Determinants Question 1
Question 1 - 2024 (01 Feb Shift 1)
If the system of equations
$2 x+3 y-z=5$
$x+\alpha y+3 z=-4$
$3 x-y+\beta z=7$
has infinitely many solutions, then $13 \alpha \beta$ is equal to
(1) 1110
(2) 1120
(3) 1210
(4) 1220
Show Answer
Answer (2)
Solution
Using family of planes
$2 x+3 y-z-5=k _1(x+\alpha y+3 z+4)+k _2(3 x-y$
$+\beta \mathbf{z}-7$ )
$2=k _1+3 k _2, 3=k _1 \alpha-k _2,-1=3 k _1+\beta k _2,-5=$
$4 k _1-7 k _2$
On solving we get
$$ \begin{gathered} k _2=\frac{13}{19}, k _1=\frac{-1}{19}, \alpha=-70, \beta=\frac{-16}{13} \\ 13 \alpha \beta=13(-70)\left(\frac{-16}{13}\right) \\ =1120 \end{gathered} $$
