If $0 ≤ [x] < 2; –1 ≤ [y] < 1$ and $1 ≤ [z] < 3$. where [*] denotes the greatest integer function, then the maximum value of the determinant $\begin{vmatrix}{[x]+1}&[y]&[z]\\{[x]}&[y]+1&[z]\\{[x]}&[y]&[z]+1\end{vmatrix}$ is
Answer & explanation
Correct answer: option 2
$∵ 0 ≤ [x] < 2 ⇒ [x] = 0,1$
$–1 ≤ [y] < 1 ⇒ [y] = –1, 0$
and $1 ≤ [z] < 3 ⇒ [z] = 1, 2$
Now, aplying in the given determinant
$R_2 → R_2 – R_1, R_3 → R_3 – R_1$, then
$\begin{vmatrix}[x]+1&[y]&[z]\\-1&1&0\\-1&0&1\end{vmatrix}$
$= ([x] + 1)(1 – 0)–[y] (–1 –0) + [z] (0+1)$
$= [x] + [y] + [z] + 1 = 1 + 0 + 2 + 1 = 4$
(∴ for maximum value $[x] = 1, [y]=0,[z]= 2$)