Find the Largest value of a third order determinant whose elements are 0 or 1.
My try: \begin{vmatrix} a_{1} & b_{1} & c_{1} \\a_{2} & b_{2} & c_{2} \\ a_{3} & b_{3} & c_{3} \end{vmatrix} I opened the determinant, and concluded that the value of detrminant cannot exceed 3. I have no idea what to do next. Please help me.
Since $$\begin{vmatrix} a_{1} & b_{1} & c_{1} \\a_{2} & b_{2} & c_{2} \\ a_{3} & b_{3} & c_{3} \end{vmatrix}=a_1b_2c_3+b_1c_2a_3+c_1a_2b_3-a_3b_2c_1-b_3c_2a_1-c_3a_2b_1.$$
Each of these terms are either $0$ or $1$ (depending on the entries chosen to be $1$ or $0$). So to maximize one may want to choose the entries with positive terms becoming $1$ and the negative terms becoming $0$. However that is NOT possible because then all entries become $1$ and the determinant takes on the value $0$. So the maximum value cannot be $3$.
Since we cannot have all the positive terms be $1$ (without having the negative terms also being equal to $1$) so we try to see if we can have two of the positive terms as $1$.
Observe that we can have the following: $$\begin{vmatrix} 1 & 1 & c_{1} \\a_{2} & 1 & 1 \\ 1 & b_{3} & 1 \end{vmatrix}=1+1+c_1a_2b_3-c_1-b_3-a_2.$$
But now to maximize we need to have $c_1=a_2=b_3=0$. This gives the maximum value of the determinant to be $2$.
It might be also good to look at the determinant as volume of the parallelepiped to convince yourself with the validity of the result.