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Matrices and Determinants question

2022 · 28 Jun · Shift 2 · Q46
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Matrices and Determinants question

2022 · 28 Jun · Shift 2 · Q46

JEE MainMathematicsMatrices and DeterminantsNumerical+4 / −1
If the system of linear equations 2x−3y=γ+52x - 3y = \gamma + 52x−3y=γ+5, αx+5y=β+1\alpha x + 5y = \beta + 1αx+5y=β+1, where α\alphaα, β\betaβ, γ∈\gamma\inγ∈ R has infinitely many solutions then the value of | 9 α\alphaα + 3 β\betaβ + 5 γ\gammaγ | is equal to ‾\underline{\hspace{2cm}}​.
Numerical answer
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Correct answer: 58

  1. For the system 2x−3y=γ+5,αx+5y=β+12x-3y=\gamma+5,\qquad \alpha x+5y=\beta+12x−3y=γ+5,αx+5y=β+1 to have infinitely many solutions, the two equations must represent the same line.

  2. Therefore, the ratios of corresponding coefficients and constants must be equal: 2α=−35=γ+5β+1\frac{2}{\alpha}=\frac{-3}{5}=\frac{\gamma+5}{\beta+1}α2​=5−3​=β+1γ+5​

  3. First use 2α=−35\frac{2}{\alpha}=\frac{-3}{5}α2​=5−3​ Cross-multiplying: 2⋅5=−3α2\cdot 5=-3\alpha2⋅5=−3α 10=−3α10=-3\alpha10=−3α α=−103\alpha=-\frac{10}{3}α=−310​

  4. Now use γ+5β+1=−35\frac{\gamma+5}{\beta+1}=\frac{-3}{5}β+1γ+5​=5−3​ Cross-multiplying: 5(γ+5)=−3(β+1)5(\gamma+5)=-3(\beta+1)5(γ+5)=−3(β+1) 5γ+25=−3β−35\gamma+25=-3\beta-35γ+25=−3β−3 3β+5γ=−283\beta+5\gamma=-283β+5γ=−28

  5. We need: ∣9α+3β+5γ∣\left|9\alpha+3\beta+5\gamma\right|∣9α+3β+5γ∣ Substitute α=−103\alpha=-\frac{10}{3}α=−310​: 9α=9(−103)=−309\alpha=9\left(-\frac{10}{3}\right)=-309α=9(−310​)=−30 Hence, 9α+3β+5γ=−30+(3β+5γ)9\alpha+3\beta+5\gamma=-30+(3\beta+5\gamma)9α+3β+5γ=−30+(3β+5γ) Using 3β+5γ=−283\beta+5\gamma=-283β+5γ=−28: 9α+3β+5γ=−30−28=−589\alpha+3\beta+5\gamma=-30-28=-589α+3β+5γ=−30−28=−58

  6. Therefore, ∣9α+3β+5γ∣=∣−58∣=58\left|9\alpha+3\beta+5\gamma\right|=|-58|=58∣9α+3β+5γ∣=∣−58∣=58

So the required integer is 585858.

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