Q 21 :    

If A and B are two non-zero n×n matrices such that A2+B=A2B, then                 [2023]

  • A2=I or B=I

     

  • A2B=BA2

     

  • AB=I

     

  • A2B=I

     

(2)

A2+B=A2BA2+B-A2B=0 

  A2-A2B+B=0A2(I-B)+B=0 

  A2(I-B)+B-I=-I(A2-I)(I-B)=-I

  (A2-I)(B-I)=IA2B-A2-B=0

  A2(B-I)=BA2=B(B-I)-1A2=B(A2-I)

  A2=BA2-BA2+B=BA2A2B=BA2



Q 22 :    

Let  A=[110310-310110] and  B=[1-i01], where i=-1.  If M=ATBA, then the inverse of the matrix AM2023AT is        [2023]

  • [1-2023i01] 

     

  • [102023i1]

     

  • [12023i01]

     

  • [10-2023i1]

     

(3)

 

 



Q 23 :    

Let A, B, C be 3×3 matrices such that A is symmetric and B and C are skew-symmetric. Consider the statements:

(S1)  A13B26-B26A13 is symmetric

(S2)  A26C13-C13A26 is symmetric

Then,                                                                        [2023]

  • Only S1 is true

     

  • Both S1 and S2 are false

     

  • Both S1 and S2 are true

     

  • Only S2 is true

     

(4)

 Since A is symmetric.  A'=A 

and B and C are skew symmetric. 

B'=-B and C'=-C

 S1:(A13B26-B26A13)'=(A13B26)'-(B26A13)'

=(B')26(A')13-(A')13(B')26  ((X'Y')=(Y'X'))

=(-B)26(A)13-(A)13(-B)26 

=B26A13-A13B26=-(A13B26-B26A13)

Which is skew symmetric.

 S1 is false.

S2:(A26C13-C13A26)'=(A26C13)'-(C13A26)'

       =(C')13(A')26-(A')26(C')13

       =(-C')13A26-A26(-C)13=-C13A26+A26C13 
 
  Which is symmetric.

 S2 is true.



Q 24 :    

Let α and β be real numbers. Consider a 3×3 matrix A such that A2=3A+αI. If A4=21A+βI, then               [2023]

  • β=-8

     

  • β=8

     

  • α=4

     

  • α=1

     

(1)

 Given, A2=3A+αI and A4=21A+βI 

Now, A2A2=(3A+αI)(3A+αI)

         =9A2+6αIA+α2I=9(3A+αI)+6αIA+α2I 

         =27A+9αI+6αA+α2I=(27+6α)A+(9α+α2)I

So, 27+6α=21α=-1 

 9α+α2=β9(-1)+(-1)2=ββ=-8



Q 25 :    

Let  A=(10004-1012-3). Then the sum of the diagonal elements of the matrix (A+I)11 is equal to:

  • 2050

     

  • 4094

     

  • 6144

     

  • 4097

     

(4)

We have, A=[10004-1012-3]

A2=[10004-1012-3][10004-1012-3]=[10004-1012-3]=A

Similarly, An=A; nN 

Now, (A+I)11= C011A11+C111A10I1+C211A9I2++C1111A0I11

=A(C011+C111++C1011)+I   [An=A and In=I]

=A(211-1)+I 

Trace of (A+I)11=1(211-1)+1+4(211-1)+1+(-3)(211-1)+1

=211+4(211)-3-3(211)+4=2×211+1

=212+1=4096+1=4097



Q 26 :    

Let A=[012a031c0], where a,c. If A3=A and the positive value of a belongs to the interval (n-1,n],

 where n, then n is equal to _________ .    [2023]



(2)

We have,

A=[012a031c0] and  A3=A

Now, A2=[012a031c0][012a031c0]

A2=[a+22c33a+3c2aac12+3c]

and A3=[a+22c33a+3c2aac12+3c][012a031c0]

A3=[2ac+3a+2+3c2a+4+6ca(a+3c)+2a3+2ac6+3a+9ca+2+3cac+c(2+3c)2ac+3]

Given that A3=A,

          2ac+3=0  ...(i)

and a+2+3c=1a+1+3c=0  ...(ii)

a+1-92a=0                   [From (i)]

2a2+2a-9=0

Since f(1)<0,f(2)>0

a(1,2]                                   [f(a)=2a2+2a-9]

Now, n-1=1n=2



Q 27 :    

Let A be a symmetric matrix such that |A|=2 and [21332]A=[12αβ]. If the sum of the diagonal elements of A is s, then βsα2 is equal to ________ .        [2023]



(5)

Given, |A|=2

Let A=[abbc]ac-b2=2  ...(i)

Also, [21332][abbc]=[12αβ]

[2a+b2b+c3a+32b3b+32c]=[12αβ]

2a+b=1  ...(ii),  2b+c=2  ...(iii)

Solving (i), (ii), and (iii) we get  b=-12,  a=34,  c=3

Now, α=3a+32b=32,  β=3b+32c=3

s=a+c=154         βsα2=3×15494=5



Q 28 :    

Let A=[aij], aijZ[0,4], 1i,j2. The number of matrices A such that the sum of all entries is a prime number p(2,13) is ________ .      [2023]



(204)

As given a+b+c+d=3 or 5 or 7 or 11

If sum = 3, then

Coefficient of x3 in (1+x+x2++x4)4 is (1-x5)4(1-x)-4

     C3=C364+3-1=20

If sum = 5, coefficient of x5 is  (1-4x5)(1-x)-4

   C54+5-1-4=C58-4=52

If sum = 7, then coefficient of x7 is  

(1-4x5)(1-x)-4C710-4×C25=80

If sum = 11 

(1-4x5+6x10)(1-x)-4C114+11-1-4C64+6-1+6C14+1-1

=C1114-4 C69+24=364-336+24=52

Total matrices=20+52+80+52=204