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Let $0\leq x <4, -2\leq y < 3$ and $-1\leq z < 5$. If $[a]$ denotes greatest integer less than or equal to $a$, and

$$\Delta = \begin{vmatrix} [x+2] & [y]& [z]\\ [x] & [y+1] & [z]\\ [x] & [y] & [z+1]\end{vmatrix}$$

then, what is the maximum value of $\Delta$?

I have learnt maximum value of a determinant is when its diagonal elements are minimum and rest are maximum. I have tried 2 cases: putting minimum values of all 3 variables so diagonal is minimum, and putting maximum values of all 3 variables so the rest of the elements are maximum. I'm getting a higher value in 2nd case but I'm not sure if we can get a higher value for some other intermediate values of the variables.

Light
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  • Have you tried to simplify $\Delta?$ I found $\Delta = 2[x] + 2[y] + 2[z]+2$ for which it is simpler to find the maximum in the given ranges. – F_M_ Oct 13 '22 at 14:45
  • Could you specify how you have simplified it please? – Light Oct 14 '22 at 06:53
  • First I used the rule $[x+2] = [x] + 2$ and similar for $[y+1] = [y] + 1, [z+1] = [z]+1.$ Then to calculate the determinant, you can subsequently do the operations row 1 minus row 3, row 2 minus row 1 and column 3 + column 2, which yield $$\begin{vmatrix}2 & 0 & -1 \ 0 & 1 & 0 \ [x] & [y] & [y] + [z]+1 \end{vmatrix} = [x] + 2[y] + 2[z]+2$$ (there was a typo in my first comment). – F_M_ Oct 14 '22 at 07:06
  • Got it, thanks a lot! – Light Oct 14 '22 at 14:19

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