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Matrices and Determinants question

2025 · 7 Apr · Shift 1 · Q36
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Matrices and Determinants question

2025 · 7 Apr · Shift 1 · Q36

JEE MainMathematicsMatrices and DeterminantsMCQ+4 / −1
Let the system of equations : 2x+3y+5z=97x+3y−2z=812x+3y−(4+λ)z=16−μ\begin{aligned} & 2 x+3 y+5 z=9 \\ & 7 x+3 y-2 z=8 \\ & 12 x+3 y-(4+\lambda) z=16-\mu \end{aligned}​2x+3y+5z=97x+3y−2z=812x+3y−(4+λ)z=16−μ​ have infinitely many solutions. Then the radius of the circle centred at (λ,μ)(\lambda, \mu)(λ,μ) and touching the line 4x=3y4 x=3 y4x=3y is :
  1. A
    75\frac{7}{5}57​
  2. B
    215\frac{21}{5}521​
  3. C
    7
  4. D
    175\frac{17}{5}517​
View written solutionFree

Correct answer: A

  1. For the system to have infinitely many solutions, the third equation must be a linear combination of the first two.

The equations are:

2x+3y+5z=9...(1)7x+3y−2z=8...(2)12x+3y−(4+λ)z=16−μ...(3)\begin{aligned} 2x+3y+5z &= 9 \quad ...(1)\\ 7x+3y-2z &= 8 \quad ...(2)\\ 12x+3y-(4+\lambda)z &= 16-\mu \quad ...(3) \end{aligned}2x+3y+5z7x+3y−2z12x+3y−(4+λ)z​=9...(1)=8...(2)=16−μ...(3)​

So, let

(3)=a(1)+b(2).(3)=a(1)+b(2).(3)=a(1)+b(2).

Then comparing coefficients:

  1. From coefficients of x,y,zx,y,zx,y,z and constants:
2a+7b=122a+7b=122a+7b=12 3a+3b=3  ⟹  a+b=13a+3b=3 \implies a+b=13a+3b=3⟹a+b=1 5a−2b=−(4+λ)5a-2b=-(4+\lambda)5a−2b=−(4+λ) 9a+8b=16−μ9a+8b=16-\mu9a+8b=16−μ
  1. Solve for a,ba,ba,b using
a+b=1  ⟹  a=1−b.a+b=1 \implies a=1-b.a+b=1⟹a=1−b.

Substitute into 2a+7b=122a+7b=122a+7b=12:

2(1−b)+7b=122(1-b)+7b=122(1−b)+7b=12 2−2b+7b=122-2b+7b=122−2b+7b=12 5b=10  ⟹  b=25b=10 \implies b=25b=10⟹b=2

Hence,

a=1−2=−1.a=1-2=-1.a=1−2=−1.
  1. Now find λ\lambdaλ:
5a−2b=−(4+λ)5a-2b=-(4+\lambda)5a−2b=−(4+λ) 5(−1)−2(2)=−(4+λ)5(-1)-2(2)=-(4+\lambda)5(−1)−2(2)=−(4+λ) −5−4=−(4+λ)-5-4=-(4+\lambda)−5−4=−(4+λ) −9=−4−λ-9=-4-\lambda−9=−4−λ λ=5.\lambda=5.λ=5.
  1. Now find μ\muμ:
9a+8b=16−μ9a+8b=16-\mu9a+8b=16−μ 9(−1)+8(2)=16−μ9(-1)+8(2)=16-\mu9(−1)+8(2)=16−μ −9+16=16−μ-9+16=16-\mu−9+16=16−μ 7=16−μ7=16-\mu7=16−μ μ=9.\mu=9.μ=9.

So the centre of the circle is (λ,μ)=(5,9)(\lambda,\mu)=(5,9)(λ,μ)=(5,9).

  1. The line 4x=3y4x=3y4x=3y can be written as
4x−3y=0.4x-3y=0.4x−3y=0.

The radius of the circle touching this line is the perpendicular distance from (5,9)(5,9)(5,9) to the line:

r=∣4(5)−3(9)∣42+(−3)2r=\frac{|4(5)-3(9)|}{\sqrt{4^2+(-3)^2}}r=42+(−3)2​∣4(5)−3(9)∣​ r=∣20−27∣16+9=75.r=\frac{|20-27|}{\sqrt{16+9}}=\frac{7}{5}.r=16+9​∣20−27∣​=57​.
  1. Therefore, the required radius is
75.\boxed{\frac{7}{5}}.57​​.

So the correct option is A.

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