Step - 1: Find the larget value of third order determinant whose elements are 0 or 1.

Let $\mathrm{△}=\mid \begin{array}{ccc}{\mathrm{a}}_1& {\mathrm{b}}_1& {\mathrm{c}}_1\\ {\mathrm{a}}_2& {\mathrm{b}}_2& {\mathrm{c}}_2\\ {\mathrm{a}}_3& {\mathrm{b}}_3& {\mathrm{c}}_3\end{array}\mid$ be a determinant of order 3.

Then, Δ = a₁b₂c₃ + a₃b₁c₂ + a₂b₃c₁ – a₁b₃c₂ – a₂b₁c₃ – a₃b₂c₁

= (a₁b₂c₃ + a₃b₁c₂ + a₂b₃c₁) – (a₁b₃c₂ + a₂b₁c₃ + a₃b₂c₁)

Since, each element of Δ is either 0 or 1.

∴ The value of the determinant cannot exceed 3.

Clearly, the value of Δ is 3 when the value of each term in the first bracket is 1.

And the value of each term in the second bracket is zero.

But a₁b₂c₃ = a₃b₁c₂ = a₂b₃c₁ = 1 implies that every element of the determinant Δ is 1 and in this case Δ = 0 so, the maximum value of Δ ≠ 3

∴ Take two of the three term as 1 and each element of the remaining term as 0.

Say, a₁b₂c₃ = a₃b₁c₂ = 1 and a₂ = b₃ = c₁ = 0

$\Rightarrow \mathrm{△}=\mid \begin{array}{ccc}1& 1& 0\\ 0& 1& 1\\ 1& 0& 1\end{array}\mid$

⇒ Δ = 2

Hence, the largest value of a third order determinant whose elements are 0 or 1 is 2.