Exams Joint Entrance Examination Mathematics Matrices and Determinants Question 55 Question 55 of 472 Medium MCQ
If the system of equations x + 4 y − z = λ , 7 x + 9 y + μ z = − 3 , 5 x + y + 2 z = − 1 x+4y-z=\lambda ,7x+9y+\mu z=-3,5x+y+2z=-1 x + 4 y − z = λ , 7 x + 9 y + μ z = − 3 , 5 x + y + 2 z = − 1 has infinitely many solutions, then ( 2 μ + 3 λ ) (2\mu +3\lambda ) ( 2 μ + 3 λ ) is equal to :
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x + 4 y − z = λ 7 x + 9 y + μ z = − 3 5 x + y + 2 z = − 1 [ [ ∣ 1 4 − 1 7 9 μ 5 1 2 ] ] ∣ [ [ ∣ x y z ] ] ∣ = [ [ ∣ λ − 3 − 1 ] ] ∣ A = [ [ ∣ 1 4 − 1 7 9 μ 5 1 2 ] ] ∣ , B = [ [ ∣ λ − 3 − 1 ] ] ∣ A X = B X = A − 1 B = a d j A ∣ A ∣ B \begin{gathered} x+4y-z=\lambda \\ 7x+9y+\mu z=-3 \\ 5x+y+2z=-1 \\ [[|\begin{array}{|c|}1 \quad 4 \quad -1 \\ 7 \quad 9 \quad \mu \\ 5 \quad 1 \quad 2\end{array}]]|[[|\begin{array}{|c|}x \\ y \\ z\end{array}]]|=[[|\begin{array}{|c|}\lambda \\ -3 \\ -1\end{array}]]| \\ A=[[|\begin{array}{|c|}1 \quad 4 \quad -1 \\ 7 \quad 9 \quad \mu \\ 5 \quad 1 \quad 2\end{array}]]|,B=[[|\begin{array}{|c|}\lambda \\ -3 \\ -1\end{array}]]| \\ AX=B \\ X={A}^{-1}B \\ \, =\frac{adjA}{|A|}B \end{gathered} x + 4 y − z = λ 7 x + 9 y + μ z = − 3 5 x + y + 2 z = − 1 [[ ∣ 1 4 − 1 7 9 μ 5 1 2 ]] ∣ [[ ∣ x y z ]] ∣ = [[ ∣ λ − 3 − 1 ]] ∣ A = [[ ∣ 1 4 − 1 7 9 μ 5 1 2 ]] ∣ , B = [[ ∣ λ − 3 − 1 ]] ∣ A X = B X = A − 1 B = ∣ A ∣ a d j A B
If ∣ A ∣ = 0 |A|=0 ∣ A ∣ = 0 and ( a d j A ) ⋅ B = 0 (adjA)\cdot B=0 ( a d j A ) ⋅ B = 0 , system has infinitely many solutions.
∣ A ∣ = 18 − μ − 4 ( 14 − 5 μ ) − 1 ( 7 − 45 ) = 0 ⇒ 18 − μ − 56 + 20 μ + 38 = 0 ⇒ 19 μ = 0 ⇒ μ = 0 = [ [ ∣ 18 − 9 9 − 14 7 − 7 − 38 19 − 19 ] ] ∣ ( A ) ⋅ B = 0 [ [ ∣ 18 − 9 9 − 14 7 − 7 − 38 19 − 19 ] ] ∣ [ [ ∣ λ − 3 − 1 ] ] ∣ = [ [ ∣ 0 0 0 ] ] ∣ 18 λ + 27 − 9 = 0 ⇒ 18 λ = − 18 ⇒ λ = − 1 ⇒ 2 μ + 3 λ = 3 ( − 1 ) = − 3 \begin{gathered} |A|=18-\mu -4(14-5\mu )-1(7-45)=0 \\ \Rightarrow 18-\mu -56+20\mu +38=0 \\ \Rightarrow 19\mu =0 \\ \Rightarrow \mu =0 \\ =[[|\begin{array}{|c|}18 \quad -9 \quad 9 \\ -14 \quad 7 \quad -7 \\ -38 \quad 19 \quad -19\end{array}]]| \\ (A)\cdot B=0 \\ [[|\begin{array}{|c|}18 \quad -9 \quad 9 \\ -14 \quad 7 \quad -7 \\ -38 \quad 19 \quad -19\end{array}]]|[[|\begin{array}{|c|}\lambda \\ -3 \\ -1\end{array}]]|=[[|\begin{array}{|c|}0 \\ 0 \\ 0\end{array}]]| \\ 18\lambda +27-9=0 \\ \Rightarrow 18\lambda =-18 \\ \Rightarrow \lambda =-1 \\ \Rightarrow 2\mu +3\lambda =3(-1)=-3 \end{gathered} ∣ A ∣ = 18 − μ − 4 ( 14 − 5 μ ) − 1 ( 7 − 45 ) = 0 ⇒ 18 − μ − 56 + 20 μ + 38 = 0 ⇒ 19 μ = 0 ⇒ μ = 0 = [[ ∣ 18 − 9 9 − 14 7 − 7 − 38 19 − 19 ]] ∣ ( A ) ⋅ B = 0 [[ ∣ 18 − 9 9 − 14 7 − 7 − 38 19 − 19 ]] ∣ [[ ∣ λ − 3 − 1 ]] ∣ = [[ ∣ 0 0 0 ]] ∣ 18 λ + 27 − 9 = 0 ⇒ 18 λ = − 18 ⇒ λ = − 1 ⇒ 2 μ + 3 λ = 3 ( − 1 ) = − 3 Answer the question to unlock