If $X, Y and Z are positive numbers such that Y and Z have respectively 1 and 0 at their unit's place and Δ is the determinant

$\begin{vmatrix} X & 4 & 1\\ Y & 0 & 1\\ Z & 1 & 0\end{vmatrix}$

If $(Δ+1)$ is divisible by 10, then X has at its unit's place

2

The correct answer is option (1) : 2

Let $X=10x+ \lambda, Y = 10 y + 1 $ and $Z= 10 z,$ where $x, y, z \in N.$ Then,

$Δ= \begin{vmatrix} X & 4 & 1\\ Y & 0 & 1\\ Z & 1 & 0\end{vmatrix}$

$⇒ Δ= \begin{vmatrix} 10x+ \lambda  & 4 & 1\\ 10y+1 & 0 & 1\\ 10z+0 & 1 & 0\end{vmatrix}= \begin{vmatrix} 10x & 4 & 1\\ 10y & 0 & 1\\ 10z & 1 & 0\end{vmatrix}+\begin{vmatrix} \lambda  & 4 & 1\\ 1 & 0 & 1\\ 0 & 1 & 0\end{vmatrix}$

$⇒ Δ= \begin{vmatrix} x & 4 & 1\\ y & 0 & 1\\ z & 1 & 0\end{vmatrix}+(1-\lambda )$

$⇒ Δ + 1= 10 \begin{vmatrix} x & 4 & 1\\ y & 0 & 1\\ z & 1 & 0\end{vmatrix}+(2-\lambda )$

$⇒ Δ+1 = 10 k +(2- \lambda), $ where $k = \begin{vmatrix} x & 4 & 1\\ y & 0 & 1\\ z & 1 & 0\end{vmatrix}$

It is given that $Δ+1 $ is divisible by 10. Therefore,

$2-\lambda = 0 $ i.e. $\lambda = 2.$

$∴X= 10x+ 2 $

$⇒2 $ is at unit's place of X.