Question
Question: Let A be the set of all \(3 \times 3\) symmetric matrices all whose entries are either 1 or 0. Five ...
Let A be the set of all 3×3 symmetric matrices all whose entries are either 1 or 0. Five of these entries are 1 and four of them are 0.
The number of matrices in A such that ∣A∣= 0, is
(A) 5
(B) 6
(C) 8
(D) None of these.
Solution
Hint – In this particular type of question use the concept that in a symmetric matrices the transpose of the matrix is equal to the original matrix (i.e. AT=A) and the determinant of the matrix is zero if the two rows or two columns are the same so use these concepts to reach the solution of the question.
Complete step-by-step answer:
Given data:
A be a 3×3 symmetric matrices all whose entries are either 1 or 0.
As we know that in symmetric matrices the transpose of the matrix is equal to the original matrix.
⇒AT=A, where T is the symbol for transpose.
Let, A = \left[ {\begin{array}{*{20}{c}}
{{a_{11}}}&{{a_{12}}} \\\
{{a_{21}}}&{{a_{22}}}
\end{array}} \right], so {A^T} = \left[ {\begin{array}{*{20}{c}}
{{a_{11}}}&{{a_{21}}} \\\
{{a_{12}}}&{{a_{22}}}
\end{array}} \right]
⇒AT=A⇒ \left[ {\begin{array}{*{20}{c}}
{{a_{11}}}&{{a_{21}}} \\\
{{a_{12}}}&{{a_{22}}}
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
{{a_{11}}}&{{a_{12}}} \\\
{{a_{21}}}&{{a_{22}}}
\end{array}} \right]
Now it is given that in these symmetric matrices five entries are 1 and four entries are zero.
So the possible case arise
(i) When all the diagonal elements are equal to one.
When all the diagonal elements are equal to one then there is three possibilities of such symmetric matrix so that it consists of 5 one and four zero which is shown below,