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Matrices and Determinants question

2019 · 12 Jan · Shift 1 · Q35
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  5. /2019 · 12 Jan · Shift 1 · Q35

Matrices and Determinants question

2019 · 12 Jan · Shift 1 · Q35

JEE MainMathematicsMatrices and DeterminantsMCQ+4 / −1
An ordered pair (α\alphaα, β\betaβ) for which the system of linear equations (1 + α\alphaα) x + β\betaβ y + z = 2 α\alphaα x + (1 + β\betaβ)y + z = 3 α\alphaα x + β\betaβ y + 2z = 2 has a unique solution, is :
  1. A
    (–3, 1)
  2. B
    (1, –3)
  3. C
    (–4, 2)
  4. D
    (2, 4)
View written solutionFree

Correct answer: D

  1. For a system of three linear equations in three variables to have a unique solution, the determinant of its coefficient matrix must be non-zero.

  2. The given system is:

(1+α)x+βy+z=2(1+\alpha)x + \beta y + z = 2(1+α)x+βy+z=2 αx+(1+β)y+z=3\alpha x + (1+\beta)y + z = 3αx+(1+β)y+z=3 αx+βy+2z=2\alpha x + \beta y + 2z = 2αx+βy+2z=2

So the coefficient matrix is

A=(1+αβ1α1+β1αβ2)A=\begin{pmatrix} 1+\alpha & \beta & 1\\ \alpha & 1+\beta & 1\\ \alpha & \beta & 2 \end{pmatrix}A=​1+ααα​β1+ββ​112​​

We need:

det⁡(A)≠0\det(A) \neq 0det(A)=0
  1. Compute the determinant.

Using row operations (which preserve the zero/non-zero nature of determinant):

Take

R1→R1−R2,R3→R3−R2R_1 \to R_1 - R_2, \qquad R_3 \to R_3 - R_2R1​→R1​−R2​,R3​→R3​−R2​

Then

R1=(1,−1,0),R2=(α,1+β,1),R3=(0,−1,1)R_1 = (1, -1, 0), \qquad R_2=(\alpha,1+\beta,1), \qquad R_3=(0,-1,1)R1​=(1,−1,0),R2​=(α,1+β,1),R3​=(0,−1,1)

So

det⁡(A)=∣1−10α1+β10−11∣\det(A)= \begin{vmatrix} 1 & -1 & 0\\ \alpha & 1+\beta & 1\\ 0 & -1 & 1 \end{vmatrix}det(A)=​1α0​−11+β−1​011​​

Expand along the first row:

det⁡(A)=1∣1+β1−11∣−(−1)∣α101∣\det(A)=1\begin{vmatrix}1+\beta & 1\\ -1 & 1\end{vmatrix}-(-1)\begin{vmatrix}\alpha & 1\\ 0 & 1\end{vmatrix}det(A)=1​1+β−1​11​​−(−1)​α0​11​​ =1((1+β)(1)−1(−1))+(α⋅1−0⋅1)=1\left((1+\beta)(1)-1(-1)\right)+\left(\alpha\cdot 1-0\cdot 1\right)=1((1+β)(1)−1(−1))+(α⋅1−0⋅1) =(1+β+1)+α=(1+\beta+1)+\alpha=(1+β+1)+α =α+β+2=\alpha+\beta+2=α+β+2

Hence the system has a unique solution when

α+β+2≠0\alpha+\beta+2 \neq 0α+β+2=0

or

α+β≠−2\alpha+\beta \neq -2α+β=−2
  1. Check each option:

  • A: (α,β)=(−3,1)(\alpha,\beta)=(-3,1)(α,β)=(−3,1) α+β=−2\alpha+\beta=-2α+β=−2 det⁡(A)=0\det(A)=0det(A)=0 Not unique.

  • B: (α,β)=(1,−3)(\alpha,\beta)=(1,-3)(α,β)=(1,−3) α+β=−2\alpha+\beta=-2α+β=−2 det⁡(A)=0\det(A)=0det(A)=0 Not unique.

  • C: (α,β)=(−4,2)(\alpha,\beta)=(-4,2)(α,β)=(−4,2) α+β=−2\alpha+\beta=-2α+β=−2 det⁡(A)=0\det(A)=0det(A)=0 Not unique.

  • D: (α,β)=(2,4)(\alpha,\beta)=(2,4)(α,β)=(2,4) α+β=6\alpha+\beta=6α+β=6 det⁡(A)=8≠0\det(A)=8 \neq 0det(A)=8=0 Unique solution exists.

  1. Therefore, the correct option is:

D (2,4)\boxed{D\,(2,4)}D(2,4)​

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