Exams Joint Entrance Examination Mathematics Matrices and Determinants Question 97 Question 97 of 472 Medium MCQ
For the system of equations
x + y + z = 6 x+y+z=6 x + y + z = 6
x + 2 y + α z = 10 x+2y+\alpha z=10 x + 2 y + α z = 10
x + 3 y + 5 z = β x+3y+5z=\beta x + 3 y + 5 z = β , which one of the following is NOT true?
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Given system of equations,
x + y + z = 6........ ( i ) x + 2 y + α z = 10........ ( i i ) x + 3 y + 5 z = β . . . . . . . . ( i i i ) \begin{gathered} x+y+z =6........(i) \\ \\ x+2y+\alpha z =10........(ii) \\ \\ x+3y+5z =\beta ........(iii) \end{gathered} x + y + z = 6........ ( i ) x + 2 y + α z = 10........ ( ii ) x + 3 y + 5 z = β ........ ( iii )
Here,
Δ = ∣ 1 1 1 1 2 α 1 3 5 ∣ = 1 ( 10 − 3 α ) − 1 ( 5 − α ) + 1 ( 3 − 2 ) = 10 − 3 α − 5 + α + 1 = 6 − 2 α \begin{gathered} \Delta \quad =|\begin{array}{|c|}1 \quad 1 \quad 1 \\ 1 \quad 2 \quad \alpha \\ 1 \quad 3 \quad 5\end{array}| \\ =1(10-3\alpha )-1(5-\alpha )+1(3-2) \\ =10-3\alpha -5+\alpha +1 \\ =6-2\alpha \end{gathered} Δ = ∣ 1 1 1 1 2 α 1 3 5 ∣ = 1 ( 10 − 3 α ) − 1 ( 5 − α ) + 1 ( 3 − 2 ) = 10 − 3 α − 5 + α + 1 = 6 − 2 α
For unique solution, Δ ≠ 0 \Delta\ne 0 Δ = 0
⇒ 6 − 2 α ≠ 0 ⇒ α ≠ 3 \Rightarrow 6-2\alpha \ne 0\Rightarrow \alpha \ne 3 ⇒ 6 − 2 α = 0 ⇒ α = 3
When, α = 3 \alpha =3 α = 3
Δ 1 = ∣ 6 1 1 10 2 3 β 3 5 ∣ = ∣ 0 1 0 − 2 2 1 β − 18 3 2 ∣ = − 1 ( − 4 − β + 18 ) = β − 14 \begin{gathered} {\Delta}_{1} \quad =|\begin{array}{|c|}6 \quad 1 \quad 1 \\ 10 \quad 2 \quad 3 \\ \beta \quad 3 \quad 5\end{array}| \\ =|\begin{array}{|c|}0 \quad 1 \quad 0 \\ -2 \quad 2 \quad 1 \\ \beta -18 \quad 3 \quad 2\end{array}|=-1(-4-\beta +18) \\ =\beta -14 \end{gathered} Δ 1 = ∣ 6 1 1 10 2 3 β 3 5 ∣ = ∣ 0 1 0 − 2 2 1 β − 18 3 2 ∣ = − 1 ( − 4 − β + 18 ) = β − 14
and
Δ 2 = ∣ 1 6 1 1 10 3 1 β 5 ∣ = 1 ( 50 − 3 β ) − 6 ( 5 − 3 ) + 1 ( β − 10 ) = 50 − 3 β − 12 + β − 10 = 28 − 2 β = 2 ( 14 − β ) \begin{gathered} {\Delta}_{2} \quad =|\begin{array}{|c|}1 \quad 6 \quad 1 \\ 1 \quad 10 \quad 3 \\ 1 \quad \beta \quad 5\end{array}| \\ =1(50-3\beta )-6(5-3)+1(\beta -10) \\ =50-3\beta -12+\beta -10 \\ =28-2\beta =2(14-\beta ) \end{gathered} Δ 2 = ∣ 1 6 1 1 10 3 1 β 5 ∣ = 1 ( 50 − 3 β ) − 6 ( 5 − 3 ) + 1 ( β − 10 ) = 50 − 3 β − 12 + β − 10 = 28 − 2 β = 2 ( 14 − β )
and
Δ 3 = ∣ 1 1 6 1 2 10 1 3 β ∣ = 1 ( 2 β − 30 ) − 1 ( β − 10 ) + 6 ( 3 − 2 ) = 2 β − 30 − β + 10 + 6 = β − 14 \begin{gathered} {\Delta}_{3} \quad =|\begin{array}{|c|}1 \quad 1 \quad 6 \\ 1 \quad 2 \quad 10 \\ 1 \quad 3 \quad \beta\end{array}| \\ =1(2\beta -30)-1(\beta -10)+6(3-2) \\ =2\beta -30-\beta +10+6 \\ =\beta -14 \end{gathered} Δ 3 = ∣ 1 1 6 1 2 10 1 3 β ∣ = 1 ( 2 β − 30 ) − 1 ( β − 10 ) + 6 ( 3 − 2 ) = 2 β − 30 − β + 10 + 6 = β − 14
Thus, at β = 14 , Δ 1 = Δ 2 = Δ 3 = 0 \beta =14,{\Delta}_{1}={\Delta}_{2}={\Delta}_{3}=0 β = 14 , Δ 1 = Δ 2 = Δ 3 = 0
⇒ α = 3 , β = 14 \Rightarrow \alpha =3,\beta =14 ⇒ α = 3 , β = 14
So, system has infinite solutions. Answer the question to unlock