If the system of equations [ aligned &x+y+z=6 &2 x+5 y+ z= &x+2 y+3 z=14 aligned ] has inf…

Question

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

Accepted Answer

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{pmatrix} 1 & 1 & 1  \
   2 & 5 & \alpha   \
   1 & 2 & 3  \

\end{pmatrix} } \right| = 0$$

$${C_1} \to {C_1} - {C_3}$$

$${C_2} \to {C_2} - {C_3}$$

$$ \Rightarrow \left| {\begin{pmatrix} 0 & 0 & 1  \
   {2 - \alpha } & {5 - \alpha } & \alpha   \
   { - 2} & { - 1} & 3  \

\end{pmatrix} } \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{pmatrix} 6 & 1 & 1  \
   \beta  & 5 & 8  \
   {14} & 2 & 3  \

\end{pmatrix} } \right| = 0$$

$${C_1} \to {C_1} - 6{C_3}$$

$${C_2} \to {C_2} - {C_3}$$

$$ \Rightarrow \left| {\begin{pmatrix} 0 & 0 & 1  \
   {\beta  - 48} & { - 3} & 8  \
   { - 4} & { - 1} & 3  \

\end{pmatrix} } \right| = 0$$

$$ \Rightarrow  - \beta  + 48 - 12 = 0$$

$$ \Rightarrow \beta  = 36$$

$$\therefore$$ $$\alpha  + \beta  = 8 + 36 = 44$$

Solutions · 1