Sign in
12thPass logo
New chatPYQ LibraryDoubtsRank report
Sign in to see Recents

Your guest activity stays on this device

Sign in to save progress →
Sign in

Matrices and Determinants question

2023 · 30 Jan · Shift 1 · Q25
Guest · filters and generic practice availableBrowsing as a guest · PYQ filters and generic practice are available. Sign in only for personalised features and saved progress.
  1. PYQ Library
  2. /JEE Main
  3. /Mathematics
  4. /Matrices and Determinants
  5. /2023 · 30 Jan · Shift 1 · Q25

Matrices and Determinants question

2023 · 30 Jan · Shift 1 · Q25

JEE MainMathematicsMatrices and DeterminantsMCQ+4 / −1
Let the system of linear equations x+y+kz=22x+3y−z=13x+4y+2z=kx+y+kz=22x+3y-z=13x+4y+2z=kx+y+kz=22x+3y−z=13x+4y+2z=k have infinitely many solutions. Then the system (k+1)x+(2k−1)y=7(2k+1)x+(k+5)y=10(k+1)x+(2k-1)y=7(2k+1)x+(k+5)y=10(k+1)x+(2k−1)y=7(2k+1)x+(k+5)y=10 has :
  1. A
    unique solution satisfying x−y=1x-y=1x−y=1
  2. B
    infinitely many solutions
  3. C
    no solution
  4. D
    unique solution satisfying x+y=1x+y=1x+y=1
View written solutionFree

Correct answer: D

  1. Interpret the first system correctly

The given system is:

x+y+kz=2,x+y+kz=2,x+y+kz=2, 2x+3y−z=1,2x+3y-z=1,2x+3y−z=1, 3x+4y+2z=k.3x+4y+2z=k.3x+4y+2z=k.

We are told this system has infinitely many solutions.

For a 3×33\times 33×3 linear system to have infinitely many solutions, the equations must be dependent, so the determinant of the coefficient matrix must be zero and the augmented system must be consistent.

The coefficient matrix is

A=(11k23−1342).A=\begin{pmatrix} 1 & 1 & k\\ 2 & 3 & -1\\ 3 & 4 & 2 \end{pmatrix}.A=​123​134​k−12​​.
  1. Find kkk using determinant condition
det⁡(A)=∣11k23−1342∣\det(A)=\begin{vmatrix} 1 & 1 & k\\ 2 & 3 & -1\\ 3 & 4 & 2 \end{vmatrix}det(A)=​123​134​k−12​​

Expanding along the first row,

=1∣3−142∣−1∣2−132∣+k∣2334∣=1\begin{vmatrix}3 & -1\\4 & 2\end{vmatrix} -1\begin{vmatrix}2 & -1\\3 & 2\end{vmatrix} +k\begin{vmatrix}2 & 3\\3 & 4\end{vmatrix}=1​34​−12​​−1​23​−12​​+k​23​34​​ =1(3⋅2−(−1)⋅4)−1(2⋅2−(−1)⋅3)+k(2⋅4−3⋅3)=1(3\cdot 2-(-1)\cdot 4)-1(2\cdot 2-(-1)\cdot 3)+k(2\cdot 4-3\cdot 3)=1(3⋅2−(−1)⋅4)−1(2⋅2−(−1)⋅3)+k(2⋅4−3⋅3) =(6+4)−(4+3)+k(8−9)=(6+4)-(4+3)+k(8-9)=(6+4)−(4+3)+k(8−9) =10−7−k=3−k.=10-7-k=3-k.=10−7−k=3−k.

For infinitely many solutions,

3−k=0  ⟹  k=3.3-k=0 \implies k=3.3−k=0⟹k=3.
  1. Check consistency for k=3k=3k=3

Substitute k=3k=3k=3 into the first system:

x+y+3z=2,x+y+3z=2,x+y+3z=2, 2x+3y−z=1,2x+3y-z=1,2x+3y−z=1, 3x+4y+2z=3.3x+4y+2z=3.3x+4y+2z=3.

Now observe:

(x+y+3z)+(2x+3y−z)=3x+4y+2z,(x+y+3z)+(2x+3y-z)=3x+4y+2z,(x+y+3z)+(2x+3y−z)=3x+4y+2z,

and on the right side,

2+1=3.2+1=3.2+1=3.

So the third equation is the sum of the first two. Hence the system is consistent and dependent, so it indeed has infinitely many solutions.

Thus,

k=3.k=3.k=3.
  1. Now solve the second system

The second system is:

(k+1)x+(2k−1)y=7,(k+1)x+(2k-1)y=7,(k+1)x+(2k−1)y=7, (2k+1)x+(k+5)y=10.(2k+1)x+(k+5)y=10.(2k+1)x+(k+5)y=10.

Substitute k=3k=3k=3:

(3+1)x+(2⋅3−1)y=7  ⟹  4x+5y=7,(3+1)x+(2\cdot 3-1)y=7 \implies 4x+5y=7,(3+1)x+(2⋅3−1)y=7⟹4x+5y=7, (2⋅3+1)x+(3+5)y=10  ⟹  7x+8y=10.(2\cdot 3+1)x+(3+5)y=10 \implies 7x+8y=10.(2⋅3+1)x+(3+5)y=10⟹7x+8y=10.

So we solve:

4x+5y=7...(1)4x+5y=7 \quad ...(1)4x+5y=7...(1) 7x+8y=10...(2)7x+8y=10 \quad ...(2)7x+8y=10...(2)
  1. Find the solution

From (1), multiply by 777:

28x+35y=4928x+35y=4928x+35y=49

From (2), multiply by 444:

28x+32y=4028x+32y=4028x+32y=40

Subtract:

3y=9  ⟹  y=3.3y=9 \implies y=3.3y=9⟹y=3.

Substitute into (1):

4x+5(3)=74x+5(3)=74x+5(3)=7 4x+15=74x+15=74x+15=7 4x=−84x=-84x=−8 x=−2.x=-2.x=−2.

Thus the unique solution is

(x,y)=(−2,3).(x,y)=(-2,3).(x,y)=(−2,3).
  1. Check the options

  • x−y=−2−3=−5≠1x-y=-2-3=-5 \ne 1x−y=−2−3=−5=1
    So A is false.

  • The system has a unique solution, not infinitely many.
    So B is false.

  • The system has a solution, so C is false.

  • x+y=−2+3=1x+y=-2+3=1x+y=−2+3=1
    So D is true.

Final Answer

The system has a unique solution satisfying x+y=1x+y=1x+y=1.

So the correct option is:

D\boxed{\text{D}}D​

PreviousNext

More from Matrices and Determinants

  • For α,β∈R, suppose the system of linear equations ​x−y+z=52x+2y+αz=83x−y+4z=β​ has infinitely many solutions. Then α and β are…2023 · MCQ
  • For the system of linear equations x+y+z=6αx+βy+7z=3x+2y+3z=14 which of the following is NOT true ?2023 · MCQ
  • Let A=​100​0412​0−1−3​​. Then the sum of the diagonal elements of the matrix (A+I)11 is equal to :2023 · MCQ
  • The number of values of α for which the system of equations : x + y + z =αα x + 2 α y + 3z = − 1 x + 3 α y + 5z = 4 is inconsistent, is2022 · MCQ
  • Let the system of linear equations x + y + α z = 2 3x + y + z = 4 x + 2z = 1 have a unique solution (x ∗, y ∗, z ∗). If (α, x ∗), (y ∗, α) and (x ∗, − y ∗) are collinear points, then the…2022 · MCQ
  • Let S={(−10​ab​);a,b∈{1,2,3,....100}} and let Tn​={A∈S:An(n+1)=I}. Then the number of elements in n=1⋂100​Tn​…2022 · Numerical
  • The number of θ∈(0,4π) for which the system of linear equations ​3(sin3θ)x−y+z=23(cos2θ)x+4y+3z=36x+7y+7z=9​ has no solution, is :2022 · MCQ
  • Let A=​211​−10−1​−1−10​​ and B=A−I. If ω=23​i−1​, then the number of elements in the set{n∈{1,2,…,100}:An+(ωB)n=A+B}…2022 · Numerical