Question

Matrices of order $3\times 3$ are formed using the elements of set $A=\{-3,-2,-1,0,1,2,3\}.$ Then the probability that matrices are either symmetric or skew-symmetric, is

• A. $\frac{1}{7^6}+\frac{1}{7^3}$
• B. $\frac{1}{7^6}+\frac{1}{7^3}-\frac{1}{7^8}$
• C. $\frac{1}{7^3}+\frac{1}{7^8}$
• D. $\frac{1}{7^3}+\frac{1}{7^6}-\frac{1}{7^9}$

Answer 1

The correct option is D $\frac{1}{7^3}+\frac{1}{7^6}-\frac{1}{7^9}$

$A=\{-3,-2,-1,0,1,2,3\}$
Let $P\left(A_1\right)$ be the probability that the matrix is symmetric.
and $P\left(A_2\right)$ be the probability that the matrix is skew-symmetric.
$P(A_1\cup A_2)=P\left(A_1\right)+P\left(A_2\right)-P(A_1\cap A_2)$

$\left(a\right)$ $P\left(A_1\right)$
For symmetric matrix, $A^T=A$
$A=\left[\begin{array}{ccc}7& 7& 7\\ 1& 7& 7\\ 1& 1& 7\end{array}\right]$
(Each entry denotes the number of ways with which it can be filled)
Favourable cases $=7^6$
Total cases $=7^9$
$P\left(A_1\right)=\frac{7^6}{7^9}=\frac{1}{7^3}$

$\left(b\right)$ $P\left(A_2\right)$
For skew-symmetric matrix, $A^T=-A$
$A=\left[\begin{array}{ccc}0& 7& 7\\ 1& 0& 7\\ 1& 1& 0\end{array}\right]$
(Each entry denotes the number of ways with which it can be filled)
Favourable cases $=7^3$
Total cases $=7^9$
$P\left(A_2\right)=\frac{7^3}{7^9}=\frac{1}{7^6}$

$\left(c\right)$ $P(A_1\cap A_2)$ (Null matrix)
$=\frac{1}{7^9}$

Thus, $P(A_1\cup A_2)=\frac{1}{7^3}+\frac{1}{7^6}-\frac{1}{7^9}$