Question

Find the largest value of a third order determinant whose elements are 0 or 1.

Answer 1

Here, we will first write the standard determinant of the third order and then write the element or constituent of the determinant. Then, assume a third order determinant whose elements are 0 or 1 and evaluate the determinant using the expansion formula of a determinant of the third order. The maximum value of a third order determinant whose elements are 0 or 1 is thus obtained.

Complete step by step answer:
Consider, the third order determinant,

{{a}_{11}} & {{a}_{12}} & {{a}_{13}} \\\ {{a}_{21}} & {{a}_{22}} & {{a}_{23}} \\\ {{a}_{31}} & {{a}_{32}} & {{a}_{33}} \\\ \end{matrix} \right|$$ A determinant of the third order can be expanded from any row or any column. During expansion from any row or any column, the successive elements of the particular row or column are multiplied by the determinants of the second order obtained by deleting the successive rows or columns through the elements. The sign in successive terms of the expansion is given by the following rule: $$\left| \begin{matrix} \+ & \- & \+ \\\ \- & \+ & \- \\\ \+ & \- & \+ \\\ \end{matrix} \right|$$ Therefore, The third order determinant can be expanded from the first row as: $$D={{a}_{11}}\left| \begin{matrix} {{a}_{22}} & {{a}_{23}} \\\ {{a}_{32}} & {{a}_{33}} \\\ \end{matrix} \right|-{{a}_{12}}\left| \begin{matrix} {{a}_{21}} & {{a}_{23}} \\\ {{a}_{31}} & {{a}_{33}} \\\ \end{matrix} \right|+{{a}_{13}}\left| \begin{matrix} {{a}_{21}} & {{a}_{22}} \\\ {{a}_{31}} & {{a}_{32}} \\\ \end{matrix} \right|$$ Now, expand the determinant of the second order, $$D={{a}_{11}}\left( {{a}_{22}}{{a}_{33}}+{{a}_{23}}{{a}_{32}} \right)-{{a}_{12}}\left( {{a}_{21}}{{a}_{21}}+{{a}_{23}}{{a}_{31}} \right)+{{a}_{13}}\left( {{a}_{21}}{{a}_{32}}+{{a}_{22}}{{a}_{31}} \right)$$ Assume, $$D=\left| \begin{matrix} 0 & 1 & 1 \\\ 1 & 0 & 1 \\\ 1 & 1 & 0 \\\ \end{matrix} \right|$$ be the third order determinant whose elements are 0 or 1 Therefore, By using the formula of expansion of a determinant of the third order, The value of the determinant is obtained as: $$\begin{aligned} & D=\left| \begin{matrix} 0 & 1 & 1 \\\ 1 & 0 & 1 \\\ 1 & 1 & 0 \\\ \end{matrix} \right| \\\ & D=0\left( 0-1 \right)-1\left( 0-1 \right)+1\left( 1-0 \right) \\\ & D=1+1 \\\ & D=2 \\\ \end{aligned}$$ Hence, the largest value of a third order determinant whose elements are 0 or 1 is 2. **Note:** A determinant is defined as an expression or a function or a number which is associated with a square matrix of order $$n\times n$$ (also called of order$$n$$). If $$A=\left[ {{a}_{ij}} \right]$$ is a square matrix of order $$n\times n$$, then $$A=\left[ {{a}_{ij}} \right]$$ is a determinant with $$n$$ rows and $$n$$ columns and total number of elements will be $${{n}^{2}}$$.