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3 and 4 .Determinants and Matrices
hard
Let the system of linear equations $x+y+k z=2$ ; $2 x+3 y-z=1$ ; $3 x+4 y+2 z=k$ , have infinitely many solutions. Then the system $( k +1) x +(2 k -1) y =7$ ; $(2 k +1) x +( k +5) y =10 \text { has : }$
A
infinitely many solutions
B
unique solution satisfying $x-y=1$
C
no solution
D
unique solution satisfying $x+y=1$
(JEE MAIN-2023)
Solution
$\left|\begin{array}{ccc}1 & 1 & k \\ 2 & 3 & -1 \\ 3 & 4 & 2\end{array}\right|=0$
$1(10)-1(7)+k(-1)-0$
$k=3$
For $k =3,2^{\text {nd }}$ system is
$4 x+5 y=7…….(1)$
and $7 x+8 y=10……..(2)$
Clearly, they have a unique solution
$(2) -(1) \Rightarrow 3 x+3 y=3$
$\Rightarrow x+y=1$
Standard 12
Mathematics
