Q 11 :

Let A be a 3×3 matrix such that A+AT=O. If A[1-10]=[332],    A2[1-10]=[-319-24], and
det(adj(2adj(A+I)))=(2)α·(3)β·(11)γ, α,β,γ are non-negative integers, then α+β+γ is equal to _________ .           [2026]



(18)

A+AT=O

A[1-10]=[332]  and  A2[1-10]=[-319-24]

A+AT=O

A is skew symmetric matrix

A=[0ab-a0c-b-c0]

Now ATQ

[0ab-a0c-b-c0][1-10]=[332]

a=-3  and  c=b+2

A2[1-10]=A[332]=[0-3b30c-b-c0][332]

=[-9+2b9+2c-3b-3c]=[-319-24]

By comparing we get

b=3,  c=5

Now A=[0-33305-3-50]

Let B=A+I=[1-33315-3-51]

det(B)=44=22×11

we find D=det(adj(2 adj(B)))

D=214·114

Compare with 2α3β11γ

α=14,  β=0,  γ=4

α+β+γ=18



Q 12 :

If  Δ=|ahghbfgfc|,  then the cofactor A21 is:

  • -(hc+fg)

     

  • fg-hc

     

  • fg+hc

     

  • hc-fg

     

(2)

Ans.    fg-hc

Explanation:

      A21=(-1)2+1M21

             =-M21=-|hgfc|

            =-(hc-fg)

            =fg-hc



Q 13 :

If  Δ=|a11a12a13a21a22a23a31a32a33| and Aij is the cofactor of aij, then the value of Δ is given by:

  • a11A31+a12A32+a13A33

     

  • a11A11+a12A21+a13A31

     

  • a21A11+a22A12+a23A13

     

  • a11A11+a21A21+a31A31

     

(4)

Ans.    a11A11+a21A21+a31A31

Explanation:

∆= sum of the products of the elements of any row (or any column) with their corresponding cofactors.

∴                Δ=a11A11+a21A21+a31A31



Q 14 :

If  A=[23-4-6],  then which of the following is true?

  • A(adjA)≠|A|I

     

  • A(adjA)≠(adjA)A

     

  • A(adjA)=(adjA)A=|A|I=[0000]

     

  • None of the above

     

(3)

Ans.     A(adjA)=(adjA)A=|A|I=[0000]

Explanation:

We know that if A is any square matrix of order n, then A(adjA)=(adjA)A=|A|·I.



Q 15 :

If A and B are invertible matrices, then which of the following is not correct?

  • adjA=|A|·A-1

     

  • det(A)-1=[det(A)]-1

     

  • (AB)-1=B-1A-1

     

  • (A+B)-1=B-1+A-1

     

(4)

Ans.    (A+B)-1=B-1+A-1

Explanation:

Since A and B are invertible matrices, we can say that

               (AB)-1=B-1A-1         ⋯(i)

We know that,

                    A-1=1|A|(adjA)

⇒          adjA=|A|·A-1

Also,  det(A)-1=[det(A)]-1

⇒    det(A)-1=1det(A)

⇒  det(A)·det(A)-1=1,

which is true.



Q 16 :

If  A=[2λ-3025113],  then A-1 exists if:

  • λ=2

     

  • λ≠2

     

  • λ=-2

     

  • None of these

     

(4)

Ans.       None of these

Explanation:

We have,              A=[2λ-3025113]

A-1 exists if  |A|≠0

Now,         |A|=2(6-5)-λ(-5)-3(-2)=8+5λ

But,           |A|≠0

            5λ+8≠0⇒5λ≠-8

⇒              λ≠-85

So, A-1 exists if and only if  λ≠-85



Q 17 :

If  A=[21031204-1], then |adj(A)| is equal to:

  • 11

     

  • 12

     

  • 225

     

  • - 225

     

(3)

Ans.       225

Explanation:

Given,      A=[21031204-1]

        adj A=[21031204-1]

Now,  A=2(-1-8)-1(-3-0)+0

              =-18+3

              =-15

Since for a 3×3 matrix,

|adj(A)|=|A|3-1=|A|2

we get  |adj(A)|=(-15)2=152=225



Q 18 :

If A and B are invertible matrices of order 3, |A|=2 and |(AB)-1|=-16, then the value of |B| is: 

  • 3

     

  • -3

     

  • 2

     

  • 16

     

(2)

Ans.      -3

Explanation:

Given,             |A|=2,

and       |(AB)-1|=-16

We know that

            |BB-1|=I  and |AA-1|=I

            |B-1|=1B  and  |A-1|=1|A|=12

            (AB)-1=B-1A-1

⇒   |B-1||A-1|=-16

⇒           1B×12=-16

⇒                      B=-3



Q 19 :

Let  A=[1-23121λ2-3].  If A-1 does not exist, then λ=

  • -2

     

  • 2

     

  • 1

     

  • -1

     

(4)

Ans.      -1

Explanation:

Given,

                   A=[1-23121λ2-3]

Since A-1 does not exist ⇒|A|=0

Now,  1(-6-2)+2(-3-λ)+3(2-2λ)=0

                                   -8-6-2λ+6-6λ=0 

                                                          -8λ-8=0

                                                                       λ=-1



Q 20 :

The sum of the products of elements of any row with the cofactors of corresponding elements is: 

  • the value of the determinant

     

  • 0

     

  • sum of cofactors

     

  • adjoint of matrix

     

(1)

Ans.         the value of the determinant

Explanation:

The sum of the products of the elements of any row with the cofactors of the corresponding elements is equal to the value of the determinant.

     Let  Δ=|a11a12a13a21a22a23a31a32a33|



Q 21 :

If A is a square matrix of order 3 such that det(A)=3 and det(adj(-4 adj(-3 adj(3 adj((2A)-1)))))=2m3n. Then m+2n is equal to:

  • 2

     

  • 3

     

  • 6

     

  • 4

     

(4)

|A|=3

|adj(-4adj(-3adj(3adj((2A)-1))))|

=|-4adj(-3adj(3adj((2A)-1)))|2

=46|adj(-3adj(3adj((2A)-1)))|2

=212·312|3adj(2A)-1|8

=212·312·324|adj(2A)-1|8

=212·336|(2A)-1|16

=212·3361|2A|16

=212·3361248|A|16

=212·3361248316

=320236=2-36·320

m=-36,  n=20

m+2n=4



Q 22 :

If A is matrix of order 3 such that |A|=5 and B=adjA, then the value of ||A-1|(AB)T| is equal to (where |A| denotes determinant of matrix A.AT denotes transpose of matrix A, A-1 denotes inverse of matrix A, adjA denotes adjoint of matrix A)

  • 5 

     

  • 1 

     

  • 25

     

  • 125

     

(2)

|A-1|=1|A|=15

|(AB)T|=|AB|=|A(adjA)|=|A|·|adjA|=5×52=53

∴  ||A-1|(AB)T|=|15(AB)T|=153|AB|=1



Q 23 :

If A is a square matrix of order n then, (adjadjadj…adjA)⏟·(n-1) times(adjadjadj…adjA)⏟n times is equal to:

  • |A n-1|(n-1)2

     

  • |A-1|(n-1)(n-1)

     

  • 1n|A|(n-1)(n-1)·In×n

     

  • |A|(n-1)(n-1)·In×n

     

(4)

A·AdjA=|A|In

(AdjA)(AdjAdjA)=|AdjA|In=|A|n-1In

(AdjAdjA)(AdjAdjAdjA)=|A|(n-1)2In

⇒(AdjAdj…A)(AdjAdjAdj…A)=|A|(n-1)(n-1)In



Q 24 :

If 2sinα sinβ+3cosβ+5cosα sinβ=38 ∀α,β∈R, then det(adj(adjA)) is equal to (where A=[tanβ-131sinα]) is equal to ______



(1)

Let p→=2i^+3j^+5k^;   q→=sinα sinβi^+cosβj^+cosα sinβk^

|q→|=sin2αsin2β+cos2β+cos2αsin2β=1

⇒ sinαsinβ=2λ;  cosβ=3λ;  cosαsinβ=5λ

1=38λ2;  λ=138

detA=[sinα sinβcosβ+13]=1



Q 25 :

Let P=[3-1-220α3-50], where α∈R. Suppose Q=[qij] is a matrix satisfying PQ=kI3 for some non-zero k∈R. If q23=-k8 and |Q|=k22, then α2+k2 is equal to

  • 13

     

  • 15

     

  • 17

     

  • 21

     

(3)

Given that PQ=kI

|P|.|Q|=k3

⇒|P|=2k≠0⇒P is an invertible matrix

∵ PQ=kI    ∴Q=kP-1I    [∵P-1P=I]

∴ Q=adjP2    ∵ q23=-k8

∴ -(3α+4)2=k8⇒12α+16=k    ...(i)

∴ |P|=2k⇒k=10+6α

From (i) and (ii) we get α=-1, k=4    

∴ α2+k2=17



Q 26 :

Consider a matrix A=[αβλα2β2λ2β+λλ+αα+β] where α,β,λ are three distinct natural numbers.

If det(adj(adj(adj(adjA))))(α-β)16(β-γ)16(γ-α)16=232×316, then the number of such 3-triples (α,β,γ) is _______ . 



(42)

Given matrix is  A=[αβγα2β2γ2β+γγ+αα+β]

R3→R3+R1

⇒|A|=|α+β+γ||αβγα2β2γ2111|

⇒|A|=(α+β+γ)(α-β)(β-γ)(γ-α)       

∵ |adjA|=|A|n-1

|adj(adjA)|=|A|(n-1)2

|adj(adj(adjA))|=|A|(n-1)3=|A|8

|adj(adj(adj(adjA)))|=|A|(n-1)4=|A|16

∴ (α+β+γ)=232·316



Q 27 :

Let A=[111–201135]. Then the sum of all elements of the matrix adj(adj(2adjA)–1)) is equal to :          [2026]

  • 3

     

  • 4

     

  • –4

     

  • –3

     

(4)

We have, A=[111–201135]

|A| = 1(–3) – 1(–10 – 1) + 2(–6) = –3 + 11 – 12 = –4≠0

Let B=2adjA)–1

Then adj(adjB)=|B|n–2·B

=|2(adjA)–1|·2(adjA)–1           [∵ n = 1]

=23|adjA|·A|A|=24|A|2×A|A|=24A16|A|=A|A|

Sum of elements of A = 12

Sum of elements of the matrix adj (adj) B) = 12–4=–3.



Q 28 :

Let A be a 3×3 matrix such that AT=[101]=[522], AT=[001]=[311], A=[101]=[344] and A=[001]=[131]. If det (A) = 1, then det(adj(A2+A)) is equal to :          [2026]

  • 16

     

  • 25

     

  • 49

     

  • 64

     

(4)

Let A=[a1a2a3a4a5a6a7a8a9]

      ⇒ A=[a1a2a3a4a5a6a7a8a9][101]=[344]

     ⇒ a1+a3=3⇒ a4+a6=4⇒ a7+a9=4          (i)

Also, [a1a2a3a4a5a6a7a8a9][001]=[131]

     ⇒ a3=1 ⇒ a1=2⇒ a6=3 ⇒ a4=1⇒ a9=1 ⇒ a7=3          [From (i)]

Now, [a1a4a7a2a5a8a3a6a9][001]=[311]

⇒ a7=3⇒ a8=1⇒ a9=1          (ii)

Also, [a1a4a7a2a5a8a3a6a9][101]=[522]

⇒ a1+a7=5 ⇒ a1=2⇒ a2+a8=2 ⇒ a2=1⇒ a3+a9=2 ⇒ a3=1          [From (ii)]

|A|=|2111a53311| ⇒ –a5+3=1 ⇒ a5=2

∴  A=|211123311|

Now, |adjA·(A+I)|=|adj(A+I)||adjA|

|A+I|=|311133312|=8

∴  |adj(A2+A)|=64×1=64



Q 29 :

Let A=[α12230045] and B=[1000–5α004α–2α]+adj(A). If det (B) = 66, then det (adj (A)) equals :          [2026]

  • 289

     

  • 361

     

  • 441

     

  • 529

     

(3)

A=[α12230045]

B=[1000–5α004α–2α]+adj(A)

adj(A)=[15–10835α–4α–643α–2]T=[153–6–105α484α3α–2]

∴  B=[1000–5α004α–2α]+[153–6–105α484α3α–2]

⇒ B=[163–6–100480α–2]

Now, det (B) = 66

⇒ –3(–10α+20–32)=66 ⇒ α=1

det(A)=|112230045|=15–10+16=21

∴  det(adjA)=|A|2=(21)2=441



Q 30 :

Consider the matrices A=[2–24–2] and B=[3913]. If matrices P and Q are such that PA = B and AQ = B, then the absolute value of the sum of the diagonal elements of 2(P + Q) is _______.          [2026]



(34)

We have, A=[2–24–2] ⇒ |A|=–4+8=4

and B=[3913] ⇒ |B|=9–9=0

∴  A–1=1|A|adj(A)=14[–22–42]

Now, PA = B

⇒ P=BA–1

     =[3913]×14[–22–42]=14[–4224–148]

Now, AQ=B ⇒ Q=A–1B

     =14[–22–42][3913]=14[–412–10–30]

Now, sum of diagonal elements of 2(P + Q)

=(–464–224)=2(–684)=–34

∴  Absolute value = 34.



Q 31 :

Let A=[–11–1101001] satisfy A2+α(adj(adj(A)))+β(adj(A)(adj(adj(A))))=[2–22–20–100–1] for some α,β∈R. Then (α–β)2 is equal to _______.          [2026]



(4)

A=[–11–1101001]

⇒ |A|=–1

∵  adj(adjA)=|A|n–2A

and AadjA=|A|I

Now, A2=[2–11–110001]

We have, A2+α(adj(adj(A))+β(adj(A)(adj(adj(A)))=[2–22–20–100–1]

⇒ A2+α(–A)+βI=[2–22–20–100–1]

⇒ [2–11–10–100–1]–α[–11–1101001]+β=[100010001]=[2–22–20–100–1]

⇒ –[2+α+β–1–α1+α–1–α1+β–α001–α+β]=[2–22–20–100–1]

By comparison on both sides, we get

     α=1 and 1+β=0 ⇒ β=–1

∴  (α–β)2=(1+1)2=22=4



Q 32 :

If  A=[2294]  and  I=[1001] then 20A-1 = ____

  • 2A-12I

     

  • A-6I

     

  • A+6I

     

  • 2A+12I

     

(1)

A-1=18-18[4-2-92],  10A-1=[-429-2]



Q 33 :

A is 3×3 matrix and |A|=4 then |adj(adjA)|=4k. Then k is _____



(4)

|adj(adjA)|=|A|(n-1)2