Q 1 :

If M=(5232-32-12), then which of the following matrices is equal to M2022?              [2022]

  • (30343033-3033-3032)

     

  • (3034-30333033-3032)

     

  • (3033-3032-3032-3031)

     

  • (30323031-3031-3030)

     

(1)

Given that M=(5232-32-12)M2=(43-3-2)

M3=(11292-92-72)M4=(76-6-5) and so on

Mn=(3n2+13n2-3n2-3n2+1)

Now, Mn=(3×20222+13×20222-3×20222-3×20222+1)

=(30343033-3033-3032)

 Option (1) is correct



Q 2 :

How many 3×3 matrices M with entries from {0,1,2} are there, for which the sum of the diagonal entries of MTM is 5?                   [2017]

  • 126

     

  • 198

     

  • 162

     

  • 135

     

(2)

Let M=[a1a2a3a4a5a6a7a8a9],    where ai{0,1,2}

Then MTM=[a1a4a7a2a5a8a3a6a9][a1a2a3a4a5a6a7a8a9]

Sum of the diagonal entries in MTM=5

(a12+a42+a72)+(a22+a52+a82)+(a32+a62+a92)=5,

which is possible when

Case I:  Five ai's areand four ai's are zero.

Which can be done in C49 ways

=9×8×7×64×3×2×1=126

Case II:  One ai is 1, one ai is 2 and the remaining seven ai's are zero

It can be done in C19×C18=9×8=72 ways

 Total number of ways=126+72=198



Q 3 :

Let P=[1004101641] and I be the identity matrix of order 3. If Q=[qij] is a matrix such that P50-Q=I, then q31+q32q21 equals                  [2016]

  • 52

     

  • 103

     

  • 201

     

  • 205

     

(2)

P=[1004101641]=I+[0004001640]=I+A,

where    A=[0004001640]A2=[0000001600]

and  A3=[000000000],     An=0, n3

Now,  P50=(I+A)50=C050I50+C150I49A+C250I48A2+0

                  =I+50A+25×49A2

   Q=P50-I=50A+25×49A2

q21=50×4=200,  q31=50×16+25×49×16=20400,

and  q32=50×4=200

  q31+q32q21=20600200=103



Q 4 :

Let β be a real number. Consider the matrix  A=(β0121-231-2).

If A7-(β-1)A6-βA5 is a singular matrix, then the value of 9β is _______.                    [2022]



(3)

A=(β0121-231-2),     |A|=-1

Given that A7-(β-1)A6-βA5 is a singular matrix.

|A5||A2-(β-1)A-βI|=0

|A5||(A+I)(A-βI)|=0

  |A5||A+I||A-βI|=0    [ |A5|=|A|5]

As |A|0|A+I|=0 or |A-βI|=0

|A+I||β+10122-231-1|=0

-1=0 (Rejected)  {|A+I|0}

  |A-βI|=|00121-β-231-2-β|=0

2-3(1-β)=02-3+3β=0

β=13

  9β=3



Q 5 :

Let z=-1+3i2, where i=-1, and r,s{1,2,3}. Let P=[(-z)rz2sz2szr] and I be the identity matrix of order 2. Then the total number of ordered pairs (r,s) for which P2=-I is                                     [2016]



(1)

z=-1+i32z3=1  and  1+z+z2=0

P2=[(-z)rz2sz2szr][(-z)rz2sz2szr]

=[z2r+z4sz2s((-z)r+zr)z2s((-z)r+zr)z4s+z2r]

For P2=-I, we should have

z2r+z4s=-1  and  z2s((-z)r+zr)=0

z2r+z4s+1=0  and  (-z)r+zr=0

r is odd and s=r but not a multiple of 3,

which is possible when s=r=1

  only one pair is there.



Q 6 :

Let M be a 3×3 matrix satisfying

M[010]=[-123], M[1-10]=[11-1],  and  M[111]=[0012]. Then the sum of the diagonal entries of M is                        [2011]



(9)

Let  M=[a1b1c1a2b2c2a3b3c3],  then  [a1b1c1a2b2c2a3b3c3][010]=[-123]

b1=-1,  b2=2,  b3=3

[a1b1c1a2b2c2a3b3c3][1-10]=[11-1]

a1-b1=1,  a2-b2=1,  a3-b3=-1

a1=0,  a2=3,  a3=2

and  [a1b1c1a2b2c2a3b3c3][111]=[0012]

a3+b3+c3=12c3=7

  Sum of diagonal elements=a1+b2+c3=0+2+7=9



Q 7 :

Let ω be a complex cube root of unity with ω1 and P=[pij] be an n×n matrix with pij=ωi+j.  Then P20, when n=                          [2013]

  • 57

     

  • 55

     

  • 58

     

  • 56

     

Select one or more options

(2, 3, 4)

For n=3,   P=[ω2ω3ω4ω3ω4ω5ω4ω5ω6]  and  P2=[000000000]

It shows  P2=0  if n is a multiple of 3.

So, for P20, n should not be a multiple of 3, hence n can take values 55, 56 and 58.