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Matrices and Determinants question

2017 · 8 Apr · Shift 1 · Q30
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Matrices and Determinants question

2017 · 8 Apr · Shift 1 · Q30

JEE MainMathematicsMatrices and DeterminantsMCQ+4 / −1
The number of real values of λ\lambdaλ for which the system of linear equations 2x + 4y −λ-\lambda−λ z = 0 4x + λ\lambdaλ y + 2z = 0 λ\lambdaλ x + 2y + 2z = 0 has infinitely many solutions, is :
  1. A
    0
  2. B
    1
  3. C
    2
  4. D
    3
View written solutionFree

Correct answer: B

  1. For a homogeneous system of 3 linear equations in 3 variables to have infinitely many solutions, the coefficient matrix must be singular, i.e. its determinant must be zero.

    The system is

    {2x+4y−λz=04x+λy+2z=0λx+2y+2z=0\begin{cases} 2x+4y-\lambda z=0\\ 4x+\lambda y+2z=0\\ \lambda x+2y+2z=0 \end{cases}⎩⎨⎧​2x+4y−λz=04x+λy+2z=0λx+2y+2z=0​

    So the coefficient matrix is

    A=(24−λ4λ2λ22).A=\begin{pmatrix} 2 & 4 & -\lambda\\ 4 & \lambda & 2\\ \lambda & 2 & 2 \end{pmatrix}.A=​24λ​4λ2​−λ22​​.

  2. Compute det⁡(A)\det(A)det(A).

    Expanding along the first row:

    det⁡(A)=2∣λ222∣−4∣42λ2∣+(−λ)∣4λλ2∣.\det(A)=2\begin{vmatrix}\lambda & 2\\2 & 2\end{vmatrix} -4\begin{vmatrix}4 & 2\\ \lambda & 2\end{vmatrix} +(-\lambda)\begin{vmatrix}4 & \lambda\\ \lambda & 2\end{vmatrix}.det(A)=2​λ2​22​​−4​4λ​22​​+(−λ)​4λ​λ2​​.

    Now evaluate each minor:

    ∣λ222∣=2λ−4,\begin{vmatrix}\lambda & 2\\2 & 2\end{vmatrix}=2\lambda-4,​λ2​22​​=2λ−4, ∣42λ2∣=8−2λ,\begin{vmatrix}4 & 2\\ \lambda & 2\end{vmatrix}=8-2\lambda,​4λ​22​​=8−2λ, ∣4λλ2∣=8−λ2.\begin{vmatrix}4 & \lambda\\ \lambda & 2\end{vmatrix}=8-\lambda^2.​4λ​λ2​​=8−λ2.

    Therefore,

    det⁡(A)=2(2λ−4)−4(8−2λ)−λ(8−λ2).\det(A)=2(2\lambda-4)-4(8-2\lambda)-\lambda(8-\lambda^2).det(A)=2(2λ−4)−4(8−2λ)−λ(8−λ2).

    Simplify:

    det⁡(A)=4λ−8−32+8λ−8λ+λ3\det(A)=4\lambda-8-32+8\lambda-8\lambda+\lambda^3det(A)=4λ−8−32+8λ−8λ+λ3 det⁡(A)=λ3+4λ−40.\det(A)=\lambda^3+4\lambda-40.det(A)=λ3+4λ−40.

  3. For infinitely many solutions,

    λ3+4λ−40=0.\lambda^3+4\lambda-40=0.λ3+4λ−40=0.

    Try rational roots. For λ=3\lambda=3λ=3:

    27+12−40=−1≠0.27+12-40=-1 \neq 0.27+12−40=−1=0.

    For λ=4\lambda=4λ=4:

    64+16−40=40≠0.64+16-40=40 \neq 0.64+16−40=40=0.

    For λ=2\lambda=2λ=2:

    8+8−40=−24≠0.8+8-40=-24 \neq 0.8+8−40=−24=0.

    Instead of factoring, examine the function

    f(λ)=λ3+4λ−40.f(\lambda)=\lambda^3+4\lambda-40.f(λ)=λ3+4λ−40.

    Its derivative is

    f′(λ)=3λ2+4>0for all real λ.f'(\lambda)=3\lambda^2+4>0 \quad \text{for all real }\lambda.f′(λ)=3λ2+4>0for all real λ.

    Hence f(λ)f(\lambda)f(λ) is strictly increasing for all real λ\lambdaλ, so it can have exactly one real root.

  4. Therefore, there is exactly one real value of λ\lambdaλ for which the system has infinitely many solutions.

  5. Hence the correct option is

    B: 1\boxed{\text{B: }1}B: 1​
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