The largest value of a third order determinant whose elements are 0 or 1, is

Solution:

$Let Δ = |\begin{array}{lll} a_{1} b_{1} c_{1} \\ a_{2} b_{2} c_{2} \\ a_{3} b_{3} c_{3} \end{array}|$ be a determinant of order 3.

Then,

$\begin{array}{l} Δ = a_{1} b_{2} c_{3} + a_{3} b_{1} c_{2} + a_{2} b_{3} c_{1} - a_{1} b_{3} c_{2} - a_{2} b_{1} c_{3} - a_{3} b_{2} c_{1} \\ Δ = (a_{1} b_{2} c_{3} + a_{3} b_{1} c_{2} + a_{2} b_{3} c_{1}) - (a_{1} b_{3} c_{2} + a_{2} b_{1} c_{3} + a_{3} b_{2} c_{1}) \end{array}$

Since each element of $∆$ is either 1 or 0. Therefore, the value of the determinant cannot exceed 3. Clearly, the value of $∆$ is maximum when the value of each term in first bracket is 1 and the value of each term in the second bracket is zero. But $a_{1} b_{2} c_{3} = a_{3} b_{1} c_{2} = a_{2} b_{3} c_{1} = 1$

implies that every element of the determinant is 1 and in that $Δ = 0$ Thus, we may have

$Δ = |\begin{array}{lll} 0 1 1 \\ 1 0 1 \\ 1 1 0 \end{array}| = 2$

Solution:

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