Question

$X,Y$ and $Z$ are positive numbers greater than $10$ such that $Y$ and $Z$ have respectively $1$ and $0$ at their unit's place and $\Delta$ is the determinant $\left|\begin{array}{ccc}X& 4& 1\\ Y& 0& 1\\ Z& 1& 0\end{array}\right|$. If $(\Delta +1)$ is divisible by $10$, then $X$ has at its unit's place

• A. $4$
• B. $3$
• C. $0$
• D. $2$

Answer 1

The correct option is D $2$

$x,y,z>0$

1 and 0,y and z have respectively

$y\to K1$

$z\to 10$

$△=\left|\begin{array}{ccc}X& 4& 1\\ Y& 0& 1\\ Z& 1& 1\end{array}\right|$

$△=-X-4(y-z)+1\left(Y\right)$

$△=-X-4y+4z+Y$

$△=-X-3Y+4z$

z end with $0$

unit place of $4z=0$

y end with $0$

Unit place of $+3Y=+3$

$(-3Y)=-3$

$△+1$it is divisible by $10$

Units place should be zero

Unit place of $X+0-3+1=0$

$X=2$.