Q 31 :    

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 λS(|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+82λ=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|=90
for λ=-2, we have no solution.

λS(|λ2|+|λ|)=(4+2)=6



Q 32 :    

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=1x=14

Δ2=|211111112|=2-1=1y=14

Δ3=|211121111|=2-1=1z=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=9x=90

Δ2=|-21111111-2|=6+3+0=9y=90

Δ3=|-2111-21111|=6+0+3=9z=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 33 :    

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+60

(N-2)(N-3)0N2,3

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

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



Q 34 :    

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=05λ=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+72517μ=-35785=-215

Thus, (λ,μ)(725,-215)



Q 35 :    

Let S1 and S2 be respectively the sets of all aR-{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 (C2C2-C1), we get

|A|=a×|11-32a+12a+13a+5-2aa+2|

=a×|11-3-a-42+2a-13a+5-2aa+2|                  Applying (R2R2-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  aR as 15a2+31a+370aR & a0

S1=R-{0}

Since |A| can never be 0. 

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

  S2=ϕ



Q 36 :    

Consider the following system of equations

αx+2y+z=1

2αx+3y+z=1

3x+αy+2z=β

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

  • It has no solution if α=-1 and β2

     

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

     

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

     

  • It has no solution for α=3 and for all β2

     

(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|0β2

Now, Δy=|α112α1132β|0 -αβ+2α0

For α=-1β2

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



Q 37 :    

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|=01(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 38 :    

For α,βR, 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 39 :    

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 α7, 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 α7, 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 40 :    

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 12λS|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=0λ=1,-13,-23

For each λ,  Δ1=|6λ-34λ226λ132λ|0

Then, 12λS|λ|=12(1+13+23)=24