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If the system of equations $$ \begin{aligned} &x+y+z=6 \\ &2 x+5 y+\alpha z=\beta \\ &x+2 y+3 z=14 \end{aligned} $$ has infinitely many solutions, then $$\alpha+\beta$$ is equal to

JEE · Math · previous-year question

  1. A.8
  2. B.36
  3. C.44correct
  4. D.48

Answer

C. 44

Explanation

Given, $$x + y + z = 6$$ ...... (1) $$2x + 5y + \alpha z = \beta $$ ..... (2) $$x + 2y + 3z = 14$$ ...... (3) System of equation have infinite many solutions. $$\therefore$$ $${\Delta _x} = {\Delta _y} = {\Delta _z} = 0$$ and $$\Delta = 0$$ Now, $$\Delta = \left| {\begin{matrix} 1 & 1 & 1 \\ 2 & 5 & \alpha \\ 1 & 2 & 3 \\ \end{matrix} } \right| = 0$$ $${C_1} \to {C_1} - {C_3}$$ $${C_2} \to {C_2} - {C_3}$$ $$ \Rightarrow \left| {\begin{matrix} 0 & 0 & 1 \\ {2 - \alpha } & {5 - \alpha } & \alpha \\ { - 2} & { - 1} & 3 \\ \end{matrix} } \right| = 0$$ $$ \Rightarrow - 2 + \alpha + 10 - 2\alpha = 0$$ $$ \Rightarrow 8 - \alpha = 0$$ $$ \Rightarrow \alpha = 8$$ Now, $$x + y + z = 6$$ $$2x + 5y + 8z = \beta $$ $$x + 2y + 3z = 14$$ $$\therefore$$ $${\Delta _x} = \left| {\begin{matrix} 6 & 1 & 1 \\ \beta & 5 & 8 \\ {14} & 2 & 3 \\ \end{matrix} } \right| = 0$$ $${C_1} \to {C_1} - 6{C_3}$$ $${C_2} \to {C_2} - {C_3}$$ $$ \Rightarrow \left| {\begin{matrix} 0 & 0 & 1 \\ {\beta - 48} & { - 3} & 8 \\ { - 4} & { - 1} & 3 \\ \end{matrix} } \right| = 0$$ $$ \Rightarrow - \beta + 48 - 12 = 0$$ $$ \Rightarrow \beta = 36$$ $$\therefore$$ $$\alpha + \beta = 8 + 36 = 44$$

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