%%

If the system of linear equations $$\begin{aligned} & x-2 y+z=-4 \\ & 2 x+\alpha y+3 z=5 \\ & 3 x-y+\beta z=3 \end{aligned}$$ has infinitely many solutions, then $$12 \alpha+13 \beta$$ is equal to

JEE · Math · previous-year question

  1. A.60
  2. B.54
  3. C.64
  4. D.58correct

Answer

D. 58

Explanation

$$\begin{aligned} & D=\left|\begin{array}{ccc} 1 & -2 & 1 \\ 2 & \alpha & 3 \\ 3 & -1 & \beta \end{array}\right| \\ & =1(\alpha \beta+3)+2(2 \beta-9)+1(-2-3 \alpha) \\ & =\alpha \beta+3+4 \beta-18-2-3 \alpha \end{aligned}$$ For infinite solutions $$\mathrm{D}=0, \mathrm{D}_1=0, \mathrm{D}_2=0$$ and $$\begin{aligned} & \mathrm{D}_3=0 \\ & \mathrm{D}=0 \end{aligned}$$ $$\alpha \beta-3 \alpha+4 \beta=17$$ ..... (1) $$\begin{aligned} & \mathrm{D}_1=\left|\begin{array}{ccc} -4 & -2 & 1 \\ 5 & \alpha & 3 \\ 3 & -1 & \beta \end{array}\right|=0 \\ & \mathrm{D}_2=\left|\begin{array}{ccc} 1 & -4 & 1 \\ 2 & 5 & 3 \\ 3 & 3 & \beta \end{array}\right|=0 \\ & \Rightarrow 1(5 \beta-9)+4(2 \beta-9)+1(6-15)=0 \\ & 13 \beta-9-36-9=0 \end{aligned}$$ $$13 \beta=54, \beta=\frac{54}{13}$$ put in (1) $$\frac{54}{13} \alpha-3 \alpha+4\left(\frac{54}{13}\right)=17$$ $$54 \alpha-39 \alpha+216=221$$ $$15 \alpha=5 \quad \alpha=\frac{1}{3}$$ Now, $$12 \alpha+13 \beta=12 \cdot \frac{1}{3}+13 \cdot \frac{54}{13}$$ $$=4+54=58$$

Practice more JEE questions

Answer thousands more real JEE previous-year questions free, get graded instantly, and climb the global ranked leaderboard.

Practice JEE free →

More JEE Math questions