Exams Joint Entrance Examination Mathematics Matrices and Determinants Question 86 Question 86 of 472 Medium MCQ
If the system of linear equations
7 x + 11 y + α z = 13 5 x + 4 y + 7 z = β 175 x + 194 y + 57 z = 361 \begin{gathered} 7x+11y+\alpha z=13 \\ 5x+4y+7z=\beta \\ 175x+194y+57z=361 \end{gathered} 7 x + 11 y + α z = 13 5 x + 4 y + 7 z = β 175 x + 194 y + 57 z = 361
has infinitely many solutions, then α + β + 2 \alpha +\beta +2 α + β + 2 is equal to :
Solutions · 1 Post your solution Official solution
Given,
7 x + 11 y + α z = 13 5 x + 4 y + 7 z = β 175 x + 194 y + 57 z = 361 \begin{gathered} 7x+11y+\alpha z=13 \\ 5x+4y+7z=\beta \\ 175x+194y+57z=361 \end{gathered} 7 x + 11 y + α z = 13 5 x + 4 y + 7 z = β 175 x + 194 y + 57 z = 361
∣ 7 11 α 5 4 7 175 194 57 ∣ = 0 \begin{vmatrix} 7 \quad 11 \quad \alpha \\ 5 \quad 4 \quad 7 \\ 175 \quad 194 \quad 57 \end{vmatrix}=0 7 11 α 5 4 7 175 194 57 = 0
⇒ ∣ 7 11 α 5 4 7 0 − 81 57 − 25 α ∣ = 0 \Rightarrow \begin{vmatrix} 7 \quad 11 \quad \alpha \\ 5 \quad 4 \quad 7 \\ 0 \quad -81 \quad 57-25\alpha \end{vmatrix}=0 ⇒ 7 11 α 5 4 7 0 − 81 57 − 25 α = 0
⇒ 81 ( 49 − 5 α ) + ( 57 − 25 α ) ( − 27 ) = 0 ⇒ 270 α = − 2430 ⇒ α = − 9 \begin{gathered} \Rightarrow 81(49-5\alpha )+(57-25\alpha )(-27)=0 \\ \Rightarrow 270\alpha =-2430\Rightarrow \alpha =-9 \end{gathered} ⇒ 81 ( 49 − 5 α ) + ( 57 − 25 α ) ( − 27 ) = 0 ⇒ 270 α = − 2430 ⇒ α = − 9
And Δ 1 = 0 {\Delta}_{1}=0 Δ 1 = 0
∣ 13 11 − 9 β 4 7 361 194 57 ∣ = 0 ⇒ β = 11 \begin{gathered} |\begin{array}{|c|}13 \quad 11 \quad -9 \\ \beta \quad 4 \quad 7 \\ 361 \quad 194 \quad 57\end{array}|=0 \\ \Rightarrow \beta =11 \end{gathered} ∣ 13 11 − 9 β 4 7 361 194 57 ∣ = 0 ⇒ β = 11
∴ α + β + 2 = 4 \therefore \alpha +\beta +2=4 ∴ α + β + 2 = 4 Answer the question to unlock