Q 11 :

Let S denote the set of all real values of λ such that the system of equations λx+y+z=1, x+λy+z=1, x+y+λz=1 is inconsistent. Then ∑λ∈S(|λ|2+|λ|) is equal to             [2023]

  • 12

     

  • 4

     

  • 2

     

  • 6

     

(4)

Since, the given system of equations is inconsistent

⇒|λ111λ111λ|=0 ⇒λ(λ2-1)-1(λ-1)+1(1-λ)=0

⇒λ3-λ-λ+1+1-λ=0 ⇒(λ-1)(λ2+λ-2)=0

⇒λ=1,  λ=-1±1+82⇒λ=1, λ=-1±32

⇒λ=1,1,-2

Now, for λ=1, we have infinite solutions as we have only one equation x+y+z=1

For λ=-2, D1=|1111-2111-2|=9≠0
⇒for λ=-2, we have no solution.

∑λ∈S(|λ2|+|λ|)=(4+2)=6



Q 12 :

For the system of linear equations αx+y+z=1,  x+αy+z=1,  x+y+αz=β, which one of the following statements is NOT correct?       [2023]

  • It has infinitely many solutions if α=2 and β=-1

     

  • x+y+z=34 if α=2 and β=1

     

  • It has infinitely many solutions if α=1 and β=1

     

  • It has no solution if α=-2 and β=1

     

(1)

|α111α111α|=0

⇒α(α2-1)-1(α-1)+(1-α)=0

⇒α3-α-α+1+1-α=0 ⇒α2(α-1)+α(α-1)-2(α-1)=0

(α-1)(α2+α-2)=0 ⇒α=1,α=-2,1

For α=2, β=1;Δ=4

Δ1=|111121112|=3-1-1=1⇒x=14

Δ2=|211111112|=2-1=1⇒y=14

Δ3=|211121111|=2-1=1⇒z=14

For α=2⇒ unique solution and x+y+z=34

For α=-2, β=1

Δ=|-2111-2111-2|=-2(4-1)-1(-2-1)+1(1+2)

=-6+3+3=0

Δ1=|1111-2111-2|=3+3+3=9⇒x=90

Δ2=|-21111111-2|=6+3+0=9⇒y=90

Δ3=|-2111-21111|=6+0+3=9⇒z=90

For α=-2⇒ no solution

For α=1 and β=1

Δ=Δ1=Δ2=Δ3=0

∴It has infinitely many solutions if α=1 and β=1

For α=2 and β=-1

Δ=0

Δ1=|111121-112|=1(3)-1(3)+1(3)=3

Δ2=|2111111-12|=2(3)-1(1)+1(-2)=3

Δ3=|21112111-1|=2(-3)-1(-2)+1(-1)=-5

Hence, it does not have infinitely many solutions if α=2 and β=-1.



Q 13 :

Let N denote the number that turns up when a fair die is rolled. If the probability that the system of equations 

x+y+z=1

2x+Ny+2z=2

3x+3y+Nz=3

has a unique solution is k6, then the sum of the value of k and all possible values of N is                [2023]

  • 18

     

  • 19

     

  • 20

     

  • 21

     

(3)

For unique solution, determinant ≠0

|1112N233N|≠0

⇒N2-5N+6≠0

⇒(N-2)(N-3)≠0⇒N≠2,3

∴Possible values of N are 1,4,5,6 and k6=46⇒k=4

Required sum=1+4+5+6+4=20



Q 14 :

If the system of equations

x+2y+3z=3

4x+3y-4z=4

8x+4y-λz=9+μ

has infinitely many solutions, then the ordered pair (λ,μ) is equal to               [2023]

  • (725,-215)

     

  • (725,215)

     

  • (-725,-215)

     

  • (-725,215)

     

(1)

x+2y+3z=3;  4x+3y-4z=4;  8x+4y-λz=9+μ

For infinite many solutions, Δ=0 and Δx=0

Δ=|12343-484-λ|=0

⇒1(-3λ+16)-2(-4λ+32)+3(16-24)=0

⇒16-3λ+8λ-64-24=0⇒5λ=72

∴   λ=725

Δx=|32343-49+μ4-λ|=0

⇒3(-3λ+16)-2(-4λ+36+4μ)+3(16-27-3μ)=0

⇒-9λ+48+8λ-72-8μ-33-9μ=0

⇒-λ-17μ=57⇒-17μ=57+λ

∴   -μ=57+72517⇒μ=-35785=-215

Thus, (λ,μ)≡(725,-215)



Q 15 :

Let S1 and S2 be respectively the sets of all a∈R-{0} for which the system of linear equations

ax+2ay-3az=1

(2a+1)x+(2a+3)y+(a+1)z=2

(3a+5)x+(a+5)y+(a+2)z=3

has a unique solution and infinitely many solutions. Then                       [2023]

  • S1=Φ and S2=R-{0}

     

  • S1 is an infinite set and n(S2)=2

     

  • S1=R-{0} and S2=Φ  

     

  • n(S1)=2 and S2 is an infinite set.

     

(3)

|A|=|a2a-3a2a+12a+3a+13a+5a+5a+2|

=a×|12-32a+12a+3a+13a+5a+5a+2|

Applying (C2→C2-C1), we get

|A|=a×|11-32a+12a+13a+5-2aa+2|

=a×|11-3-a-42+2a-13a+5-2aa+2|                  Applying (R2→R2-R3)

=a×[2a2+6a+4-2a-3a-5+(a+4)(a+2)-3[(a+4)2a-(2+2a)(3a+5)]]

=a×(2a2+6a+4-2a-3a-5+a2+6a+8)-3(2a2+8a-6a2-16a-10)

|A|=a(15a2+31a+37)

|A|≠0  ∀a∈R as 15a2+31a+37≠0 ∀a∈R & a≠0

S1=R-{0}

Since |A| can never be 0. 

There cannot be infinitely many solutions for any value of a.  

∴  S2=ϕ



Q 16 :

Consider the following system of equations

αx+2y+z=1

2αx+3y+z=1

3x+αy+2z=β

for some α,β∈ℝ. Then which of the following is NOT correct?               [2023]

  • It has no solution if α=-1 and β≠2

     

  • It has no solution for α=-1 and for all β∈ℝ

     

  • It has a solution for all α≠-1 and β=2

     

  • It has no solution for α=3 and for all β≠2

     

(2)

Given system of equations has no solution if

|α212α313α2|=0

⇒α(6-α)-2(4α-3)+(2α2-9)=0 ⇒α=3,-1

For no solution, any one of Δx,Δy,Δz should be non-zero.

Now, Δx=|211311α2β|≠0⇒β≠2

Now, Δy=|α112α1132β|≠0 ⇒-αβ+2α≠0

For α=-1⇒β≠2

So, option (2) is incorrect as it has no solution for α=-1 and β∈R-{2}.



Q 17 :

Let the system of linear equations

x+y+kz=2

2x+3y-z=1

3x+4y+2z=k

have infinitely many solutions. Then the system

(k+1)x+(2k-1)y=7

(2k+1)x+(k+5)y=10

has                                                                      [2023]

  • infinitely many solutions

     

  • unique solution satisfying x-y=1

     

  • no solution

     

  • unique solution satisfying x+y=1

     

(4)

We have system of linear equations,

x+y+kz=2,

2x+3y-z=1,

3x+4y+2z=k

System has infinitely many solutions, then |A|=0 and (adj A)B=0

Here A=[11k23-1342]

⇒|A|=0⇒1(6+4)-1(4+3)+k(8-9)=0

⇒3-k=0,  k=3

For k=3, (adj A)B=0

Now, new system will be 4x+5y=7, 7x+8y=10

After solving these two equations, we get x=-2, y=3

⇒x-y=-2-(3)=-5 and x+y=-2+3=1



Q 18 :

For α,β∈R, suppose the system of linear equations

x-y+z=5

2x+2y+αz=8

3x-y+4z=β

has infinitely many solutions. Then α and β are the roots of               [2023]

  • x2-18x+56=0

     

  • x2+14x+24=0

     

  • x2-10x+16=0

     

  • x2+18x+56=0

     

(1)

Given, x-y+z=5,  2x+2y+αz=8,  3x-y+4z=β

It can be written as AX=B

Where, A=[1-1122α3-14], B=[58β], X=[xyz]

Condition for infinite solutions |A|=0 and (Adj A)B=0

∴   |A|=8+α+8-3α-8=0 ⇒α=4

Adj A=[123-641-2-8-24]

(Adj A)B=[123-641-2-8-24][58β]=0

⇒60+24-6β=0 ⇒β=14   ∴   α+β=4+14=18,   αβ=4×14=56

∴   Required equation is given by  x2-18x+56=0



Q 19 :

For the system of linear equations x+y+z=6, αx+βy+7z=3, x+2y+3z=14, which of the following is NOT true?          [2023]

  • If α=β and α≠7, then the system has a unique solution 

     

  • There is a unique point (α,β) on the line x+2y+18=0 for which the system has infinitely many solutions 

     

  • For every point (α,β)≠(7,7) on the line x-2y+7=0, the system has infinitely many solutions 

     

  • If α=β=7, then the system has no solution

     

(3)

We have, x+y+z=6,  αx+βy+7z=3,  x+2y+3z=14

Now, let 
A=[111αβ7123],   B=[6314]

⇒|A|=1(3β-14)-1(3α-7)+1(2α-β) =2β-α-7

If α=β and α≠7, then |A|≠0, so the system has unique solutions.  

Hence (1) is true.

If α=β=7, then |A|=0 and the system of equations is:

x+y+z=6  ...(i)

7x+7y+7z=3  ...(ii)

x+2y+3z=14  ...(iii)

From equations (i) and (ii), two rows in matrix A will get identical; therefore, the system has no solutions.  

Hence (1) is true.

For every point (α,β)≠(7,7) on the line x-2y+7=0,  

|A|≠0. Hence, the system has unique solutions. Therefore, (3) is false.

(α,β)≡(3,3) does not satisfy x+2y+18=0, then 

Δ1=|6113β71423|=6(3β-14)-1(9-98)+1(6-14β)

=4β+11

Δ2=|161α371143|=1(9-98)-6(3α-7)+1(14α-3)

=-4α-50

Δ3=|116αβ31214|=1(14β-6)-1(14α-3)+6(2α-β)

=8β-2α-3

For infinite solution, |A|=0 and Δ1=Δ2=Δ3=0, then

2β-α=7,  β=-114,  α=-252, and 8β-2α=3

Hence, there is a unique (α,β) on the line x+2y+18=0 for which the system has infinitely many solutions.  

Hence (2) is true.

 



Q 20 :

Let S be the set of values of λ, for which the system of equations

6λx-3y+3z=4λ2,

2x+6λy+4z=1,

3x+2y+3λz=λ has no solution. Then 12∑λ∈S|λ| is equal to ___________ .              [2023]



(24)

For no solution, Δ=0

⇒|6λ-3326λ4323λ|=0

⇒6λ(18λ2-8)+3(6λ-12)+3(4-18λ)=0

⇒18λ3-14λ-4=0⇒λ=1, -13, -23

For each λ,  Δ1=|6λ-34λ226λ132λ|≠0

Then, 12∑λ∈S|λ|=12(1+13+23)=24



Q 21 :

System of linear equations  5x+2y=4, 7x+3y=5 has:

  • unique solution

     

  • infinitely many solutions

     

  • no solution

     

  • none of the above

     

(1)

Ans.      Unique solution

Explanation:

Given,

               5x+2y=4

               7x+3y=5

Here,

             a1a2=57,   b1b2=23

Since,  a1a2≠b1b2

Thus, given system of linear equations has a unique solution.



Q 22 :

The following system of linear equations

                        3x+5y=7

                        2x-5y=9

is:

  • consistent

     

  • inconsistent

     

  • data incomplete

     

  • can't be said anything

     

(1)

Ans.       Consistent

Explanation:

Given,

                    3x+5y=7

                    2x-5y=9

The coefficient matrix is

                  A=[352-5]

Now,   |A|=3(-5)-2(5)=-15-10=-25≠0

∴  A-1 can be calculated

∴  System of linear equations is consistent.



Q 23 :

The value of x and y, if the system of linear equations is 2x-y=2 and 3x+4y=3 is:

  • x=1, y=0

     

  • x=1, y=-1

     

  • x=0, y=1

     

  • x=-1, y=-1

     

(1)

Ans.     x=1,  y=0

Explanation:

Given,

                 2x-y=2

                 3x+4y=3

The coefficient matrix is

             A=[2-134]

             B=[23],  X=[xy]

Now,    |A|=2(4)-(-1)(3)=8+3=11

     adj(A)=[41-32]

Therefore,   A-1=111[41-32]

Hence,  X=A-1B

         [xy]=111[41-32][23]=111[8+3-6+6]

Thus,  x=1111=1,    y=011=0

Hence,  x=1,    y=0



Q 24 :

The number of distinct real values of 'K' such that the system of equations x+2y+z=1,  x+3y+4z=K,  x+5y+10z=K2 has infinitely many solutions is:

  • 0

     

  • 4

     

  • 2

     

  • 3

     

(3)

D=|1211341510|=0

D1=|121K34K2510|=5(K2-3K+2)=5(K-1)(K-2)

D2=|1111K41K210|=-3(K2-3K+2)=-3(K-2)(K-1)

D3=|12113K15K2|=K2-3K+2=(K-2)(K-1)



Q 25 :

The system of equations

(1-λ)x+3y-4z=0

x-(3+λ)y+5z=0

3x+y-λz=0

possesses non-trivial solutions for

  • exactly three distinct real values of λ

     

  • exactly two distinct real values of λ

     

  • exactly one real value of λ

     

  • infinite number of distinct real values of λ

     

(2)

Put  |1-λ3-41-(3+λ)531-λ|=0

⇒λ(λ+1)2=0⇒λ=-1,0



Q 26 :

If the system of equations

x+y+z=16

2x+5y+αz=β

x+2y+3z=14

has infinitely many solutions, then α+β is equal to



(34)

It has infinitely many solutions.



Q 27 :

Let α,β∈R be such that the system of linear equations x + 2y + z = 5; 2x+y+αz=5; 8x+4y+βz=18 has no solution. Then βα is equal to :          [2026]

  • –4

     

  • 4

     

  • 8

     

  • –8

     

(2)

Given system of linear equations: x + 2y + z = 5, 2x+y+αz=5, 8x+4y+βz=18

Since the given system of linear equations has no solution, so, △=0 

⇒ △=|12121α84β|=12α–3β=0

⇒ β=4α

    △z=|1252158418|=6≠0

⇒ z=△z△ ⇒ z=60 = not defined

Hence, βα=4



Q 28 :

If the system of equations x + 5y + 6z = 4, 2x + 3y + 4z = 7, x + 6y + az = b has infinitely many solutions, then the point (a, b) lies on the line          [2026]

  • y – x = 3

     

  • x – y = 3

     

  • x + y = 11

     

  • x + y = 12

     

(2)

△=|15623416a|=0

⇒ (3a–24)–5(2a–4)+6(9)=0

⇒ –7a+50=0 ⇒ a=507

Now, △z==0 ⇒ |15423716b|=0

⇒ (3b–42)–5(2b–7)+4(9)=0

⇒ –7b+29=0 ⇒ b=297

Now, a – b = 3

∴  x – y = 3



Q 29 :

If the system of equations: x + y + z = 5; x + 2y + 3z = 9; x + 2y + λz = μ has infinitely many solutions, then the value of λ+μ is:          [2026]

  • 16

     

  • 18

     

  • 19

     

  • 21

     

(2)

Since, given system equations have infinitely many solutions.

∴  |11112313λ|=0

⇒ 1(2λ–9)–1(λ–3)+1(3–2)=0

⇒ 2λ–9–(λ–3)+1=0 ⇒ λ–5=0

⇒ λ=5

∴  △3=|11512913μ|=0

⇒ 1(2μ–27)–1(μ–9)+5(3–2)=0

⇒ (2μ–27)–(μ–9)+5(1)=0

⇒ μ–13=0 ⇒ μ=13

So, λ+μ=5+13=18



Q 30 :

The sum of all possible values of θ∈[0,2π], for which the system of equations:

xcos3θ–8y–12z=0;

xcos2θ+3y+3z=0;

x + y + 3z = 0

has a non-trivial solution, is equal to:          [2026]

  • π

     

  • 2π

     

  • 3π

     

  • 4π

     

(4)

Since the given system of equations has non-trivial solution,

△=|cos3θ–8–12cos2θ33113|

Expanding along C1

⇒ cos3θ|3313|–cos2θ|–8–1213|+1|–8–1233|=0

⇒ 6cos3θ–(–12)cos2θ+12=0

⇒ cos3θ+2cos2θ+2=0

Since, cos2θ=2cos2θ–1 and cos3θ=4cos3θ–3cosθ

∴  (4cos3θ–3cosθ)+4cos2θ–2+2=0

⇒ 4cos3θ+4cos2θ–3cosθ=0

⇒ cosθ(4cos2θ+4cosθ–3)=0

⇒ cosθ(2cosθ–1)(2cosθ+3)=0

⇒ cosθ=0, cosθ=12 and cosθ=–32 (no real soluion)

If cosθ=0, θ=π2, 3π2

If cosθ=12, θ=π3, 3π3

∴  Required sum = π2+3π2+π3+5π3

                                    =2π+2π=4π



Q 31 :

If the system of linear equations:

x + y + z = 6

x + 2y + 5z = 10

2x + 3y + λz = μ

has infinitely many solutions, then the value of λ+μ equals:          [2026]

  • 12

     

  • 16

     

  • 22

     

  • 28

     

(3)

x + y + z = 6

x + 2y + 3z = 10

2x + 3y + λz = μ

For infinitely many solutions, △=0

⇒ |11112523λ|=0

⇒ (2λ–15)–(λ–10)+(3–4)=0

⇒ λ=6

Also, △1=0

⇒ |6111025μ36|=0

⇒ 6(12–15)–(60–5μ)+(30–2μ)=0

⇒ –18–60+5μ+30–2μ=0

⇒ 3μ=48 ⇒ μ=16

∴  λ+μ=22



Q 32 :

If the system of equations x+y+z=5, x+2y+3z=9, x+3y+αz=β has infinitely many solutions, then β-2α equals:

  • 21

     

  • 3

     

  • 18

     

  • 5

     

(2)

x+y+z=5,    x+2y+3z=9,    x+3y+αz=β

D=|11112313α|=0

⇒(2α-9)+(3-α)+(3-2)=0

⇒α=5

Now,  D3=|11512913β|=0

⇒2β-27+9-β+5(3-2)=0

⇒β=13

⇒at α=5, β=13



Q 33 :

If A and B are the two real values of k for which the system of equations x+2y+z=1, x+3y+4z=k, x+5y+10z=k2 is consistent,

then A2+B2 is _______



(5)

Here Δ=0 and Δ1=Δ2=Δ3=0⇒k=1,2 then A2+B2=1+4=5