Determinants Question 9
Question 9 - 2024 (31 Jan Shift 1)
If the system of linear equations
$x-2 y+z=-4$
$2 x+\alpha y+3 z=5$
$3 x-y+\beta z=3$
has infinitely many solutions, then $12 \alpha+13 \beta$ is equal to
(1) 60
(2) 64
(3) 54
(4) 58
Show Answer
Answer (4)
Solution
$\mathrm{D}=\left|\begin{array}{ccc}1 & -2 & 1 \ 2 & \alpha & 3 \ 3 & -1 & \beta\end{array}\right|$
$=1(\alpha \beta+3)+2(2 \beta-9)+1(-2-3 \alpha)$
$=\alpha \beta+3+4 \beta-18-2-3 \alpha$
For infinite solutions $D=0, D_{1}=0, D_{2}=0$ and
$D_{3}=0$
$D=0$
$\alpha \beta-3 \alpha+4 \beta=17 \ldots$
$D_{1}=\left|\begin{array}{ccc}-4 & -2 & 1 \ 5 & \alpha & 3 \ 3 & -1 & \beta\end{array}\right|=0$
$D_{2}=\left|\begin{array}{ccc}1 & -4 & 1 \ 2 & 5 & 3 \ 3 & 3 & \beta\end{array}\right|=0$
$\Rightarrow 1(5 \beta-9)+4(2 \beta-9)+1(6-15)=0$
$13 \beta-9-36-9=0$
$13 \beta=54, \beta=\frac{54}{13}$ put in (1)
$\frac{54}{13} \alpha-3 \alpha+4\left(\frac{54}{13}\right)=17$
$54 \alpha-39 \alpha+216=221$
$15 \alpha=5 h \alpha=\frac{1}{3}$
Now, $12 \alpha+13 \beta=12 \cdot \frac{1}{3}+13 \cdot \frac{54}{13}$
$=4+54=58$
