Question

The number of different n$\times$n symmetric matrices with each elements being either 0 or 1 is

• A. $2^n$
• B. $2^{n^2}$
• C. $2^{\frac{n^2+n}{2}}$
• D. $2^{\frac{n^2-n}{2}}$

Answer 1

The correct option is C $2^{\frac{n^2+n}{2}}$

Let A= $\left[\begin{array}{cccc}{a}_{11}& a_{12}& \cdots & a_{1n}\\ a_{21}& a_{22}& \cdots & a_{2n}\\ ⋮& ⋮& ⋮& ⋮\\ ⋮& ⋮& ⋮& ⋮\\ a_{n1}& a_{n2}& \cdots & a_{nn}\end{array}\right]$

Then to form symmetric matrix we have to put

$a_{ij}=a_{ji}$ $\forall$ $i\ne j$

i.e. A= $\left[\begin{array}{cccc}{a}_{11}& a_{12}& \cdots & a_{1n}\\ a_{12}& a_{22}& \cdots & a_{2n}\\ a_{13}& a_{23}& a_{33}& a_{3n}\\ ⋮& ⋮& ⋮& ⋮\\ a_{1n}& a_{2n}& \cdots & a_{nn}\end{array}\right]$

Now it is symmetric matrix.

($\because$ it is symmetrical about leading diagonal)

So, no. of different elements in A

= n + (n-1) + (n - 2) + ..... + 3 + 2 + 1

= $\frac{n(n+1)}{2}$

$\because$ Each element is filled by two ways (either 0 or 1)

so total no. of different symmetric matrices = $2^{\frac{n(n+1)}{2}}$