Matrices and Determinants 4 Question 11
11. If the system of equations
$ \begin{gathered} x+y+z=5 \\ x+2 y+3 z=9 \\ x+3 y+\alpha z=\beta \end{gathered} $
has infinitely many solutions, then $\beta-\alpha$ equals
(2019 Main, 10 Jan I)
(a) 8
(b) 18
(c) 21
(d) 5
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Answer:
Correct Answer: 11. (a)
Solution:
- Since, the system of equations has infinitely many solution, therefore $D=D_{1}=D_{2}=D_{3}=0$
Here,
$ \begin{aligned} D & =\left|\begin{array}{ccc} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 3 & \alpha \end{array}\right|=1(2 \alpha-9)-1(\alpha-3)+1(3-2) \\ & =\alpha-5 \\ \text { and } D_{3} & =\left|\begin{array}{lll} 1 & 1 & 5 \\ 1 & 2 & 9 \\ 1 & 3 & \beta \end{array}\right|=1(2 \beta-27)-1(\beta-9)+5(3-2) \end{aligned} $
$=\beta-13$
$ \begin{aligned} & \text { Now, } \quad D=0 \\ & \Rightarrow \quad \alpha-5=0 \Rightarrow \alpha=5 \\ & \text { and } \quad D_{3}=0 \Rightarrow \beta-13=0 \\ & \Rightarrow \quad \beta=13 \\ & \therefore \quad \beta-\alpha=13-5=8 \end{aligned} $
