JEE Mains · Maths · STD 12 - 3 and 4 . metrices and determinant
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 \(x+y=1\)
Answer & Solution
Correct Answer
(D) unique solution satisfying \(x+y=1\)
Step-by-step Solution
Detailed explanation
\(\left|\begin{array}{ccc}1 & 1 & k \\ 2 & 3 & -1 \\ 3 & 4 & 2\end{array}\right|=0\) \(1(10)-1(7)+k(-1)-0\) \(k=3\) 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…
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