Exams Joint Entrance Examination Mathematics Matrices and Determinants Question 53 Question 53 of 472 Medium MCQ
Let λ , μ ∈ R \lambda ,\mu \in R λ , μ ∈ R . If the system of equations
3 x + 5 y + λ z = 3 7 x + 11 y − 9 z = 2 97 x + 155 y − 189 z = μ \begin{gathered} 3x+5y+\lambda z=3 \\ 7x+11y-9z=2 \\ 97x+155y-189z=\mu \end{gathered} 3 x + 5 y + λ z = 3 7 x + 11 y − 9 z = 2 97 x + 155 y − 189 z = μ
has infinitely many solutions, then μ + 2 λ \mu +2\lambda μ + 2 λ is equal to :
Solutions · 1 Post your solution Official solution
3 x + 5 y + λ z = 3 7 x + 11 y − 9 z = 2 97 x + 155 y − 189 z = μ \begin{gathered} 3x+5y+\lambda z=3 \\ 7x+11y-9z=2 \\ 97x+155y-189z=\mu \end{gathered} 3 x + 5 y + λ z = 3 7 x + 11 y − 9 z = 2 97 x + 155 y − 189 z = μ
[ [ ∣ 3 5 λ 7 11 − 9 97 155 − 189 ] ] ∣ [ [ ∣ x y z ] ] ∣ = [ [ ∣ 3 2 μ ] ] ∣ A X = B X = A − 1 B X = a d j A ∣ A ∣ B \begin{gathered} [[|\begin{array}{|c|}3 \quad 5 \quad \lambda \\ 7 \quad 11 \quad -9 \\ 97 \quad 155 \quad -189\end{array}]]|[[|\begin{array}{|c|}x \\ y \\ z\end{array}]]|=[[|\begin{array}{|c|}3 \\ 2 \\ \mu\end{array}]]| \\ AX=B \\ X={A}^{-1}B \\ X=\frac{adjA}{|A|}B \end{gathered} [[ ∣ 3 5 λ 7 11 − 9 97 155 − 189 ]] ∣ [[ ∣ x y z ]] ∣ = [[ ∣ 3 2 μ ]] ∣ A X = B X = A − 1 B X = ∣ A ∣ a d j A B
For Infinitely many solution
∣ A ∣ = 0 ( a d j A ) B = 0 ∣ 3 5 λ 7 11 − 9 97 155 − 189 ∣ = 0 − 2052 + 2250 + 18 λ = 0 ⇒ λ = − 11 a d j A = [ [ ∣ − 684 − 760 76 450 500 − 50 18 20 − 2 ] ] ∣ ( a d j A ) B = [ [ ∣ 684 − 760 76 450 500 − 50 18 20 − 2 ] ] ∣ [ [ ∣ 3 2 μ ] ] ∣ = [ [ ∣ 0 0 0 ] ] ∣ ⇒ 54 + 40 − 2 μ = 0 ⇒ 2 μ = 94 ⇒ μ = 47 ⇒ μ + 2 λ = 25 \begin{gathered} |A|=0(adjA)B=0 \\ |\begin{array}{|c|}3 \quad 5 \quad \lambda \\ 7 \quad 11 \quad -9 \\ 97 \quad 155 \quad -189\end{array}|=0 \\ -2052+2250+18\lambda =0 \\ \Rightarrow \lambda =-11 \\ adjA=[[|\begin{array}{|c|}-684 \quad -760 \quad 76 \\ 450 \quad 500 \quad -50 \\ 18 \quad 20 \quad -2\end{array}]]| \\ (adjA)B=[[|\begin{array}{|c|}684 \quad -760 \quad 76 \\ 450 \quad 500 \quad -50 \\ 18 \quad 20 \quad -2\end{array}]]|[[|\begin{array}{|c|}3 \\ 2 \\ \mu\end{array}]]|=[[|\begin{array}{|c|}0 \\ 0 \\ 0\end{array}]]| \\ \Rightarrow 54+40-2\mu =0 \\ \Rightarrow 2\mu =94 \\ \Rightarrow \mu =47 \\ \Rightarrow \mu +2\lambda =25 \end{gathered} ∣ A ∣ = 0 ( a d j A ) B = 0 ∣ 3 5 λ 7 11 − 9 97 155 − 189 ∣ = 0 − 2052 + 2250 + 18 λ = 0 ⇒ λ = − 11 a d j A = [[ ∣ − 684 − 760 76 450 500 − 50 18 20 − 2 ]] ∣ ( a d j A ) B = [[ ∣ 684 − 760 76 450 500 − 50 18 20 − 2 ]] ∣ [[ ∣ 3 2 μ ]] ∣ = [[ ∣ 0 0 0 ]] ∣ ⇒ 54 + 40 − 2 μ = 0 ⇒ 2 μ = 94 ⇒ μ = 47 ⇒ μ + 2 λ = 25 Answer the question to unlock