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
$2 \mathrm{x}+3 \mathrm{y}-\mathrm{z}-5=\mathrm{k}{1}(\mathrm{x}+\alpha \mathrm{y}+3 \mathrm{z}+4)+\mathrm{k}{2}(3 \mathrm{x}-\mathrm{y}$
$+\beta \mathbf{z}-7$ )
$2=\mathrm{k}{1}+3 \mathrm{k}{2}, 3=\mathrm{k}{1} \alpha-\mathrm{k}{2},-1=3 \mathrm{k}{1}+\beta \mathrm{k}{2},-5=$
$4 \mathrm{k}{1}-7 \mathrm{k}{2}$
On solving we get
$$ \begin{gathered} k_{2}=\frac{13}{19}, k_{1}=\frac{-1}{19}, \alpha=-70, \beta=\frac{-16}{13} \ 13 \alpha \beta=13(-70)\left(\frac{-16}{13}\right) \ =1120 \end{gathered} $$
