Matrices and Determinants 4 Question 9
9. If the system of linear equations
$ \begin{aligned} & 2 x+2 y+3 z=a \\ & 3 x-y+5 z=b \\ & x-3 y+2 z=c \end{aligned} $
where $a, b, c$ are non-zero real numbers, has more than one solution, then
(2019 Main, 11 Jan I)
(a) $b-c-a=0$
(b) $a+b+c=0$
(c) $b-c+a=0$
(d) $b+c-a=0$
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Answer:
Correct Answer: 9. (a)
Solution:
- We know that, if the system of equations
$ \begin{aligned} & a _1 x+b _1 y+c _1 z=d _1 \\ & a _2 x+b _2 y+c _2 z=d _2 \\ & a _3 x+b _3 y+c _3 z=d _3 \end{aligned} $
has more than one solution, then $D=0$ and $D _1=D _2=D _3=0$. In the given problem,
$ \begin{array}{cc} & D _1=0 \Rightarrow\left|\begin{array}{ccc} a & 2 & 3 \\ b & -1 & 5 \\ c & -3 & 2 \end{array}\right|=0 \\ \Rightarrow & a(-2+15)-2(2 b-5 c)+3(-3 b+c)=0 \\ \Rightarrow & 13 a-4 b+10 c-9 b+3 c=0 \\ \Rightarrow & 13 a-13 b+13 c=0 \\ \Rightarrow & a-b+c=0 \Rightarrow b-a-c=0 \end{array} $
