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 · 1 Feb · Shift 1 · Q26
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 · 1 Feb · Shift 1 · Q26

Matrices and Determinants question

2023 · 1 Feb · Shift 1 · Q26

JEE MainMathematicsMatrices and DeterminantsMCQ+4 / −1
Let SSS denote the set of all real values of λ\lambdaλ such that the system of equations λx+y+z=1x+λy+z=1x+y+λz=1\lambda x+y+z=1x+\lambda y+z=1x+y+\lambda z=1λx+y+z=1x+λy+z=1x+y+λz=1 is inconsistent, then ∑λ∈S(∣λ∣2+∣λ∣)\sum_{\lambda \in S}\left(|\lambda|^{2}+|\lambda|\right)λ∈S∑​(∣λ∣2+∣λ∣) is equal to
  1. A
    12
  2. B
    2
  3. C
    4
  4. D
    6
View written solutionFree

Correct answer: D

  1. Write the system in matrix form

The given system is

λx+y+z=1,\lambda x+y+z=1,λx+y+z=1, x+λy+z=1,x+\lambda y+z=1,x+λy+z=1, x+y+λz=1.x+y+\lambda z=1.x+y+λz=1.

Its coefficient matrix is

A=(λ111λ111λ),A=\begin{pmatrix} \lambda & 1 & 1\\ 1 & \lambda & 1\\ 1 & 1 & \lambda \end{pmatrix},A=​λ11​1λ1​11λ​​,

and the RHS vector is

b=(111).\mathbf{b}=\begin{pmatrix}1\\1\\1\end{pmatrix}.b=​111​​.

We need the values of λ\lambdaλ for which the system is inconsistent.

  1. When can a system be inconsistent?

A linear system can be inconsistent only if the coefficient matrix is singular, i.e.

det⁡(A)=0.\det(A)=0.det(A)=0.

So first compute det⁡(A)\det(A)det(A).

For a 3×33\times 33×3 matrix with diagonal entries λ\lambdaλ and off-diagonal entries 111, the determinant is

det⁡(A)=(λ−1)2(λ+2).\det(A)=(\lambda-1)^2(\lambda+2).det(A)=(λ−1)2(λ+2).

Hence,

det⁡(A)=0  ⟺  λ=1 or λ=−2.\det(A)=0 \iff \lambda=1 \text{ or } \lambda=-2.det(A)=0⟺λ=1 or λ=−2.

So only these two values need to be checked for inconsistency.

  1. Check λ=1\lambda=1λ=1

Then the system becomes

x+y+z=1,x+y+z=1,x+y+z=1, x+y+z=1,x+y+z=1,x+y+z=1, x+y+z=1.x+y+z=1.x+y+z=1.

This is clearly consistent (infinitely many solutions). So λ=1∉S\lambda=1 \notin Sλ=1∈/S.

  1. Check λ=−2\lambda=-2λ=−2

Then the system becomes

−2x+y+z=1,-2x+y+z=1,−2x+y+z=1, x−2y+z=1,x-2y+z=1,x−2y+z=1, x+y−2z=1.x+y-2z=1.x+y−2z=1.

Now add all three equations:

(−2x+y+z)+(x−2y+z)+(x+y−2z)=1+1+1.(-2x+y+z)+(x-2y+z)+(x+y-2z)=1+1+1.(−2x+y+z)+(x−2y+z)+(x+y−2z)=1+1+1.

Left side simplifies to

0,0,0,

so we get

0=3,0=3,0=3,

which is impossible.

Hence the system is inconsistent for λ=−2\lambda=-2λ=−2. Therefore,

S={−2}.S=\{-2\}.S={−2}.
  1. Compute the required sum

We need

∑λ∈S(∣λ∣2+∣λ∣).\sum_{\lambda\in S}\left(|\lambda|^2+|\lambda|\right).λ∈S∑​(∣λ∣2+∣λ∣).

Since S={−2}S=\{-2\}S={−2},

∣−2∣2+∣−2∣=22+2=4+2=6.|{-2}|^2+|{-2}|=2^2+2=4+2=6.∣−2∣2+∣−2∣=22+2=4+2=6.

So the required value is

6.6.6.
  1. Compare with the stored correct answer

Stored correct answer: D

Our derived answer is 6, which corresponds to Option D.

So the stored answer is correct.

PreviousNext

More from Matrices and Determinants

  • 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 · MCQ
  • If A=21​[1−3​​3​1​], then :2023 · MCQ
  • If the system of equations x+y+az=b2x+5y+2z=6x+2y+3z=3 has infinitely many solutions, then 2a+3b is equal to :2023 · MCQ
  • Let A=[aij​]2×2​, where aij​eq0 for all i,j and A2=I. Let a be the sum of all diagonal elements of A and b=∣A∣…2023 · MCQ
  • Let P be a square matrix such that P2=I−P. For α,β,γ,δ∈N, if Pα+Pβ=γI−29P and Pα−Pβ=δI−13P, then α+β+γ−δ is equal to :2023 · MCQ
  • For the system of equations x+y+z=6x+2y+αz=10x+3y+5z=β, which one of the following is NOT true?2023 · MCQ
  • Let A=​210​12−1​0−12​​. If ∣adj(adj(adj2A))∣=(16)n, then n is equal to :2023 · MCQ
  • Let P=[23​​−21​​21​23​​​],A=[10​11​] and Q=PAPT. If PTQ2007P=[ac​bd​]…2023 · MCQ