If the system of equations []x + y + z = 5, x + 2y + 3z = 4, x + 3y + 5z = [] has infinite…

Question

If the system of equations
$$\alpha$$x + y + z = 5, x + 2y + 3z = 4, x + 3y + 5z = $$\beta$$
has infinitely many solutions, then the ordered pair ($$\alpha$$, $$\beta$$) is equal to :

Accepted Answer

Given system of equations

$$\alpha x + y + z = 5$$

$x + 2y + 3z = 4$, has infinite solution

$$x + 3y + 5z = \beta $$

$$\therefore$$ $$\Delta  = \left| {\begin{pmatrix} \alpha  & 1 & 1  \
   1 & 2 & 3  \
   1 & 3 & 5  \

\end{pmatrix} } \right| = 0 \Rightarrow \alpha (1) - 1(2) + 1(1) = 0$$

$$ \Rightarrow \alpha  = 1$$

and $${\Delta _1} = \left| {\begin{pmatrix} 5 & 1 & 1  \
   4 & 2 & 3  \
   \beta  & 3 & 5  \

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

$$ \Rightarrow 5(1) - 1(20 - 3\beta ) + 1(12 - 2\beta ) = 0$$

$$ \Rightarrow \beta  = 3$$

and $${\Delta _2} = \left| {\begin{pmatrix} 1 & 5 & 1  \
   1 & 4 & 3  \
   1 & \beta  & 5  \

\end{pmatrix} } \right| = 0 \Rightarrow (20 - 3\beta ) - 5(2) + 1(\beta  - 4) = 0$$

$$ \Rightarrow  - 2\beta  + 6 = 0$$

$$ \Rightarrow \beta  = 3$$

Similarly,

$$ \Rightarrow {\Delta _3} = \left| {\begin{pmatrix} 1 & 1 & 5  \
   1 & 2 & 4  \
   1 & 3 & \beta   \

\end{pmatrix} } \right| = 0 \Rightarrow \beta  = 3$$

$$\therefore$$ ($$\alpha$$, $$\beta$$) = (1, 3)

Solutions · 1