Question

The number of $3\times 3$ matrices that can be formed by using the elements $0,1,2,3$ such that all the principal diagonal elements are different is

• A. $4^6.6$
• B. $4^7.6$
• C. $3^7.6$
• D. $4^7$

Answer 1

Answer: (B)

$4^7.6$

Since the matrix is $3\times 3$, there will be $9$ elements out of which $3$ are diagonal elements.
Since the diagonal elements are kept to be different, so number of ways $=4\times 3\times 2$ and number of ways of arranging other elements $4^6$.
$\therefore$ Required number of ways $4^6\times 4\times 3\times 2=4^7\times 6$