Matrices and Determinants 4 Question 12
12. If the system of linear equations
$ x-4 y+7 z=g, 3 y-5 z=h,-2 x+5 y-9 z=k $
is consistent, then
(2019 Main, 9 Jan II)
(a) $2 g+h+k=0$
(b) $g+2 h+k=0$
(c) $g+h+k=0$
(d) $g+h+2 k=0$
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Answer:
Correct Answer: 12. (a)
Solution:
- (a) Here, $D=\left|\begin{array}{ccc}1 & -4 & 7 \\ 0 & 3 & -5 \\ -2 & 5 & -9\end{array}\right|$
$ =1(-27+25)+4(0-10)+7(0+6) $
[expanding along $R _1$ ]
$ =-2-40+42=0 $
$\therefore$ The system of linear equations have infinite many solutions.
$[\because$ system is consistent and does not have unique solution as $D=0]$
$\Rightarrow \quad D _1=D _2=D _3=0$
Now, $D _1=0 \Rightarrow\left|\begin{array}{ccc}g & -4 & 7 \\ h & 3 & -5 \\ k & 5 & -9\end{array}\right|=0$
$\Rightarrow g(-27+25)+4(-9 h+5 k)+7(5 h-3 k)=0$
$\Rightarrow-2 g-36 h+20 k+35 h-21 k=0$
$\Rightarrow-2 g-h-k=0 \Rightarrow 2 g+h+k=0$
