Question

The total number of $3\times 3$ matrices $A$ having entries from the set $\{0,1,2,3\}$ such that the sum of all the diagonal entries of $A{A}^T$ is $9$, is equal to

• A. 766.00
• B. 766
• C. 766.0

Answer 1

Answer: (A), (B), (C)

$A{A}^T=\left[\begin{array}{ccc}x& y& z\\ a& b& c\\ d& e& f\end{array}\right]\left[\begin{array}{ccc}x& a& d\\ y& b& e\\ z& c& f\end{array}\right]=\left[\begin{array}{ccc}{x}^2+y^2+z^2& ax+by+cz& dx+ey+fz\\ ax+by+cz& a^2+b^2+c^2& ad+be+cf\\ dx+ey+fz& ad+be+cf& d^2+e^2+f^2\end{array}\right]$

$Tr\left(A{A}^T\right)=x^2+y^2+z^2+a^2+b^2+c^2+d^2+e^2+f^2=9$
All $1$
$=1$

one $3$ rest $0$
$=\frac{9!}{8!}=9$

two $2$ one $1$ & rest $0$
$=\frac{9!}{2!6!}=63\times 4=252$

one $2$, five $1$, rest $0$
$\frac{9!}{5!3!}=63\times 8=504$

$\therefore$ Total $=766$