
Find the largest value of a third order determinant whose elements are 0 or 1.
Answer
511.5k+ views
Hint: 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,
\[D=\left| \begin{matrix}
{{a}_{11}} & {{a}_{12}} & {{a}_{13}} \\
{{a}_{21}} & {{a}_{22}} & {{a}_{23}} \\
{{a}_{31}} & {{a}_{32}} & {{a}_{33}} \\
\end{matrix} \right|\]
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:
\[\begin{align}
& D=\left| \begin{matrix}
0 & 1 & 1 \\
1 & 0 & 1 \\
1 & 1 & 0 \\
\end{matrix} \right| \\
& D=0\left( 0-1 \right)-1\left( 0-1 \right)+1\left( 1-0 \right) \\
& D=1+1 \\
& D=2 \\
\end{align}\]
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 \[{{n}^{2}}\].
Complete step by step answer:
Consider, the third order determinant,
\[D=\left| \begin{matrix}
{{a}_{11}} & {{a}_{12}} & {{a}_{13}} \\
{{a}_{21}} & {{a}_{22}} & {{a}_{23}} \\
{{a}_{31}} & {{a}_{32}} & {{a}_{33}} \\
\end{matrix} \right|\]
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:
\[\begin{align}
& D=\left| \begin{matrix}
0 & 1 & 1 \\
1 & 0 & 1 \\
1 & 1 & 0 \\
\end{matrix} \right| \\
& D=0\left( 0-1 \right)-1\left( 0-1 \right)+1\left( 1-0 \right) \\
& D=1+1 \\
& D=2 \\
\end{align}\]
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 \[{{n}^{2}}\].
Recently Updated Pages
Master Class 12 Business Studies: Engaging Questions & Answers for Success

Master Class 12 English: Engaging Questions & Answers for Success

Master Class 12 Economics: Engaging Questions & Answers for Success

Master Class 12 Social Science: Engaging Questions & Answers for Success

Master Class 12 Maths: Engaging Questions & Answers for Success

Master Class 12 Chemistry: Engaging Questions & Answers for Success

Trending doubts
Which are the Top 10 Largest Countries of the World?

Why is insulin not administered orally to a diabetic class 12 biology CBSE

a Tabulate the differences in the characteristics of class 12 chemistry CBSE

Why is the cell called the structural and functional class 12 biology CBSE

The total number of isomers considering both the structural class 12 chemistry CBSE

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE
