Question

The only integral root of the equation $ \begin{vmatrix}2-y&2&3\\ 2&5-y&6\\ 3&4&10-y\end{vmatrix} =0 $ is

• A. $y = 3$
• B. $y = 2$
• C. $y = 1$
• D. $None\, of\, these$

Answer 1

The Correct Option is C

We have, $\begin{vmatrix}2-y&2&3\\ 2&5-y&6\\ 3&4&10-y\end{vmatrix} = 0 $ $\Rightarrow\left(2 - r\right)\left[\left(5-y\right)\left(10-y\right)-24\right]-2\left[2\left(10-y\right)-18\right]$
$+3\left[8 - 3\left(5-y\right)\right] = 0$
$\Rightarrow\left(2 -y\right)\left[50 - 15y + y^{2} -24\right]-2\left[20 - 2y -18\right]$
$ + 3\left[8-15 + 3y\right] = 0 $
$ \Rightarrow\left(2 -y\right)\left[y^{2} -15y +26\right]$
$-2\left[2-2y\right]+ 3\left[3y-7\right] = 0$
$ \Rightarrow 2y^{2} -30 y + 52 -y^{3} + 15 y^{2} - 26 y - 4 + 4y $
$+ 9 y - 21 = 0 $
$ \Rightarrow -y^{3} + 17 y^{2} - 43 y + 27 = 0 $
$ \Rightarrow y^{3} - 17y^{2} + 43y -27 = 0 $
$ \Rightarrow\left(y-1\right)\left(y^{2}-16y+27\right) = 0 $
$\Rightarrow y -1 = 0$ or $y^{2 } - 16 y + 27 = 0$
$\Rightarrow y = 1$