Question

If $0 \leq[x]<2,-1 \leq[y] < 1$ and $1 \leq[z]<3$ where $[\,\,.\,\,]$ denotes greatest integral function then the maximum value of the determinant. $D=\begin{vmatrix}{[x]+1} & {[y]} & {[z]} \\ {[x]} & {[y]+1} & {[z]} \\ {[x]} & {[y]} & {[z]+1}\end{vmatrix}$ is

Answer 1

$\begin{vmatrix} 1 & -1 & 0 \\ 0 & 1 & -1 \\ {[x]} & {[y]} & {[z]+1} \end{vmatrix}$
solving $=[x]+[y]+[z]+1$
taking maximum value we get $4$, note that $[x]$ is always an integer