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Matrices and Determinants question

2016 · Shift 0 · Q28
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  5. /2016 · Shift 0 · Q28

Matrices and Determinants question

2016 · Shift 0 · Q28

JEE MainMathematicsMatrices and DeterminantsMCQ+4 / −1
The system of linear equations x+λy−z=0λx−y−z=0x+y−λz=0\begin{matrix} {x + \lambda y - z = 0} \\ {\lambda x - y - z = 0} \\ {x + y - \lambda z = 0} \\ \end{matrix}x+λy−z=0λx−y−z=0x+y−λz=0​ has a non-trivial solution for :
  1. A
    infinitely many values of λ.\lambda .λ.
  2. B
    exactly one value of λ.\lambda .λ.
  3. C
    exactly two values of λ.\lambda .λ.
  4. D
    exactly three values of λ.\lambda .λ.
View written solutionFree

Correct answer: D

  1. For a homogeneous system of linear equations to have a non-trivial solution, the determinant of its coefficient matrix must be zero.

  2. Write the coefficient matrix:

A=(1λ−1λ−1−111−λ)A = \begin{pmatrix} 1 & \lambda & -1 \\ \lambda & -1 & -1 \\ 1 & 1 & -\lambda \end{pmatrix}A=​1λ1​λ−11​−1−1−λ​​

We need:

det⁡(A)=0\det(A)=0det(A)=0
  1. Compute the determinant by expanding along the first row:
det⁡(A)=1∣−1−11−λ∣−λ∣λ−11−λ∣+(−1)∣λ−111∣\det(A)=1\begin{vmatrix}-1 & -1 \\ 1 & -\lambda\end{vmatrix} -\lambda\begin{vmatrix}\lambda & -1 \\ 1 & -\lambda\end{vmatrix} +(-1)\begin{vmatrix}\lambda & -1 \\ 1 & 1\end{vmatrix}det(A)=1​−11​−1−λ​​−λ​λ1​−1−λ​​+(−1)​λ1​−11​​

Now evaluate each minor:

∣−1−11−λ∣=(−1)(−λ)−(−1)(1)=λ+1\begin{vmatrix}-1 & -1 \\ 1 & -\lambda\end{vmatrix}=(-1)(-\lambda)-(-1)(1)=\lambda+1​−11​−1−λ​​=(−1)(−λ)−(−1)(1)=λ+1 ∣λ−11−λ∣=λ(−λ)−(−1)(1)=−λ2+1=1−λ2\begin{vmatrix}\lambda & -1 \\ 1 & -\lambda\end{vmatrix}=\lambda(-\lambda)-(-1)(1)=-\lambda^2+1=1-\lambda^2​λ1​−1−λ​​=λ(−λ)−(−1)(1)=−λ2+1=1−λ2 ∣λ−111∣=λ(1)−(−1)(1)=λ+1\begin{vmatrix}\lambda & -1 \\ 1 & 1\end{vmatrix}=\lambda(1)-(-1)(1)=\lambda+1​λ1​−11​​=λ(1)−(−1)(1)=λ+1

So,

det⁡(A)=(λ+1)−λ(1−λ2)−(λ+1)\det(A)=(\lambda+1)-\lambda(1-\lambda^2)- (\lambda+1)det(A)=(λ+1)−λ(1−λ2)−(λ+1)

Since the first and third terms cancel,

det⁡(A)=−λ(1−λ2)=λ(λ2−1)\det(A)=-\lambda(1-\lambda^2)=\lambda(\lambda^2-1)det(A)=−λ(1−λ2)=λ(λ2−1)

Thus,

det⁡(A)=λ(λ−1)(λ+1)\det(A)=\lambda(\lambda-1)(\lambda+1)det(A)=λ(λ−1)(λ+1)
  1. For non-trivial solutions,
λ(λ−1)(λ+1)=0\lambda(\lambda-1)(\lambda+1)=0λ(λ−1)(λ+1)=0

Hence,

λ=0,  1,  −1\lambda=0,\;1,\;-1λ=0,1,−1

So there are exactly three values of λ\lambdaλ.

  1. Check options:
  • A: infinitely many values of λ\lambdaλ — false
  • B: exactly one value — false
  • C: exactly two values — false
  • D: exactly three values — true

Therefore, the correct option is D.

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