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Matrices and Determinants question

2019 · 10 Apr · Shift 1 · Q36
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Matrices and Determinants question

2019 · 10 Apr · Shift 1 · Q36

JEE MainMathematicsMatrices and DeterminantsMCQ+4 / −1
If the system of linear equations x + y + z = 5 x + 2y + 2z = 6 x + 3y + λ\lambdaλ z = μ\muμ, (λ\lambdaλ, μ∈\mu \inμ∈ R), has infinitely many solutions, then the value of λ\lambdaλ+μ\muμ is :
  1. A
    10
  2. B
    9
  3. C
    7
  4. D
    12
View written solutionFree

Correct answer: A

  1. Write the system:
{x+y+z=5x+2y+2z=6x+3y+λz=μ\begin{cases} x+y+z=5 \\ x+2y+2z=6 \\ x+3y+\lambda z=\mu \end{cases}⎩⎨⎧​x+y+z=5x+2y+2z=6x+3y+λz=μ​

For infinitely many solutions, the third equation must be dependent on the first two, and the system must be consistent.

  1. Subtract the first equation from the second:
(x+2y+2z)−(x+y+z)=6−5(x+2y+2z)-(x+y+z)=6-5(x+2y+2z)−(x+y+z)=6−5 y+z=1y+z=1y+z=1
  1. Use the first equation:
x+y+z=5x+y+z=5x+y+z=5

Since y+z=1y+z=1y+z=1, we get

x+1=5⇒x=4x+1=5 \Rightarrow x=4x+1=5⇒x=4

So from the first two equations,

x=4,y+z=1x=4, \quad y+z=1x=4,y+z=1

This already gives infinitely many solutions in one parameter. For the full system to also have infinitely many solutions, the third equation must not impose any new independent condition.

  1. So the third equation must be a linear combination of the first two.

Let

α(x+y+z)+β(x+2y+2z)=x+3y+λz\alpha(x+y+z)+\beta(x+2y+2z)=x+3y+\lambda zα(x+y+z)+β(x+2y+2z)=x+3y+λz

Comparing coefficients:

  • Coefficient of xxx: α+β=1\alpha+\beta=1α+β=1
  • Coefficient of yyy: α+2β=3\alpha+2\beta=3α+2β=3
  • Coefficient of zzz: α+2β=λ\alpha+2\beta=\lambdaα+2β=λ

From the first two equations,

(α+2β)−(α+β)=3−1(\alpha+2\beta)-(\alpha+\beta)=3-1(α+2β)−(α+β)=3−1 β=2\beta=2β=2

Then

α+2=1⇒α=−1\alpha+2=1 \Rightarrow \alpha=-1α+2=1⇒α=−1

Hence,

λ=α+2β=−1+4=3\lambda=\alpha+2\beta=-1+4=3λ=α+2β=−1+4=3
  1. For consistency, the constant term must also follow the same combination:
μ=α⋅5+β⋅6\mu=\alpha\cdot 5+\beta\cdot 6μ=α⋅5+β⋅6 μ=−1⋅5+2⋅6=7\mu=-1\cdot 5+2\cdot 6=7μ=−1⋅5+2⋅6=7
  1. Therefore,
λ+μ=3+7=10\lambda+\mu=3+7=10λ+μ=3+7=10

So the correct option is:

10\boxed{10}10​

This matches option A.

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