Skip to content

GATE CS 2004 Q26 — Linear Algebra: Matrices, Determinants, System of Linear Equations

Explain with AI

The number of different n×nn \times n symmetric matrices with each element being either 0 or 1 is: (Note: power(2,X)\text{power} \left(2, X\right) is same as 2X2^X)

Choose one option
View worked solution

Answer: C

For an n×nn \times n matrix AA to be symmetric, it must satisfy A=ATA = A^T, which means aij=ajia_{ij} = a_{ji} for all 1≤i,j≤n1 \le i, j \le n.

This symmetry constraint implies that the elements on the main diagonal and the elements in either the upper triangular or lower triangular part uniquely determine the entire matrix. The elements in the remaining part are automatically determined by symmetry.

Let's count the number of independent elements we can choose:

  • Main diagonal elements: There are nn such elements (a11,a22,…,anna_{11}, a_{22}, \dots, a_{nn}).
  • Strictly upper triangular elements: There are n2−n2\frac{n^2 - n}{2} such elements.

The total number of independent elements is:

n+n2−n2=2n+n2−n2=n2+n2n + \frac{n^2 - n}{2} = \frac{2n + n^2 - n}{2} = \frac{n^2 + n}{2}

Since each of these n2+n2\frac{n^2 + n}{2} independent positions can be filled with either 00 or 11 (which gives 22 choices per position), the total number of different symmetric matrices is:

2n2+n2=power(2,n2+n2)2^{\frac{n^2 + n}{2}} = \text{power}\left(2, \frac{n^2 + n}{2}\right)

Therefore, the correct option is C.

Free practice · Your answers on this page are not saved.

Published
Report a correction
Keep exploringMore Linear Algebra: Matrices, Determinants, System of Linear Equations questions