Determinants Question 4
Question 4 - 30 January - Shift 1
Let the system of linear equations
$x+y+kz=2$
$2 x+3 y-z=1$
$3 x+4 y+2 z=k$
have infinitely many solutions. Then the system
$(k+1) x+(2 k-1) y=7$
$(2 k+1) x+(k+5) y=10$ has :
(1) infinitely many solutions
(2) unique solution satisfying $x-y=1$
(3) no solution
(4) unique solution satisfying $x+y=1$
Show Answer
Answer: (4)
Solution:
Formula: System of equations with 3 variables, consistency of solutions: infinite solutions
$ \begin{vmatrix} 1 & 1 & k \\ 2 & 3 & -1 \\ 3 & 4 & 2\end{vmatrix} =0$
$\Rightarrow$ 1(10) $-1(7)+k(-1)-0$
$\Rightarrow k=3$
For $k=3,2^{\text{nd }}$ system is
$4 x+5 y=7$
and $7 x+8 y=10$
Clearly, they have a unique solution
(2) $-(1) \Rightarrow 3 x+3 y=3$
$\Rightarrow x+y=1$
