An n times n matrix is formed using 0, 1 and | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) For a matrix $A = [a_{ij}]$ to be skew-symmetric, it must satisfy the condition $a_{ij} = -a_{ji}$ for all $i, j$ and $a_{ii} = 0$ for all $i$.$1$

Vedclass · Accepted Answer

$3^{n(n-1)/2}$

Vedclass · Answer

(A) For a matrix $A = [a_{ij}]$ to be skew-symmetric, it must satisfy the condition $a_{ij} = -a_{ji}$ for all $i, j$ and $a_{ii} = 0$ for all $i$.$1$. The diagonal elements $a_{ii}$ must be $0$. There is only $1$ choice for each of the $n$ diagonal elements.$2$. For the off-diagonal elements, we only need to choose the elements $a_{ij}$ where $i $3$. The number of pairs $(i, j)$ such that $i $4$. Each of these $\frac{n(n-1)}{2}$ positions can be filled with any of the $3$ values: $\{0, 1, -1\}$.$5$. Therefore, the total number of such skew-symmetric matrices is $3^{\frac{n(n-1)}{2}}$.

An $n \times n$ matrix is formed using $0, 1$ and $-1$ as its elements. The number of such matrices which are skew-symmetric is

Similar Questions

Find the values of $x$ and $y$ from the following equation:
$2\begin{bmatrix} x & 5 \\ 7 & y-3 \end{bmatrix} + \begin{bmatrix} 3 & -4 \\ 1 & 2 \end{bmatrix} = \begin{bmatrix} 7 & 6 \\ 15 & 14 \end{bmatrix}$

If $A^2 = A$,then $(I + A)^3 - 7A =$ . . . . . . ,where $A$ is a square matrix.

$\cos \theta \begin{bmatrix} \cos \theta & \sin \theta \\ - \sin \theta & \cos \theta \end{bmatrix} + \sin \theta \begin{bmatrix} \sin \theta & - \cos \theta \\ \cos \theta & \sin \theta \end{bmatrix} = $

The matrix $A = \frac{1}{3}\begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & -2 \\ -2 & 2 & -1 \end{bmatrix}$ is

Construct a $2 \times 2$ matrix,$A = [a_{ij}]$,whose elements are given by $a_{ij} = \frac{i}{j}$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module