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Matrices and Determinants question

2019 · 10 Apr · Shift 2 · Q43
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  5. /2019 · 10 Apr · Shift 2 · Q43

Matrices and Determinants question

2019 · 10 Apr · Shift 2 · Q43

JEE MainMathematicsMatrices and DeterminantsMCQ+4 / −1
Let λ\lambdaλ be a real number for which the system of linear equations x + y + z = 6, 4x +λ\lambdaλ y – λ\lambdaλ z = λ\lambdaλ– 2, 3x + 2y – 4z = – 5 has infinitely many solutions. Then λ\lambdaλ is a root of the quadratic equation:
  1. A
    λ\lambdaλ 2 + λ\lambdaλ - 6 = 0
  2. B
    λ\lambdaλ 2 - λ\lambdaλ - 6 = 0
  3. C
    λ\lambdaλ 2 - 3 λ\lambdaλ - 4 = 0
  4. D
    λ\lambdaλ 2 + 3 λ\lambdaλ - 4 = 0
View written solutionFree

Correct answer: B

  1. For the system to have infinitely many solutions, the three equations must be consistent and dependent.

    The system is: x+y+z=6x+y+z=6x+y+z=6 4x+λy−λz=λ−24x+\lambda y-\lambda z=\lambda-24x+λy−λz=λ−2 3x+2y−4z=−53x+2y-4z=-53x+2y−4z=−5

  2. Write the augmented matrix:

    \begin{array}{ccc|c} 1&1&1&6\\ 4&\lambda&-\lambda&\lambda-2\\ 3&2&-4&-5 \end{array} \right]$$

  3. For infinitely many solutions in a system of 3 variables, we need: det⁡(A)=0\det(A)=0det(A)=0 where

    1&1&1\\ 4&\lambda&-\lambda\\ 3&2&-4 \end{pmatrix}$$

  4. Compute the determinant:

    \begin{vmatrix} 1&1&1\\ 4&\lambda&-\lambda\\ 3&2&-4 \end{vmatrix}$$ Expanding along the first row: $$\det(A)=1\begin{vmatrix}\lambda&-\lambda\\2&-4\end{vmatrix} -1\begin{vmatrix}4&-\lambda\\3&-4\end{vmatrix} +1\begin{vmatrix}4&\lambda\\3&2\end{vmatrix}$$ $$=1\big(\lambda(-4)-(-\lambda)(2)\big)-\big(4(-4)-(-\lambda)(3)\big)+\big(4\cdot2-\lambda\cdot3\big)$$ $$=(-4\lambda+2\lambda)-(-16+3\lambda)+(8-3\lambda)$$ $$=-2\lambda+16-3\lambda+8-3\lambda$$ $$=24-8\lambda$$ So, $$\det(A)=24-8\lambda=8(3-\lambda)$$ Hence, $$\det(A)=0 \Rightarrow \lambda=3$$

  5. Now check consistency at λ=3\lambda=3λ=3.

    The system becomes: x+y+z=6x+y+z=6x+y+z=6 4x+3y−3z=14x+3y-3z=14x+3y−3z=1 3x+2y−4z=−53x+2y-4z=-53x+2y−4z=−5

    From the first equation: x=6−y−zx=6-y-zx=6−y−z

    Substitute into the second: 4(6−y−z)+3y−3z=14(6-y-z)+3y-3z=14(6−y−z)+3y−3z=1 24−4y−4z+3y−3z=124-4y-4z+3y-3z=124−4y−4z+3y−3z=1 −y−7z=−23-y-7z=-23−y−7z=−23 y+7z=23...(i)y+7z=23 \quad ...(i)y+7z=23...(i)

    Substitute into the third: 3(6−y−z)+2y−4z=−53(6-y-z)+2y-4z=-53(6−y−z)+2y−4z=−5 18−3y−3z+2y−4z=−518-3y-3z+2y-4z=-518−3y−3z+2y−4z=−5 −y−7z=−23-y-7z=-23−y−7z=−23 y+7z=23...(ii)y+7z=23 \quad ...(ii)y+7z=23...(ii)

    Equations (i) and (ii) are identical, so the system is consistent and dependent. Therefore, it has infinitely many solutions for: λ=3\lambda=3λ=3

  6. Now check which quadratic equation has λ=3\lambda=3λ=3 as a root.

    • Option A: λ2+λ−6=0\lambda^2+\lambda-6=0λ2+λ−6=0 32+3−6=9+3−6=6≠03^2+3-6=9+3-6=6\ne032+3−6=9+3−6=6=0
    • Option B: λ2−λ−6=0\lambda^2-\lambda-6=0λ2−λ−6=0 32−3−6=9−3−6=03^2-3-6=9-3-6=032−3−6=9−3−6=0
    • Option C: λ2−3λ−4=0\lambda^2-3\lambda-4=0λ2−3λ−4=0 9−9−4=−4≠09-9-4=-4\ne09−9−4=−4=0
    • Option D: λ2+3λ−4=0\lambda^2+3\lambda-4=0λ2+3λ−4=0 9+9−4=14≠09+9-4=14\ne09+9−4=14=0

  7. Therefore, the correct option is: B\boxed{\text{B}}B​

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