

     

(1)

a. |A|=2⇒|2A-1|=23/|A|=4

b. |adj(adj(2A))|=|2A|4=212|A|4=212/212=1

c. (A+B)2=A2+B2⇒AB+BA=O

    ⇒|AB|=|-BA|=-|BA|=-|AB|⇒|AB|=0⇒|B|=0

d.  Product ABC is not defined.